For $n\geq 1$, $f\_n\in\mathcal{C}^1([0,1],\mathbb{R})$ such that $f\_n(x)\geq\sqrt{x}$ for $x\in[0,1]$, and


$$\lim\limits\_{n\to+\infty}\sup\_{x\in[0,1]}\big|f\_n(x)-\sqrt{x}\big|= 0.$$


Let $y\_n$ be the unique solution of


$$\begin{cases}
y\_n(0)=0 \\
y\_n'=f\_n(y\_n) \text{ on [0,1]}.
\end{cases}$$


**Question:** Is there a function $y\in\mathcal{C}^1([0,1],\mathbb{R})$ such that


$$\lim\limits\_{n\to+\infty}\sup\_{x\in[0,1]}\big|y\_n(x)-y(x)\big|= 0$$


which is solution of the system (which has itself an infinity of solutions)


$$\begin{cases}
y(0)=0 \\
y'=\sqrt{y} \text{ on [0,1]}
\end{cases}$$


and satisfies the condition: $y(x)>0$ for $x\in\,]0,1]$ (which, I hope, permits to characterize $y$).



Forcing the uniqueness of a solution of an ODE
$\newcommand\ep\varepsilon$First, the conditions that $f\_n\in\mathcal{C}^1([0,1],\mathbb{R})$ and $f\_n(x)\ge\sqrt{x}$ for $x\in[0,1]$ imply $f\_n(0)>0$. Since
\begin{equation\*}
 \begin{cases}
y\_n(0)=0, \\
y\_n'=f\_n(y\_n) \text{ on [0,1]}, 
\end{cases}
\tag{2}\label{2}
\end{equation\*}
we see that $y\_n>0$ in a right neighborhood of $0$. Since $y'\_n=f\_n(y\_n)\ge0$, we see that $y\_n>0$ on $(0,1]$.
Letting then $u\_n:=\sqrt{y\_n}$, we get $2u\_n u'\_n=f\_n(u\_n^2)\ge u\_n$, whence $u'\_n\ge1/2$, $u\_n(x)\ge x/2$, and
\begin{equation\*}
 y\_n(x)\ge x^2/4 \tag{3}\label{3}
\end{equation\*}
for all $x\in[0,1]$.


Next, for $\ep\in(0,1)$, let $z\_\ep$ be the unique solution of the ODE
\begin{equation\*}
 z'\_\ep=\sqrt{z\_\ep}+\ep
\end{equation\*}
on $[0,1]$ with the initial condition $z\_\ep(0)=0$. It is not hard to see that
\begin{equation\*}
 z\_\ep(x)\to x^2/4 \tag{4}\label{4}
\end{equation\*}
uniformly in $x\in[0,1]$ as $\ep\to0$. (See details on this at th end of this answer.)


Let now
\begin{equation\*}
 \ep\_n:=\sup\_{x\in[0,1]}\big|f\_n(x)-\sqrt{x}\big|
 =\sup\_{x\in[0,1]}\big(f\_n(x)-\sqrt{x}\big), 
\end{equation\*}
so that $\ep\_n\to0$ (as $n\to\infty$), and then let
\begin{equation\*}
 w\_n:=z\_{\ep\_n+1/n}. 
\end{equation\*}
So, $y\_n(0)=0=w\_n(0)$,
\begin{equation\*}
 y\_n'\le\sqrt{y\_n}+\ep\_n,\quad w\_n'=\sqrt{w\_n}+\ep\_n+1/n. 
\end{equation\*}
Suppose that
\begin{equation\*}
 x\_n:=\sup\{x\in[0,1]\colon y\_n\le w\_n\text{ on }[0,x]\}<1. 
\end{equation\*}
Then $x\_n>0$ and $w\_n(x\_n)=y\_n(x\_n)$, and hence $y'\_n(x\_n)\ge w'\_n(x\_n)$, so that
\begin{equation\*}
 \sqrt{w\_n(x\_n)}+\ep\_n=\sqrt{y\_n(x\_n)}+\ep\_n \\ 
 \ge y'\_n(x\_n)\ge w'\_n(x\_n)=\sqrt{w\_n(x\_n)}+\ep\_n+1/n,
\end{equation\*}
a contradiction. So, $x\_n=1$ and hence, in view of \eqref{3},
\begin{equation\*}
 x^2/4\le y\_n(x)\le w\_n(x)=z\_{\ep\_n+1/n}(x)\to x^2/4
\end{equation\*}
uniformly in $x\in[0,1]$, by \eqref{4}.


On the other hand, the only solution $y$ of the system  

\begin{equation\*}
 \begin{cases}
y(0)=0 \\
y'=\sqrt{y} \text{ on [0,1]}
\end{cases}
\end{equation\*}
such that $y>0$ on $(0,1]$ is given by the formula $y(x)=x^2/4$.


Thus, $y\_n\to y$ uniformly on $[0,1]$, as desired.




---


**Details on \eqref{4}:** Letting $t\_\ep:=\sqrt{z\_\ep}$, rewrite
$z'\_\ep=\sqrt{z\_\ep}+\ep$ as $2t\_\ep t'\_\ep=t\_\ep+\ep$. "Separating the variables", we find that $t\_\ep=g^{-1}\_\ep$, where
\begin{equation\*}
 g\_\ep(t):=2t-2\ep\ln\frac{t+\ep}\ep. 
\end{equation\*}
We have $g'\_\ep(t):=2-\frac{2\ep}{t+\ep}>0$ for $t>0$, so that the inverse function $g^{-1}\_\ep$ is well defined. Using the inequalities $\ln(t+\ep)\le t+\ep-1<t$, we get
\begin{equation\*}
 g\_\ep(t)\ge (2-2\ep)t+2\ep\ln\ep 
\end{equation\*}
for $t\ge0$, whence for $\ep\to0$ we have $t\_\ep(x)=g^{-1}\_\ep(x)\le\frac{x-2\ep\ln\ep}{2-2\ep}\to x/2$ and
\begin{equation\*}
 z\_\ep(x)=t\_\ep(x)^2\le(1+o(1))x^2/4
\end{equation\*}
uniformly in $x\in[0,1]$. Also, similarly to \eqref{3}, $z\_\ep(x)\ge x^2/4$ for $x\in[0,1]$. Now \eqref{4} follows.


