Being a novice with differential equations, I recently learned that if $\mu$ is an integrating factor for $\frac{dy}{dx}f(x,y)+ g(x,y)=0$, then the corresponding 1-form, $\mu fdy+\mu g dx$, is exact.

My question is, is there some constructive proof that given a 1-form, $fdy+gdx$, we can find a scalar function $\mu$ so that $\mu fdy+\mu gdx$ is exact, and under what conditions can we do so? The only proofs for the existence of integrating factors I've found so far use existence theorems for solutions to ODEs and then prove the there is an integrating factor based on the existence of a solution.

To be clear, I would be very satisfied with an answer that looked like: subdivide a region into squares; linearly interpolate a scalar function to ensure the integral around each square is zero; cut each square into smaller squares; repeat. I would also be satisfied (though less so) with a less-constructive fixed-point theorem method. But, I'd like to stay in the context of 1-forms if possible.

  • $\begingroup$ The Poincare lemma lets you convert the exactness condition to a closedness condition. If you call your one form $\omega$ then you are asking for $d \mu \wedge \omega + \mu d\omega = 0$ which is saying that $d\omega$ can be written as $-\frac{1}{\mu} d\mu \wedge \omega$, at least, where $\mu$ is not zero. So it's like saying $d \omega$ factors as the derivative of a logarithm wedge the original form $\omega$. $\endgroup$ Jan 6, 2016 at 17:54
  • $\begingroup$ I suppose you can turn that into a procedure to compute $\ln(\mu)$, at least on convex sets. It's under-determined. You should probably be able to write $d(\mu)$ as a combination of contractions and duals of $d \omega$. $\endgroup$ Jan 6, 2016 at 18:01

2 Answers 2


You are basically asking for an «algorithm» deciding whether a first-order ODE (with, say, analytic coefficients $f, g$) has a (meromorphic,say) first-integral. This question (which Poincaré asked when $f, g$ are polynomials) is difficult and still open. I'll try to give here examples of undecidability from the sole knowledge of Taylor series representation of $f$ and $g$.

First, I don't know what you exactly mean by «constructive», as a lot depends on your coefficients field. For instance the $1$-form $$\omega_\lambda:=\lambda x\mathrm{d}y-y\mathrm{d}x$$ has an integrating factor if, and only if, $\lambda\in \mathbb Q$. But in most reasonable computation models (ex: BSS) you can't decide if a given (complex/real) number is rational.

That being said, even in a «perfect» computational model you still can't determine this fact algorithmically in all generality. Take the example of the focus/center problem: being given the $1$-form $$\omega:=x\mathrm{d}x+y(1+h(x,y))\mathrm{d}y,$$ is it possible to read in a finite jet of $h$ at $(0,0)$ if there exists a local integrating factor? The answer is «no» in general. If $h=0$ the foliation integrating the distribution $\ker\omega$ has a center singularity at $(0,0)$ (all orbits are closed loops) with first-integral $x^2+y^2$, for $\mathrm{d}(x^2+y^2)=2\omega$. In general though the singularity is a focus (no closed orbits). To see this, fix the Poincaré section $\Sigma:=\{x<0, y=0\}$) and take the partial orbit $\gamma_z$ starting from $(z,0)\in\Sigma$, making one turn around the singularity and coming back at $\Sigma$ at a point $(P(z),0)$. Then $\gamma_z$ is closed if, and only if, $P(z)=z$, that is $$\int_{\gamma_z} \frac{yh(x,y)}{x^2+y^2}\mathrm{d}y = 0.~~~~~(*)$$ Gory computations shows you all the Taylor coefficients of $h$ mixed up with (in general) non-zero coefficients in a big infinite sum. You can always arrange this sum to be $0$ or not, by perturbing a sufficiently far away Taylor term of $h$ (meaning that you can't decide if it sums to $0$ or not without encompassing the whole infinite collection of terms).

If $h$ is polynomial of given degree $d$ (or more generally if $h$ belongs to a finitely-determined functional space) then condition $(*)$ determines an analytic hypersurface. I'm almost sure that for degrees $d$ as low as $4$ the equation of this hypersurface is not known. I don't know if it is computable (in a reasonable model) or not, but I'm sure brute-forcing the equation is not an option here.

  • $\begingroup$ Why this $\omega_\lambda$ has an integrating factor if, and only if, $\lambda\in \mathbb Q$? It's equal to $\lambda .x^{(\lambda+1)/\lambda}.d(yx^{-1/\lambda})$. It has to do with being well defined around the origin? $\endgroup$ Nov 10, 2021 at 21:49
  • 1
    $\begingroup$ I mean meromorphic integrating factor. $\endgroup$ Nov 11, 2021 at 15:15

This seems indeed equivalent to the classical question treated by Darboux, Painleve, Poincare and others. Not surprisingly it can have several different formulations.

Classical formulation is the following: Given a polynomial equation $f(x,y)dx+g(x,y)dy=0$, to find with finitely many operations an integrating factor, or to prove it does not exist. The next question is "what kind of integrating factor"? The simplest case is a rational integrating factor (though more general factors were considered in the classical papers). If we look for a rational integrating factor of degree at most $n$, we can just write it with undetermined coefficients, and the question is reduced to solving a finite system of algebraic equations. There is an algorithm of doing this, if the original equation is over rational (or algebraic) numbers. So the real question is: can we bound the degree of the factor?

Then the question is "bound in terms of what"? A relatively recent breakthrough in this problem is the counterexample: one cannot bound the degree of the factor in terms of degrees of $f,g$, and the types of singularities: there is a one-parametric family of equations, with singularities of fixed type (that is the type is independent of parameter) such that for a countable dense set of values of parameter it has integrating factors, but their degrees tend to infinity.

Lins Neto, Alcides, Some examples for the Poincaré and Painlevé problems. Ann. Sci. École Norm. Sup. (4) 35 (2002), no. 2, 231–266.

This paper also contains a survey of previous results.

However it remains a possibility that the degree of the integrating factor can be estimated in terms of some arithmetic properties of the coefficients.

Here is a survey of the topic: MR2166493 Llibre, Jaume Integrability of polynomial differential systems. Handbook of differential equations, 437–532, Elsevier/North-Holland, Amsterdam, 2004.

All this was about the global aspect of the problem (as I said it has many formulations). The local aspect is usually called the Center-Focus Problem: for the same kind of system, to determine whether an equilibrium point is a center or a focus. There is an enormous literature on this, and this problem is much better understood. Just look at the "center-focus problem" as a keyword.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.