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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 $y$).

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    $\begingroup$ If you separate the variables and integrate you get $x=\int_0^{y} \frac{ds}{f_n(s)}:=G_n(y)$. By dominate convergence $G_n(y) \to 2 \sqrt y$ uniformly in bounded intervals and then $y_n(x)=G_n^{-1}(x) \to x^2/4$ at least pointwise. To get uniform convergence you can trap $\sqrt y \leq f_n \leq \sqrt y+\epsilon_n$ and proceed as before, I guess. $\endgroup$ Sep 8 at 16:06

2 Answers 2

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$\newcommand\ep\varepsilon$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.

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    $\begingroup$ $f_n(0)$ must be positive since $f_n(x) \geq \sqrt x $ and $f_n$ is differentiable at 0. $\endgroup$ Sep 8 at 18:51
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    $\begingroup$ @GiorgioMetafune : I forgot about this condition. Anyhow, to take this condition into account, the answer only needed a slight modification, which is now done. Thank you for your comment. $\endgroup$ Sep 8 at 19:38
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This question would be possibly at a better place on MathStack Exchange.

Yet, once the statement of the question is corrected (the functions $y_n$ need to be defined on $\mathbb{R_+}$ and not only on $[0,1]$, and they should converge uniformly to $\sqrt{\cdot}$ on compact sets), the answer will be positive.

For strictly positive functions $y'=\sqrt{y}$ is equivalent to $y'/\sqrt{y}=1$, namely $2\sqrt{y}-x$ constant (by integration). Hence the only positive solution of the ODE $y'=\sqrt{y}$ on $[0,1]$ which vanishes only at $0$ is $y_0 : x \mapsto (x/2)^2$.

In a same way, the only positive solution of the ODE $y'=1+\sqrt{y}$ on $\mathbb{R}_+$ which vanishes at $0$ is $G^{-1}$, where $G$ is the strictly (increasing) function defined by $$G(y) := \int_0^y \frac{\mathrm{d}z}{1+\sqrt{z}}.$$

Since the $(f_n)_{n \ge 1}$ are $\mathcal{C}^1$ and bounded below by $\sqrt{}$, they are strictly positive everywhere (including at $0$), so the solutions $y_n$ of the Cauchy problem $y'=f_n(y)$, $y(0)=0$ are (strictly) increasing.

Let $A:=G^{-1}(1)$. We assume that $(f_n)_{n \ge 1}$ converges uniformly to $\sqrt{}$ on $[0,A]$. Hence, whenever $n$ is greater to some positive integer $N$, we have $\sqrt{} \le f_n \le 1+\sqrt{}$ on $[0,A]$. While $y_n$ remains in $]0,A]$, we derive $$\frac{y_n'}{1+\sqrt{y_n}} \le 1 \le \frac{y_n'}{\sqrt{y_n}}.$$ Since $y_n(0)=0$, we get by integrating $$G(y_n(x)) \le x \le 2\sqrt{y_n(x)},$$ so $y_0(x) = (x/2)^2 \le y_n(x) \le G^{-1}(x)$. As a result, for every $n \ge N$, $y_n$ remains on $[0,A]$ on the time interval $[0,1]$ and $y_n'$ is uniformly bounded (namely $0 \le y_n' \le 1+\sqrt{A}$ on the time interval $[0,1]$. By Ascoli-Arzela's theorem, the sequence $(y_n)_{n \ge N}$ is relatively compact in $\mathcal{C}([0,1],\mathbb{R})$.

Thus, we have to check that its only limit point is $y_0$. Indeed, if $y$ is a subsequential limit, then taking limits along this subsequence in the equalities $$y_n(x) = \int_0^x f_n(y_n(t)) \mathrm{d}t$$ yields $$y_n(x) = \int_0^x \sqrt{y_n(t)} \mathrm{d}t$$ But $y_n$ is bounded below by $y_0$. Hence $y_n=y_0$ by the uniqueness argument viewed at the beginning. We are done.

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    $\begingroup$ You wrote: "once the statement of the question is corrected (the functions $y_n$ need to be defined on $\mathbb{R_+}$ and not only on $[0,1]$, and they should converge uniformly to $f$ on compact sets), the answer will be positive." -- It is unclear what your $f$ is. It also appears that, instead of $y_n$, you meant $f_n$. If so, then $f_n$ does not need to be defined on $\mathbb{R_+}$. It even suffices that $f_n$ be defined on $[0,1/4+\delta]$ for some $\delta>0$ and all large enough $n$. So, it is actually more than enough that $f_n$ be defined on $[0,1]$ for all large enough $n$. $\endgroup$ Sep 9 at 16:23
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    $\begingroup$ I corrected the typo, by replacing $f$ with $\sqrt{}$. Anyway, you do not know at the beginning what the range of the $y_n$ can be. It has no reason to be contained in $[0,1]$, that is why I wrote that $f_n$ should be defined on $\mathbb{R}_+$ (and we still can have explosion in finite time for small values of $n$). As you note, it is sufficient to assume that $f_n$ converges uniformly to $\sqrt{}$ on some interval $[0,1/4+\delta]$, so the interval $[0,1]$ works. It is because we are lucky. In general, it is not always possible to take the `spatial' interval equal to the time interval. $\endgroup$ Sep 9 at 18:42
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    $\begingroup$ There are at least two ways to deal with this: (i) You yourself extend (almost arbitrarily) $f_n$ to $[0,\infty)$ or (ii) just assume what is already assumed in the OP: that the ODE (makes sense and) satisfies the listed conditions (initially not caring whether such assumptions are contradictory in themselves) and finally justifying the OP setting after the proof is complete. $\endgroup$ Sep 9 at 19:03

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