Characterization of solutions to $f' = f(1-f)$

Question

The sigmoid function $f(x) = \frac{1}{1+e^{-x}}$ has the property that $$f'(x) = f(x)(1-f(x))~~~ and ~~~f(0) = \frac 12$$

My question: is $f$ the unique function from $\mathbb R$ to $(0,1)$, perhaps up to some kind of scaling, that satisfies $f' = f(1-f)$?

I don't have much experience with differential equations so a nonlinear one like this is beyond anything I've done before.

In case it helps, my motivation for this is that this property makes the log likelihood a lot easier in a logistic regression, and I'm wondering if assuming that the inverse link function satisfies this property is equivalent to just taking it to be $f$.

Answer 1

Note that since $f(0)=1/2$ then, $f\not\equiv 1$ and , $f\not\equiv 0$ hence a solution cannot an equilibrium $$f'=f(1-f)\Longleftrightarrow\frac{f'}{f} +\frac{f'}{1-f} =1\Longleftrightarrow \ln (|f|)- \ln(|1-f|) =x+c$$

But, $f(x)\in (0,1)$ we obtain $$ \ln (f)- \ln(1-f) =x+c\Longleftrightarrow \frac{f}{1-f} =ke^x \Longleftrightarrow f(x) =1-\frac{1}{1+ ke^x}$$

I am sure you can the $k$ by yourself.

Answer 2

writing your equation in the form $$-\frac{\frac{df(x)}{dx}}{(f(x)-1)(f(x)}=1$$ and by integrating $$-\int \frac{\frac{df(x)}{dx}}{(f(x)-1)f(x)}dx=\int 1dx$$ doing this we obtain $$-\log(-f(x)+1)+\log(f(x))=x+C$$