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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).

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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.

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  • $\begingroup$ It is not that simple. Let $\displaystyle F(y,s)= \frac{w'(y-s)}{w'(y-s)+g(y)v'(s)}$, $0\leq s\leq y\leq 1$. Due to $w'(0)=v'(0)=\infty$ (for $a,b\in(0,1)$), one has $F(y,0)=0$, $F(y,y)=1$ for $0<y<1$. That is, there is a singularity exactly at the point $(y,s)=(0,0)$, where the evolution starts. $\endgroup$
    – ifw
    Aug 26, 2015 at 11:13
  • $\begingroup$ This sounds promising, but a bit unclear. It is safe to assume for my purposes that $w$, $v$ and $g$ are all continuously differentiable, but indeed $w'(0)=v'(0)=\infty$. If I understand correctly, to apply Peano's theorem, I should define $f$ in an open set (for example $(0,1)$) and have an initial condition for $x_0$ in this set? $\endgroup$
    – TomH
    Aug 26, 2015 at 11:32
  • $\begingroup$ Assuming $g\in C^1([0,1])$, Picard-Lindelöf can be applied in the interior of the triangle $\Delta=\{0\leq s\leq y\leq1\}$. For instance, fix $0<\delta<1$ and denote by $f_\lambda$, $\lambda\in(0,1)$, the unique solution of $f'(y) = F(y,f(y))$, $f(\delta)= \lambda \delta$ (defined on a maximal existence interval). Then there is a non-empty interval $\Sigma$, closed in $(0,1)$, such that $f_\lambda$ for $\lambda\in \Sigma$ is defined for $0\leq y\leq 1$, and $f_\lambda(0)=0$, $\ldots$ $\endgroup$
    – ifw
    Aug 26, 2015 at 14:30
  • $\begingroup$ $\ldots$, while, solving backwards, $f_\lambda$ for $\lambda\in (0,1)\setminus\Sigma$ terminates at either the lower or the upper face of $\Delta$. So, $\{f_\lambda\mid \lambda\in\Sigma\}$ is the set of solutions, and unique solvability holds iff $\Sigma$ consists of a single point. $\endgroup$
    – ifw
    Aug 26, 2015 at 14:32

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