A direct answer to your question appears difficult, as the implicit function theorem is not effective. Thus one could only try to compute a series approximation and then to prove that it actually converges (implying that y(z) is analytic in some neighbourhood of the chosen expansion point).

However, I would argue differently. The right hand side of your differential equation is analytic (and thus trivially satisfy a local Lipshitz condition). One approach to prove the standard existence and uniqueness theorem for ODEs, the Picard-Lindelöf theorem, is via the Banach fixed point theorem. Usually, one considers there the space of C1 functions as underlying Banach space. But you may also consider the Banach space of analytic functions, as you are dealing with an analytic equations, and then the standard argument tells you that all solutions of your equations are locally analytic.

Regards Werner

A complex autonomic ODE $y' = f(y)$ has non-constant solutions  $F(y) = z-z_o$  where  $F$  is a primitive of $1/f$ on the domain  $D = \{u :f(u) \neq 0\}$, in other words,. $F' = 1/f$  on $D$. There are two question;

1) Does a holomorphic primitive function for $1/f$  exist on $D$?

2) Is  $F = \int(1/f)$  invertible by a meromorphic function?

In your example, $f(u) = 1-u^3 = (1-u)(w-u)(\bar w - u)$  with  $w = e^{2\pi i/3}$. Thus $1/f$ has 1st order poles at $1, w, \bar w$ and the answer to 1) is No.

But still the answer to 2) could be Yes. Consider a neighbourhood of $1$ in $C$. Put $s = u-1$ and let g(s) = f(u) = f(s+1). We have $g(s) = s h(s))$  where  $h(s) = a + bs + \dots$ with  $a\neq 0$. Hence  $\int(1/g) = \int(a/s + b + \dots) = a \ln(s) + k(s)$ (all functions g,h,k, etc. are holomorphic). Thus we have  $a \ln(y-1) + k(y-1) =  z-z_o$. Exponentiating we have  $(y-1)^a e^{k(y-1)} = e^{z-z_o}$. The left hand side can be invertible as a function of $y$ only if the power $a$ is rational (this condition is necessary but not sufficient). But in our case,  $a = -(w-1)(\bar w -1) = -|w-1|^2$ which is an irrational number. Therefore, non-constant solutions are not meromorphic. Hope that everything is correct.

Best regards

Jost

Thanks to you, Jost and Werner.

I will make use of your ideas to better figure out this case.

The origin of my question is the theorem of Malmquist-Yosida on Riccati equations :

Let $w'=R(z,w)$ be a diff. eq. where $R$ is a rational function of z and w. If the ODE has a solution which is transcendental, meromorphic  and has infinitely many poles, then $R(z,w)$ is a polynomial of degree 2 in $w$.

The classical example is $y'=1+y^2$ and $y(z)=\tan z$.

So when $R$ is not of degree 2, (here d°R=3 in $w$,   as $w' = 1-w^3$)

so either a solution is not meromorphic or it is, and has finitely many poles (unless it is not transcendental). 

That's why I have to study the first integral below:

$$-\ln(abs(y-1))+1/2 \ln(1+y+y^2)+\sqrt(3)\atan((2y+1)/sqrt(3)) - y_0 = z - z_0$$

.

Gabriel

Hello Jost,

 I performed again the computation and cannot get the value $a=-(w-1)/(\bar w -1)$.

Using $g(s) = s h(s)$ where $h(s) = a + bs + \dots$, I find out that $a$ is a rational number.

Best wishes,

  Gabriel

Hello,

I thought (after reading Hille's book on differential equations)

to apply the so-called "power test" : if there is a polar singularity at $z=z_1$,

we write the leading term $W(z)\sim a(z-z_1)^(-b)$ and so $W'(z)\sim -ab(z-z_1)^(-b-1)$.

$a\neq 0$ and $b>0$.

After substituting in the equation $w'=1-w^3$ we get $b=1/2$ so that there is no meromorphic map, solution to this equation.

Gabriel