I would like to obtain $g$ by solving the following integral equation
$$ \int_s^T R(u) dg(u) + f(s,T)\int_s^T g(u)du =0$$
where $f,R:\mathbb R _+ ^*\rightarrow \mathbb R _+ $and $g: \mathbb R _+ \rightarrow[0,1]$ is
- non-decreasing continuous function,
- $g(0)=1$,
- $g(T)<1$
- $\lim _{s\to \infty}g(s)=0$.
One can think about g as being such that $g= 1- F$ where $F$ is the law of an absolute continuous random variable.
If we assume that $g$ is differentiable then we have
$$ \int_s^T \left(f(s,T)g(u)+R(u) g'(u)\right) ~du =0, \quad \forall s \in [0,T] $$
So I am tempted to conclude that
$$ f(s,T)g(u)+R(u) g'(u)=0, \quad \forall u \in [s,T]$$ therefore $$g(s) = \exp\left(-\int _0^s \frac{f(\tau,T)}{R(\tau)}~d\tau\right), \quad \forall u \in [0,T]$$
Is my approach correct or have I made a mistake when I assumed that the integrand is zero as the integral is zero for each $s$ ?
I have no restrictions on $f$ and $R$ for the moment so we can assume any necessary condition about $f$ as necessary to solve it.
Could anybody give me an opinion please? Please leave a comment. All advices are appreciated.
Edit
A friend wisely advised me to take a look at Volterra integral equation which is exactly what we have here after integrating by parts the first integral and inverting the time as follows ( after that point I use the notation abuse $R(u): =R(-u), g(u):=g(-u) \text{ and } f(t,T) = f(-t,T) $):
$$g(t) = \alpha + \int_{-T }^t K_T(t,u))g(u) du $$
where $\alpha := (Rg)(-T)$ and $$K_T(t,u):=\frac{R(u) + f(t,T)}{R(t)}$$
Many thanks
Since this question haven't received any answer or comment but some upvotes I posted it at mathoverflow too
Your mistake in deriving your solution is that in
$$f(s,T)g(u) + R(u)g'(u) = 0$$
you have two arguments: $u$ and $s$ and you cannot just take $u=s$ as $u$ is the integration variable ranging over $[s,T]$ for a given $s$.
To derive an equation for $g$ start by taking the derivative of
$$\int_s^T g(u)du + f(s,T)\int_s^T R(u)g'(u)du = 0$$
we obtain
$$g(s) - \frac{df}{ds}\int_s^T R(u)g'(u)du + fRg'(u) = 0$$
which equals (using the first equation to simplify)
$$g(s) + fRg'(u) = \frac{d\log f}{ds}\int_s^T g(u)du $$
Now taking the derivative again gives
$$g'(s) + \frac{dfR}{ds}g'(u) + fR g''(s) = \frac{d^2\log f}{ds^2}\int_s^T g(u)du - \frac{d\log f}{ds}g(s)$$
Which simplifies (by using the equation above) to
$$g''(s) + g'(s)\left(\frac{ff' + 2ff'^2R + f^2 f'R' - f^2f''R}{ff'R}\right) + \left(\frac{2f'^2 - f''f}{f^2f'R}\right)g(s) = 0$$
I highly doubt this equation has a general closed form solution, it just too general.