Dear Gabriel,

I think you are confused here by the use of the word "component" in differential algebra, in particular in the context of singular integrals. People do not mean prime components here.

In the most classical example for the appearance of a singular integral, the Clairaut equation with F=y-xy'+(y')2/4, you are right: if you look at the prolongation DxF=-(x+y'/2)y'', then it factors. The first factor yields the singular integral and the second factor the general integral. But I think that this is a pecularity of this example. Consider another example due to Ritt: F=(y')2-4y3 with the singular integral y=0. Looking at the prolongations, I do not think that anything factors and thus I conjecture that {F} is actually prime (although I do not really know how to prove it rigorously).

Decompositions that identify singular integrals are of a different nature than prime decomposition. Two of the best known ones are the one by Boulier et al and the Thomas decomposition going back to the thirties. They decompose into systems comprising equations and inequations. Furthermore, not the differential ideal is decomposed, but the solution set. In the case of the Thomas decomposition, you even obtain a disjoint decomposition of the solutions.

But even if singular integrals are related to prime components in some cases, it would be a bit strong to consider them as artifical. The real problem of singular integrals is that generic differential equations do not have any: if you perturb an equation with a singular integral only a tiny bit, the singular integral will most probably disappear. Thus one may argue that they are not relevant for equations coming from applications where at least some coefficients stem from measurements. Only when you consider a whole family of differential equations, you may not be able to avoid that certain members of the family possess singular integrals.

Regards Werner

Hello Werner, this is the point :

"In the most classical example for the appearance of a singular integral, the Clairaut equation with F=y-xy'+(y')2/4, you are right: if you look at the prolongation DxF=-(x+y'/2)y'', then it factors. The first factor yields the singular integral and the second factor the general integral. But I think that this is a pecularity of this example. Consider another example due to Ritt: F=(y')2-4y3 with the singular integral y=0. Looking at the prolongations, I do not think that anything factors and thus I conjecture that {F} is actually prime (although I do not really know how to prove it rigorously). "

Any such polynomial example I found in literature factors ( here F' = 2y'y"-12y'y^2 is reducible). I tried all the ones proposed in the article from C.N. Srinivasiengar 1935 and after some effort you find a facrotizable derivate.

I am convinced that what we call singular integrals "have to be" joined to the differential ideal, at a certain depth of derivation !

Thanks for your clear answer, this is still a debate for me. Ritt is not very clear on the subject.

Gabriel

Hi Gabriel,

you are of course right (it seems that I have to practise differentiating a bit more)! What about a slight modification of Ritt's example: (u'')2=u3. Now, it is not so obvious that you get a factorisation at some higher order. But independent of whether or not this is example like you look for, it is interesting that apparently in (almost?) all cases the singular integral corresponds to a prime component.

I am not aware of any algorithm for determining a prime decomposition of a radical differential ideal. The traditional algorithms of commutative algebra cannot be applied here, as algebraically we are always dealing with infinitely generated ideals. But coming back to your original question, I do not see an obvious reason why {F, sep(F)} should be a prime component of {F}; its is simply a super ideal of {F}. If you think of classical algebraic geometry and take an ideal I, then for any point on the variety of I the corresponding maximal ideal is a super ideal of I, but very rarely it will be a prime component.

By the way, what is this article by Srinivasiengar that you mention? I have never heard of it, but it looks interesting. Does it discusses what possibilities there are for singular integrals? In Ritt, two cases appear: the classical envelope as in the Clairaut example and "adherence" as in Ritt's example where the singular integral is a kind of asymptotic for the members of the general integral. I have always wondered whether there are further possibilities and whether it is possible to predict in which case one is without actually computing the singular/general integral?

Regards Werner

Hello Werner,

Here's the place to get Srivaniengar's paper (134).

https://www.ias.ac.in/article/fulltext/seca/001/10/0668-0693

Regards,

Gabriel

Hi Gabriel,

thanks for the file! It looks like I will need some time to digest it. I have several upcoming conferences which must be prepared. Thus I will be able to come back to your question only some time in October. Have you taken a look at my modified Ritt example and have you found there a factorisation? In general, I would be interested in seeing a concrete example where the factorisation appears only after several differentiations. In our work on singular initial value problems we are also studying prolongations to make statements about the regularity of solutions and I wonder whether there is a relation between these questions. Thus I would like to study such an example from our geometric point of view.

Regards Werner

Hello Werner,

I think there is such a case in the paper of Srinivangar, I have to check in my papers.

Regards,

Gabriel