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Consider the functional differential equation $$4f\left(\frac x2-f(x)\right)+4~\epsilon~ f(x)f'\left(\frac x2-f(x)\right)=f(x)$$ for all $x\geq0$ together with the initial condition $f(0)=0$ and the additional constraint $f(x)\geq0$ for all $x$. This equation arises as a first-order optimality condition in a problem that I am studying. $~\epsilon~$ is a parameter in an interval around $0$.

For all $~\epsilon~$ the equation has the trivial solution $~f(x)=0~$ for all $~x~$, which is not of any interest for me.

The equation also has a linear solution $$f(x)=\frac x{4(1-\epsilon)},$$ which is valid for all $\epsilon<1$.

Finally, for $\epsilon=0$, there exists the non-linear but smooth solution $$f_0(x)=\frac{1+2x-\sqrt{1+4x}}8.$$

My question is whether you have any idea how I can embed the solution $f_0$ for the singular case $\epsilon=0$ into a whole family of solutions $f_\epsilon$ for values of $\epsilon$ close to $0$.

I understand that this is a singular perturbation problem. Although I am primarily interested in an analytical solution, I would already be happy about an existence proof.

A numerical solution is not what I am looking for but it may be useful to get an idea about how the solution family could look like.

I would appreciate any hints or references.

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Hint:

Let $g(x)=\dfrac{x}{2}-f(x)$ ,

Then $f(x)=\dfrac{x}{2}-g(x)$

$f\left(\dfrac{x}{2}-f(x)\right)=\dfrac{\dfrac{x}{2}-f(x)}{2}-g\left(\dfrac{x}{2}-f(x)\right)=\dfrac{g(x)}{2}-g(g(x))$

$f'\left(\dfrac{x}{2}-f(x)\right)=\dfrac{g'(x)}{2}-g'(x)g'(g(x))$

$\therefore2g(x)-4g(g(x))+4\epsilon\left(\dfrac{x}{2}-g(x)\right)\left(\dfrac{g'(x)}{2}-g'(x)g'(g(x))\right)=\dfrac{x}{2}-g(x)$

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    $\begingroup$ I do not understand the fourth line of your hint. Shouldn't it be $f'(x)=(1/2)-g'(x)$? But more importantly, I do not see how your hint helps me to answer my question. $\endgroup$ – Gerhard S. Jul 12 at 9:26

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