Logistic function

The logistic function characterizes a continuous one-dimensional probability distribution and is a functional representation of saturation processes from the class of the so-called sigmoid functions with unlimited time extension.

Until the 20th century, the logarithm was occasionally given the Italian name of the logistic curve ( curva logistica ). Today the name is clearly assigned to the S function .

description

Logistic function just in case ${\ displaystyle G = 1 \,; \, k = 1 \,; \, f (0) = 0 {,} 5}$

The logistic function, as it results from the discrete logistic equation , describes the connection between the elapsing time and a growth, for example an ideal bacterial population or (approximately) the spread of a disease in the context of an epidemic. For this purpose, the model of exponential growth is modified by a resource that consumes itself with growth - the idea behind it is, for example, a bacteria culture medium of limited size or a decreasing number of people who behave carelessly during an epidemic and thus expose themselves to the risk of infection.

Example of an epidemic: illnesses and deaths (black) in the course of the Ebola fever epidemic in West Africa until July 2014 (almost logistical functions)

At the beginning the function value is not 0, but it applies . ${\ displaystyle f (0)> 0}$

The following applies to the bacterial example:

• The limited habitat forms an upper limit for the number of bacteria .${\ displaystyle G}$${\ displaystyle f (t)}$
• The bacterial growth is proportional to: ${\ displaystyle f '(t)}$
• the current stock ${\ displaystyle f (t)}$
• the remaining capacity ${\ displaystyle Gf (t)}$

This development is therefore represented by a Bernoulli differential equation of the form

${\ displaystyle f '(t) = k \ cdot f (t) \ cdot \ left (Gf (t) \ right)}$

described with a proportionality constant . Solving this differential equation gives: ${\ displaystyle k}$

${\ displaystyle f (t) = G \ cdot {\ frac {1} {1 + e ^ {- k \ cdot G \ cdot t} \ left ({\ frac {G} {f (0)}} - 1 \ right)}}}$

The graph of the function describes an S-shaped curve, a sigmoid . In the beginning the growth is small because the population and thus the number of reproducing individuals is small. In the middle of the development (more precisely: at the turning point ) the population grows most strongly until it is slowed down by the exhausting resources.

Other uses

The logistic equation describes a connection that occurs very often and is used far beyond the idea of ​​describing a population of living things. The life cycle of a product in the store can also be modeled with the logistic function. Further areas of application are growth and decay processes in language ( Language Change Act , Piotrowski Law ) and the development in the acquisition of the mother tongue ( Language Acquisition Act ). The logistic function is also used in the SI model of mathematical epidemiology.

Solution of the differential equation

Be . is steady. It is important to solve the differential equation . ${\ displaystyle F \ colon \ mathbb {R} \ to \ mathbb {R}, t \ mapsto kt (Gt)}$${\ displaystyle F}$${\ displaystyle {\ frac {\ mathrm {d} f} {\ mathrm {d} t}} (t) = F (f (t))}$

The differential equation can be solved with the " separation of variables " procedure . It applies to everyone , so the mapping to the real numbers is well-defined. ${\ displaystyle F (t) \ neq 0}$${\ displaystyle t \ in \ mathbb {R}}$${\ displaystyle \ Phi (t): = \ int _ {0} ^ {t} {\ frac {1} {F (s)}} \, {\ mathrm {d}} s}$

After separating the variables, the solution of the above differential equation is identical to the solution of the differential equation ${\ displaystyle f}$

${\ displaystyle \ int _ {f (0)} ^ {f (t)} {\ frac {1} {F (s)}} \, \ mathrm {d} s = t}$.

The decomposition of partial fractions results in

${\ displaystyle \ int _ {f (0)} ^ {f (t)} {\ frac {1} {F (s)}} \, \ mathrm {d} s = \ int _ {f (0)} ^ {f (t)} {\ frac {1} {ks (Gs)}} \, \ mathrm {d} s = {\ frac {1} {kG}} \ int _ {f (0)} ^ { f (t)} \ left ({\ frac {1} {s}} + {\ frac {1} {Gs}} \ right) \, \ mathrm {d} s}$

According to the main law of differential and integral calculus , the above is integral

${\ displaystyle = \ left. {\ frac {1} {kG}} {\ bigg (} \ ln (s) - \ ln (Gs) {\ bigg)} \ right | _ {s = f (0)} ^ {s = f (t)} = {\ frac {1} {kG}} {\ bigg (} \ ln {\ big (} {\ frac {f (t)} {Gf (t)}} {\ big)} + c {\ bigg)}}$

in which

${\ displaystyle c: = - \ ln {\ frac {f (0)} {Gf (0)}} = \ ln {\ big (} {\ frac {G} {f (0)}} - 1 {\ big)}}$

So the function equation applies

${\ displaystyle \ ln {\ frac {f (t)} {Gf (t)}} = kGt-c}$

to solve as long as they lie between and , which can be assumed because of the prerequisite . Where is the natural logarithm . The application of the exponential function on both sides leads to ${\ displaystyle f (t)}$${\ displaystyle 0}$${\ displaystyle G}$${\ displaystyle 0 ${\ displaystyle \ ln}$

${\ displaystyle e ^ {kGt-c} \, = \, {\ frac {f (t)} {Gf (t)}}}$.

and subsequent reciprocal value to

${\ displaystyle (*) \ quad \ quad \ quad e ^ {- kGt + c} \, = \, {\ frac {Gf (t)} {f (t)}} \, = \, {\ frac { G} {f (t)}} - 1}$.

We now bring the to the left, then form the reciprocal value again, and finally get ${\ displaystyle 1}$

${\ displaystyle {\ frac {f (t)} {G}} \, = \, {\ frac {1} {1 + e ^ {- kGt + c}}}}$

and it

${\ displaystyle (**) \ quad \ quad \ quad f (t) \, = \, G \ cdot {\ frac {1} {1 + e ^ {- kGt + c}}}}$

If we insert the definition of into the found solution (**), we come to the solution of the logistic differential equation claimed above: ${\ displaystyle c}$

${\ displaystyle f (t) \, = \, G \ cdot {\ frac {1} {1 + e ^ {- kGt + c}}} \, = \, G \ cdot {\ frac {1} {1 + e ^ {- kGt} e ^ {c}}} \, = \, G \ cdot {\ frac {1} {1 + e ^ {- kGt} \ left ({\ frac {G} {f (0 )}} - 1 \ right)}}}$

It is easy to see from this functional equation that the values ​​are always between and , which is why the solution applies to all . This can of course also be confirmed in retrospect by inserting it into the differential equation. ${\ displaystyle 0}$${\ displaystyle G}$${\ displaystyle - \ infty

Calculation of the turning point

To determine the turning point of the solution function , we first determine the derivatives using the product rule${\ displaystyle f}$

{\ displaystyle {\ begin {aligned} f '(t) & = k \ cdot f (t) \ cdot (Gf (t)) \\ f' '(t) & = k \ cdot f' (t) \ cdot (Gf (t)) + k \ cdot f (t) \ cdot (-f '(t)) \\ & = k \ cdot f' (t) \ cdot (Gf (t) -f (t)) = k \ cdot f '(t) \ cdot (G-2 \ cdot f (t)) \ end {aligned}}}

and determine the zero of the second derivative: ${\ displaystyle t_ {W}}$

${\ displaystyle f '' (t_ {W}) = k \ cdot f '(t_ {W}) \ cdot (G-2 \ cdot f (t_ {W})) = 0}$
${\ displaystyle G-2 \ cdot f (t_ {W}) = 0}$
${\ displaystyle G = 2 \ cdot f (t_ {W})}$
${\ displaystyle f (t_ {W}) = {\ tfrac {G} {2}}}$

With this we know the function value at the inflection point and determine that the population in the inflection point just exceeds half the saturation limit. To determine , we use the solution formula and calculate as follows: ${\ displaystyle t_ {W}}$${\ displaystyle f (t_ {W}) = {\ tfrac {G} {2}}}$

${\ displaystyle G \ cdot {\ frac {1} {1 + e ^ {- k \ cdot G \ cdot t_ {W}} \ cdot e ^ {c}}} = {\ frac {G} {2}} }$
${\ displaystyle 1 + e ^ {- k \ cdot G \ cdot t_ {W}} \ cdot e ^ {c} = 2}$
${\ displaystyle e ^ {- k \ cdot G \ cdot t_ {W}} \ cdot e ^ {c} = 1 = e ^ {0}}$
${\ displaystyle -k \ cdot G \ cdot t_ {W} + c = 0}$
${\ displaystyle t_ {w} = {c \ over k \ cdot G}}$;

For follows with further: ${\ displaystyle \ textstyle G> f (0)}$${\ displaystyle \ textstyle e ^ {c} = {\ frac {G} {f (0)}} - 1 \ Rightarrow c = ln \ left ({\ frac {G} {f (0)}} - 1 \ right)}$

${\ displaystyle t_ {W} = {\ frac {\ ln \ left ({\ frac {G} {f (0)}} - 1 \ right)} {k \ cdot G}}}$

With this the turning point is completely determined and there is only this one. Substituting in the first derivative gives the maximum growth rate: ${\ displaystyle f (t_ {W}) = {\ tfrac {G} {2}}}$

${\ displaystyle f '(t_ {W}) = k \ cdot {\ frac {G} {2}} \ cdot \ left (G - {\ frac {G} {2}} \ right) = k \ cdot { \ frac {G} {2}} \ cdot {\ frac {G} {2}}}$
${\ displaystyle f '(t_ {W}) = {\ frac {k \ cdot G ^ {2}} {4}}}$

Further representations

${\ displaystyle f (t) = G \ cdot {\ frac {1} {1 + e ^ {- k \ cdot G \ cdot t} \ left ({\ frac {G} {f (0)}} - 1 \ right)}}}$
${\ displaystyle = G \ cdot {\ frac {1} {1 + e ^ {- k \ cdot G \ cdot t} \ cdot {\ frac {G} {f (0)}} - e ^ {- k \ cdot G \ cdot t}}} \ cdot {\ frac {f (0)} {f (0)}} = {\ frac {G \ cdot f (0)} {f (0) + e ^ {- k \ cdot G \ cdot t} \ cdot Ge ^ {- k \ cdot G \ cdot t} \ cdot f (0)}}}$
${\ displaystyle = {\ frac {G \ cdot f (0)} {f (0) + \ left (Gf (0) \ right) \ cdot e ^ {- k \ cdot G \ cdot t}}}}$

or:

${\ displaystyle f (t) = {\ frac {G} {2}} \ cdot \ left (\ tanh \ left ({\ frac {kG} {2}} (t-t_ {W}) \ right) + 1 \ right)}$, where the turning point calculated above is:${\ displaystyle t_ {W}}$${\ displaystyle t_ {W} = {\ frac {\ ln \ left ({\ frac {G} {f (0)}} - 1 \ right)} {k \ cdot G}}}$