How to prove that a differential equation has a solution I want to prove that there exists $f : [0,1] \to [0,1]$ such that
$$
  w(y - f(y)) 
  =
  \int_0^y g(x) dv(f(x)),
$$
where $w : [0,1] \to [0,1]$ and $v : [0,1] \to [0,1]$ are continuous strictly increasing functions with $w(0)=v(0)=0$ and $v(1)=w(1)=1$, and $g : [0,1] \to (0,\infty)$ is a continuous function.
For example, when $v(x)=w(x)=x$, then simply $f(y) = \int_0^y \frac{1}{1+g(x)} dx$.
But I don't know how to approach this problem with more general $w$ and $v$ functions. I would appreciate any hints. Also, any hints, how to (numerically) characterize $f$ for known functional forms would be very useful.

You have a very non-trivial problem. Your integral equation is equivalent to the differential equation (obtained by differentiating your equation)
$$
\frac{d w(y-f(y))}{dy} = g(y) \frac{dv(z)}{dz}|_{z= f(y)} \frac{df(y)}{df}
$$
with the initial condition $w(0-f(0)) = 0$, from your context (I guess) you have $f(0) =0$ as an additional condition. The problem is that if $v(z)$ above is not a linear or simple enough function, you end up with a nonlinear differential equation. I am not very well informed, but I have heard that it is very hard in general to even prove the existence of solution in nonlinear differential equations, although some tricks might work for special instances, as far as I know it is a highly nontrivial general task. 
