# Analytic solutions to algebraic differential equation

Dear Colleagues and Friends,

Here I need to find some good reference on a subject that seems very much studied: sorry, if the rest of this question is too naive.

I believe that it's known that if a function $f(z)$ satisfies an equation $P(z, f(z)) = 0$ with $P(z, \xi) \in \mathbb{C}[[z, \xi]]$ a non-zero analytic function with $P(0, 0)=0$, then $f(z)$ is analytic in some neighbourhood of zero (for my purposes $P(z, \xi)$ can be taken to be just a polynomial).

What happens if instead of $P$ above we have, say, $P(z, \xi_0, \xi_1, \dots, \xi_n)$ (also a polynomial in $\mathbb{C}[z,\xi_0,\dots,\xi_n]$) and the equation now becomes $P(z, f(z), f'(z), ..., f^{(n)}(z)) = 0$?

Is anything known about $f(z)$ being analytic in this case, and under which conditions on $P$?

Any discussion and especially a good reference will be much appreciated!

Sasha

• Is $z$ a complex variable? Is $f$ assumed analytic on a dense open set, or meromorphic, or continuously differentiable, or smooth, or a distribution, or in one of the usual function spaces? – Ben McKay Jul 11 '17 at 16:54
• Your premise is false. $f(z) = \sqrt{z}$ satisfies $P(z, f(z)) = z - f(z)^2 = 0$ but is not analytic in any neighbourhood of $0$. – Robert Israel Jul 11 '17 at 17:13
• I had assumed that $f$ was supposed to be continuous in a neighborhood of $0$; otherwise, the OP's first assertion is absurd. If one does assume continuity, then $\sqrt{z}$ does not provide a counterexample to the initial statement, as it cannot be made a continuous function (i.e., a single-valued relation) in a neighborhood of $0$. For the second part, as soon as one assumes that $f'$ exists (i.e., that the function is complex differentiable), it must analytic. OTOH, interpreting $f^{(k)}(z)$ as $\partial^k f/\partial z^k$, it's false: $f = \bar z$ satisfies $\partial f/\partial z=0$. – Robert Bryant Jul 12 '17 at 9:20

Your belief is not correct, as Robert Israel pointed in his comment. Same for differential equations: $yy'-(3/2)z^2=0$ has a solution $y(z)=z^{3/2}$ which is not analytic at $0$. The correct theorem is: If $P(z_0,y_0,y_1,\ldots,y_{n})=0$, where $P$ is a polynomial, (or an analytic function in a neighborhood of $(z_0,\ldots,y_n)$), and $\partial P/\partial y_{n}\neq 0$ at the same point, then there exists a unique analytic function $y(z)$ in a neighborhood of $z_0$, which satisfies $P(z,y,y',\ldots,y^{(n)})=0$ in this neighborhood, and $y(z_0)=y_0,\; y'(z_0)=y_0,\ldots,y^{(n)}(z_0)=y_{n}$. One does not need to assume a priori that $y(z)$ is analytic. If it is $n$ times differentiable in a neighborhood of $z_0$, it is automatically analytic. For the proof, see, for example H. Cartan, Elementary theory of analytic functions of one and several complex variables, or Coddington and Levinson, Ordinary differential equations.
EDIT. The second question (asked in the comment) seems to be: if $f(z)$ is a formal power series satisfying a polynomial equation, does it follow that $f$ is convergent (so $f$ is analytic). The answer is yes in the case of implicit function theorem ($P(z,f(z))=0$), and no in the case of differential equations: the divergent series $$f(z)=\sum_{n=1}^\infty (n-1)!z^n$$ satisfies the Euler equation $z^2f'(z)-f(z)-z=0$, and $f(0)=0$.
• Would the situation change if I ask for $f(z)$ in $P(z, f(z))=0$ to be a formal power series first $f(z) \in \mathbb{C}[[z]]$, and then assert it should be analytic (may be the only one such solution is identically $0$)? I'd believe that if $P$ involves also derivatives of $f$ then it's a wrong statement (Riccati's equation is of this type and it has formal series solutions that are divergent except $z=0$). However I wonder what is known in general about this type of equations (hopefully I don't ask anything silly again). – SashaKolpakov Jul 12 '17 at 5:05
• Sorry for being unclear. Indeed, you construed my question correctly. Basically, I think that one can show that if a divergent formal power series satisfies some differential equation $y^{\prime} = Q(y(z))$, with $Q\in\mathbb{C}[z][\xi]$ of degree $\geq 2$ in $\xi$, then $y$ cannot satisfy a simpler, linear differential equation. May be some other conditions are necessary. – SashaKolpakov Jul 12 '17 at 14:12
• The general line of thought (again, can be blurry): if $y(z)$ satisfies a linear diff equation (with coefficient polynomials in $z$), then we can substitute in it all the derivatives of $y(z)$ with polynomials in $\xi = y(z)$ and get a polynomial equation $P(z,y(z))=0$. Then the implicit function theorem says that $y(z)$ must be analytic, though it cannot be, since it's represented by a divergent everywhere (except $z=0$) series. – SashaKolpakov Jul 12 '17 at 14:41
• Knowing this will make sure that some sequences represented by coefficients of certain divergent series do not have "simple" recursive formulas behind them - that's the motivation. However, when I'm confronted with a case when $y(z)$ satisfies some equation $y^{(n)}(z)=Q(y^{(n−1)},y^{(n−2)},…,y′,z)$, with $n\geq 2$, the above reasoning would not work. That's why I was asking for clarification and references to make sure I'm getting things right. For now I've learnt a few things from your answers, and my problem has got a clearer shape. Thank you again! – SashaKolpakov Jul 12 '17 at 14:43