Skip to main content
added the implicit continuity assumption
Source Link
SashaKolpakov
  • 1.3k
  • 1
  • 11
  • 22

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 continuous 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

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

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 continuous 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

Source Link
SashaKolpakov
  • 1.3k
  • 1
  • 11
  • 22

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