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A possibly soft question for you guys and gals. Say a system of analytic PDEs has been completed to involution (in the sense that it's geometric symbol has a Pommaret basis, or has vanishing Spencer/Koszul (co)homology). Is its linearization locally solvable?

Just to include why I'm asking: Lie-Pseudo-groups are determined by an involutive system and their Lie-algebroids by its linearization. If the answer to the question is positive, then since the linearized equation has the same symbol module (at the identity) it also is locally solvable. I have not been able to find this (possible) fact in the literature.

edit: I changed the title of the question as it was pretty bad.

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  • $\begingroup$ I see that you're a graduate student at Minnesota. Surely, Peter Olver would know the answer to this? $\endgroup$
    – Deane Yang
    Commented Mar 23, 2015 at 17:59

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If I understand correctly, you're asking if a nonlinear system of PDE's is involutive, then is the corresponding linearized system also involutive? The answer is yes, but I'm not sure that it's obvious. It is true, as you say, that the symbol, which is the same for both systems, is involutive. However, involutivity also requires conditions on the lower order terms, and these need to be checked. I believe (but am not sure) that this is demonstrated in my monograph "Involutive Hyperbolic Differential Systems" for quasilinear systems.

ADDED: Here's an argument that the homogeneous linearized equation is involutive: Let $$ \Phi(x, u, \partial u) = 0 $$ be a first order involutive system of PDE's. (If the PDE's are higher order, you can always rewrite the system as a first order one)

Given an involutive symbol, you can count the dimension of the space of formal second order solutions there are. In terms of Cartan characters $s_1, \dots, s_n$ (where $n$ is the number of independent variables), it's given by $s_1 + 2s_2 + \cdots ns_n$.

Since the homogeneous linearized equation has the same symbol, it is involutive, if it has the same dimension of formal second order solutions (which is now a linear subspace of the space of all second order Taylor series for the unknown function).

Given any formal $1$-parameter family $u_t$ of 2nd order solutions to the nonlinear system, you have $$ \Phi(x, u_t, \partial u_t) = 0. $$ Differentiating this with respect to $t$, you get $$ \Phi'(x, u_0, \partial u_0)\dot{u} = 0. $$ This implies that the dimension of the space of 2nd order solutions $\dot{u}$ to the linearized system is equal to the dimension of the space of 2nd order solutions to the nonlinear system. By Cartan's test for involutivity (I'm sure there's an equivalent condition in the other definitions of involutivity), the homogeneous linearized system is involutive.

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    $\begingroup$ Unfortunately, by now I find my own monograph unreadable. $\endgroup$
    – Deane Yang
    Commented Mar 23, 2015 at 20:22
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    $\begingroup$ I agree that this is pretty easy in the homogeneous case, and when I work through the proof of Cartan-Kahler I can't see where the local solvability for the linearized system fails, but I wanted to double check on MO. $\endgroup$
    – Orn Arnaldsson
    Commented Mar 23, 2015 at 20:26
  • $\begingroup$ The inhomogeneous case requires only that the inhomogeneous term satisfy compatibility conditions, which are equivalent to saying that the system has at least one formal 2nd order solution. $\endgroup$
    – Deane Yang
    Commented Mar 23, 2015 at 23:14
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    $\begingroup$ The step in the Cartan-Kahler theorem that I didn't find obvious is the dimension of the space of formal 2nd order solutions to the linearized system. The last paragraph of my answer shows how to get this from the involutivity of the nonlinear system. $\endgroup$
    – Deane Yang
    Commented Mar 24, 2015 at 1:21
  • $\begingroup$ Many thanks for taking the time to work through this. Maybe I can now convince Peter and his collaborator on the pseudogroup paper to drop the extra condition. $\endgroup$
    – Orn Arnaldsson
    Commented Mar 24, 2015 at 20:42
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Take a look at the book Exterior Differential Systems page 71 theorem 2.2 for a precise statement of the Cartan--Kaehler theorem, which says that involutive real analytic torsion-free exterior differential systems have local real analytic integral manifolds. I don't know what a Pommaret basis is, so I can't be sure that involutivity means the same in your sense and in Cartan's sense.

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  • $\begingroup$ The Pommaret theory of involution and the Spencer cohomology one in EDS are equivalent, see Werner Seiler's book, "Involution". There is a new formulation of pseudo-groups by Olver and Pohjanpelto (math.umn.edu/~olver/mf_/psa.pdf) where they basically say that it is not clear wheather the linearization of an involutive system is still involutive and make this an extra condition on the pseudo-group (calling it "tameness" of the pseudo-group). Do you agree that since the involutive system and its linearization have the same symbol module, then both are involutive? $\endgroup$
    – Orn Arnaldsson
    Commented Mar 23, 2015 at 17:23
  • $\begingroup$ In my lecture notes, I describe the linearization of exterior differential systems, and prove that the tableau is invariant under linearization, so involutive systems have involutive linearization: arxiv.org/pdf/1706.09697.pdf, p. 48. $\endgroup$
    – Ben McKay
    Commented Apr 29, 2020 at 18:25
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With the crucial hypothesis of analyticity, the answer is Yes. I believe that, when the PDEs are presented as an exterior differential system, the result is known as the Cartan-Kähler theorem.

Without recasting the PDEs in terms of exterior differential systems, that is, in the language that you are using, you can find the linear version of this result in Section 2.1 of the review by Spencer (1969) on overdetermined PDE systems. The non-linear version can be found in this paper by Goldschmidt (1967).

Without analyticity, there are further obstructions than lack of involutivity to having smooth solutions, as demonstrated by the famous example by Lewy.

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  • $\begingroup$ Thanks much for the answer, see my comment to the other answer for a follow up question. $\endgroup$
    – Orn Arnaldsson
    Commented Mar 23, 2015 at 17:24
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    $\begingroup$ That's an interesting question. I never really thought hard about it myself. I would defer to Deane's opinion. $\endgroup$
    – Igor Khavkine
    Commented Mar 23, 2015 at 19:07

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