I want to prove that there exists $f:[0,1] \to [0,1]$ such that $f(0)=0$,
$$
\frac{d w(y-f(y))}{d y} = g(y) \frac{d v(f(y))}{d y}, \forall y \in [0,1],
$$
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.


If it simplifies the problem, I am particularly interested in the case where
$$
 w(x)=\frac{x^a}{[x^a+(1-x)^a]^{1/a}},
 v(x)=\frac{x^b}{[x^b+(1-x)^b]^{1/b}},
$$
for some $a,b \in (0,1]$.


For example, if $a=b=1$, then $v(x)=w(x)=x$, and the solution is $f(y) = \int\_0^y \frac{1}{1+g(x)} dx$.


I don't know how to approach this problem with nonlinear $w$ and $v$ functions. I only need to prove that $f$ exists (and if there is a way to compute it numerically, it would be even better).



How to prove that a non-linear differential equation has a solution
Before attempting anything complicated (such as the theory of Differential Algebraic Equations), I would try to put it in a standard first-order ODE form. By the chain rule:
$$w'(y-f(y))(1-f'(y))=g(y) v'(f(y)) f'(y)$$
Now, since $w'$, $v'$, and $g$ are positive, you can write an explicit ODE:
$$f'(y)=\frac{w'(y-f(y))}{w'(y-f(y))+g(y) v'(f(y))}$$
and you only have to prove Lipschitz continuity of the right-hand side to apply a standard existence and uniqueness theorem. (If $g$ were differentiable, Lipschitz continuity would be straightforward).


Now you can put your equation into any numerical code to get a numerical solution.


