Is there a non constant entire function $\gamma(t)=(x(t),y(t)): \mathbb{C} \to \mathbb{C}^{2}$ which satisfy the following Vander pol dififferential equation?

$$\begin{cases}\dot{x}=y-x^{3}\\\dot y=-x\end{cases}$$

For a related question see the last part of the following post:

The error in Petrovski and Landis' proof of the 16th Hilbert problem


No, there is not. Eliminate $y$ to get $x''=-x-3x^2x'$. There is no entire function $x(t)$ which satisfies this. Idea of the proof. Let $$M(r,f)=\max_{|t|\leq r}|f(t)|.$$ At the point $w$, $|w|=r$ where this maximum is achieved we have $\log|x''(w)|\sim \log|x'(w)|\sim\log|x(w)|$ for most values of $r$. This gives a contradiction: the last term is much larger than the rest.

This is called Wiman-Valiron theory. See MR1438606
Hayman, Walter K. The growth of solutions of algebraic differential equations. Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei (9) Mat. Appl. 7 (1996), no. 2, 67–73.

or Valiron's book Fonctions analytiques.

  • $\begingroup$ Prof. Eremenko Thank you very much for your answer. Can I ask you to elaborate? $\endgroup$ May 3 '16 at 12:42
  • $\begingroup$ What do you want me to elaborate? Just look to the references I gave. $\endgroup$ May 3 '16 at 12:54
  • $\begingroup$ Am I mistaken to think that the same argument should work for $x"=-x-x'$ rather than $x"=-x-x^2x'$? $\endgroup$ Sep 11 '17 at 7:28
  • 1
    $\begingroup$ @Ali Targavi: no, of course not. The equation you wrote is linear, therefore all solutions are entire. In the equation in my answer, there is a single term of higher degree then all other terms. This is crucial. $\endgroup$ Sep 11 '17 at 11:57

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