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Let $U \subseteq \mathbb R^{n+2}$ be an open set for some $n \geq 0$, and let $f: U \to \mathbb R$ be an analytic function. Then we say the equation $f(x,y,y',\ldots,y^{(n)})=0$ is an analytic differential equation.

Let $g(x)=y$ be a solution to an analytic differential equation.

Is $g$ necessarily analytic on some open set $V \subseteq \mathbb R$? What about if $U \subseteq \mathbb R^\infty$?

It is known that the answer to a more general question is no, namely, the solutions to a functional analytic differential equations need not be analytic anywhere. For example, the equation $f'(x)=2f(2x)$ uniquely defines the Fabius function, which is nowhere analytic.

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    $\begingroup$ It depends on the properties of the function $f(x,y,y',\ldots,y^{(n)})$. Can you solve your equation for the highest derivatives $f^{(n)} = h(x,y,y',\ldots,y^{(n-1)})$ with analytic $h$ (even if not uniquely) on an open subset $V\subseteq U$? Or are you specifically concerned with more degenerate cases? $\endgroup$
    – Igor Khavkine
    Commented May 13 at 6:51
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    $\begingroup$ Just an explanation on @IgorKhavkine'scomment: if it is possible to solve the equation respect to the highest order derivative, then the Cauchy part of the standard Cauchy-Kovalevskaya theorem applies and locally there is existence, uniqueness and analyticity of a solution. $\endgroup$
    – Daniele Tampieri
    Commented May 13 at 8:08
  • $\begingroup$ @DanieleTampieri: Thank you, this is partially what I was looking for. $\endgroup$
    – cubicquartic
    Commented May 14 at 0:14
  • $\begingroup$ @IgorKhavkine: To be clear, do you mean $y^{(n)}=h(x,\dots,y^{(n-1)})$ or $f^{(n)}$? The former seems to be what the Cauchy-Kovalevskaya theorem requires unless I'm reading incorrectly. If it's the former, wouldn't $f$ necessarily have a locally analytic inverse somewhere? $\endgroup$
    – cubicquartic
    Commented May 14 at 0:14
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    $\begingroup$ An ODE $y^{n}=g(x,y,y',\dots,y^{n-1})$ is said to be in normal form (note the keyword). For ODE in general form $f(x,y,\dots,y^{n})=0$, the first step is usually (but not always) looking for equivalent ODE in normal form, e.g. via the implicit function theorem $\endgroup$
    – Pietro Majer
    Commented May 14 at 6:39

1 Answer 1

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No. Take for instance the following ODE where $0$ is a regular singular point, $$x y'=\lambda y.$$ The functions $cx_+^{\lambda}$ are solutions and are not analytic. If $\lambda$ is not a non-negative integer, they are not even the restriction to $[0,+\infty)$ of an analytic function. On the top of that, you have a non-uniqueness phenomenon for the Cauchy problem at 0, with plenty of solutions vanishing at the origin.

There are worse situations. Consider the following ODE where $0$ is an irregular singular point, $$ x^2 y'=y. \tag{$\sharp$}$$ Looking for solutions of type $e^a$, you get the equation $ x^2 a'=1, $ showing that $e^{-1/x}H(x)$ solves $(\sharp)$ (here $H$ stands for the indicatrix of $\mathbb R_+$). This is worse than the previous example, since here you do get a (non-unique) solution which is not analytic in any neighborhood of 0.

P.S. Of course if you have a non-singular point with an equation (or a system) such as $$ y'=f(x,y), $$ with $f$ analytic, you will be able to find a unique local solution for the Cauchy problem and that solution will be analytic.

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    $\begingroup$ But the question asks if the solution is analytic on 'some' open set. So a counterexample would have to be a nowhere analytic function. Or am I misinterpreting something? $\endgroup$
    – Michael Bächtold
    Commented May 14 at 7:19
  • $\begingroup$ The solution of $(\sharp)$ given above is a smooth function which is not analytic in any neighborhood of the origin. Since $0$ is the only singular point (at a finite distance) of that equation, any solution will be analytic on $\mathbb R^*$ (consequence of the P.S.), so in that sense, this counterexample is optimal. $\endgroup$
    – Bazin
    Commented May 14 at 13:37

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