Proof of Peano's existence theorem for ODEs

Question

Theorem: Given a continuous function $f:(a,b)\times (c,d)\to \mathbb R$, the problem $$\begin{cases} y'(t)=f(t,y(t))\\ y(t_0)=y_0\end{cases}$$ has a local solution in some neighbourhood of a given point $t_0$.

A "standard" proof of Peano existence theorem makes use of the Ascoli-Arzelà theorem. I suspect that the following proof, which doesn't, is therefore wrong. But where?

Consider, instead of the open rectangle $(a,b)\times (c,d)$, a closed rectangle contained in it (since the thesis is local, this makes no difference), say $[a',b']\times[c',d']$. Consider polynomials $P_n(t,y)$ which converge uniformly to $f$ on the closed rectangle. These exist because of the Weiertrass theorem. Now the problems $$\begin{cases} y_n'(t)=P_n(t,y_n(t))\\ y_n(t_0)=y_0\end{cases}$$ have a unique global solution in $[a,b]$, since $P_n$ is a polynomial, hence globally Lipschitz on the rectangle (recall that globally Lipschitz implies global existence by gluing local solutions).

Now we say that the function $P_n$ has an uniform limit $f$, and therefore the functions $t\to y_n'(t)$ have an uniform limit, which we shall denote by $g(t)$. (This is the point which I am less sure of.)

Now we have a sequence of functions $y_n$ which take the same value in $t_0$, which are $C^1([a',b;])$ and whose derivatives converge uniformly to a function $g$. Therefore a limit for $y_n$ exists, is differentiable in a neighbourhood of $t_0$ and its derivative is $g$. (This is a theorem of elementary analysis: $f_n(a)\to l$ for a given $a$, $f'_n\to g$ uniformly, then $f_n$ has a limit $f$ which satisfies $f(a)=l$, $f'=g$.)

Thank you in advance for helping me finding the (possible) error.

Answer 1

I am not sure the proof is wrong or correct; however here are is one unwarranted conclusions that is drawn.

You say that the problems have a unique global solution in $[a,b]$. I don't see that hold. They have a unique ${\bf local}$ solution, but the interval where the uniqueness holds may depend on $n$ and may well shrink to 0. To check that this does or does not happen you need to unpack the proof of existence and uniqueness and see if you can guarantee a lower bound to the width of the existence-uniqueness intervals.

The argument you use that "globally Lipshitz etc" is not quite correct. Your polynomial approximates $f$ in the given rectangle. The solution may leave that rectangle. Moreover the Lipshitz constants for your polynomials $P_n$ is not uniformly bounded with respect to $n$ (in general).

As an example consider the simple equation

$ y' =y^2,\ \ \ y(0)=1$. The solution is $y(t)= \frac {1}{1-t}$ which has no global existence on [0,2] (for example), although the ODE is nice and polynomial (and even time-independent).

Answer 2

In your ode you should have $y_n'(t)=P_n(t,y_n(t))$. And you have to extract from the sequence of $y_n$ a converging subsequence. For this Ascoli helps a lot. After that your argument works to show that the limit of the subsequence verifies the ode.

Answer 3

I think that one of the major problems proof-verifying your question is that there are missing details. I don't know if you agree with me in this point but it was the feeling I got first time I read your question. In particular, the objections made by @user490850 made sense to me at that time and I won't repeat them here.

But the basic idea of using Weierstrass' approximation is good and it dates backs according to my research to Constantin Corduneanu who developed this proof in his article "Sur les inégalités différentielles" (it wasn't the main theme of the article). I'm pretty sure that he quoted Differential Equations by Birkhoff and Rota in its edition of the 30s.

Your theorem is a particular case. You could very well take $\mathbb{R}^n$ instead of $\mathbb{R}$ as in Proving Peano's Existence Theorem by approximating with $C^{\infty}$ functions using Weierstrass' Theorem. I haven't considered the latter as a duplicate for this reason. You can check my answer there as well as the linked document that contains the notations that may be strange and other details. On the other hand, the tough part is the $\mathbb{R}$ case...