Finding minimal $\gamma$ that satisfies the integral equation

Question

I have a function $f(z) \in [0,\infty]$ that satisfies $-1 < f(z) < 0$.

I would like to find the minimal $\gamma$ that satisfies:

$$ \int_0^{\gamma} f(z)dz = \log(1+f(0)).$$

Clearly, I cannot bound such $\gamma$ since $f$ can go to 0 so I can further relax the requirement:

find minimal $\gamma$ such that there exist some $\xi$ and:

$$ \int_{\xi}^{\gamma} f(z)dz = \log(1+f(\xi)).$$

That is, if $f$ increases, I can update the goal.

So the worst function $f$ that maximizes $\gamma$ should satisfy:

$$ \int_{\xi}^{\gamma} f(z)dz = \log(1+f(\xi))$$

for all $\xi \in [0,\gamma]$.

Using $\int_{\xi}^{\gamma} = \int_{0}^{\gamma}-\int_0^{\xi}$ and $\int_{0}^{\gamma} = \log(1+f(0))$, I need to find $f$ such that:

$$ \int_0^{\xi} f(z)dz = \log(1+f(0))-\log(1+f(\xi)).$$

So if I would find such $f$ where $f(0)$ is the boundary condition, it would be the worst function and the $\gamma$ for that function is the greatest such $\gamma$ for all function.

So, finally, the question is: how can I solve this integral equation? or, is it possible to bound the function in some sense?

Clearly, the bound on $\gamma$ should depend on $f(0)$.

Answer 1

Differentiate your integral equation to get the differential equation

$$ f(\xi) = - \frac{f'(\xi)}{1+f(\xi)} $$

Solve this with initial condition $f(0) = f_0$ to get

$$ f(\xi) = \frac{f_0}{(1+f_0) e^\xi - f_0}$$