My question is about computability in principle. You're right about real numbers containing infinite data, so I don't mind restricting the question to polynomials with rational coefficients, and the model of computability being (say) a Turing machine. But perhaps the question with real coefficients can also make sense, for example maybe for polynomials of a given maximal degree of $p,q$ one can define some polynomial expression in the coefficients of $p,q$ which will be non-negative if and only if the zero sets of $p,q$ are homeomorphic.

Thank you, this is very interesting. Let me add that my question is motivated by particular systems of ODE's for which one can show that assumptions (1)-(3) hold, and one would like to prove global stability. In these systems, the ODE's are polynomial ones. So I wonder if it is possible to get a positive answer by restricting the class of vector fields to polynomial ones. Since a "plugging" construction cannot be performed with polynomial vector fields, maybe there is a chance..

For the "pushing in" part - don't you need some restrictions on the topology of M so that you can contract the whole of $R^n$ to it?

I'd like to note that the answer to my question is positive in dimensions $n=1$ (trivially) and $n=2$ (using the Poincare-Bendixon theorem), so the counterexample can only work in dimension 3 or higher

Locally asymptotically stable means that there is a neighborhood of the stationary point which such that starting in this neighborhood you approach the stationary point as $t\rightarrow \infty$. Globally asymptotically stable means that this holds for starting from any point. The stationary point must be in the absorbing ball because if it were outside it, when you start at the stationary point you would remain there, contradicting the definition of absorbing ball.

Thank you for the pointer! Is it clear that the results for discrete dynamical systems extend to continuous-time ones? Is there any chance of constructing an explicit ODE system (say with polynomial nonlinearities) which will exhibit this behavior?

@Wadim: I understand the general intuition is that in any case where you have an entire solution, an arbitrarily small perturbation will destroy this. This can be stated as the following conjecture: consider the space of polynomials $P$ in $n+2$ variables with degree $d\geq 2$. Give it the natural topology. Then the set of $P$'s for which an entire solution exists is nowhere dense (doesn't contain any open set).

Thank you kaminoite and Wadim for the very interesting connection with the question of integrability. A quick search led me to a review paper by Martin D.Kruskal, Nalini Joshi, Rod Halburd arxiv.org/abs/solv-int/9710023 There, the authors state that "There is strong evidence [60, 61] that the integrability of a nonlinear sytem is intimately related to the singularity structure admitted by the system in its solutions." It seems from this formulation that the relation is not entirely established. If you can recommend other sources that clarify this, I would be most grateful.

@Wadim: My question was not prompted by a specific example. Rather, I was thinking about the fact that writing a polynomial ODE with initial conditions is an explicit way to define a holomorphic function, so that it is natural to wonder what properties of the function can be deduced from this representation. It seems that indeed it is harder than I suspected. Can you give a hint as to how you show that any perturbation of the example you gave leads to a non-entire solution?

Is the criterion you are talking about really an if-and-only-if one? Does the lack of singularities imply integrability in the sense you refer to?

Could it be that the Borel problem you mentioned relates to the more general equation $P(z,f(z),f'(z),...f^{(n)}(z))=0$, and has a known solution in the case that the equation is of the form $f^{(n)}(z)=P(z,f(z),f'(z),...f^{(n-1)}(z))$? Just wondering.. (of course my original question can also be formulated for the more general form of the differential equation).

It's also possible to formulate a somewhat more general question for which the linear equation issue will (probably) not be relevant, by allowing $P$ to be a rational function, in which case there are probably many examples of entire functions which are not solutions of linear equations. E.g. $f(z)=e^{e^z}$ solves $f''(z)=\frac{(f'(z))^2}{f(z)} +f'(z)$, but I would guess it doesn't solve any linear differential equation with coefficients that are polynomials in $z$ (proof?).

Leonid's question and fedja's response brings to mind the question whether any entire solution of such a differential equation is also a solution of linear equation. I don't know the answer. Even if it is positive, it will lead to the solution of the decision problem only if it can be decided whether the solution of such a nonlinear equation is also the solution of a linear one.