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In the study of partial differential equations, it is often considered enough to analyze the principal symbols and their characteristic variety (see for example, http://www.sciencedirect.com/science/article/pii/S0001870802000993, and https://en.wikipedia.org/wiki/Symbol_of_a_differential_operator). In the first link, only the principal symbol is used to count the number of solutions to a holonomic system. But there are easy examples for which this information does not appear to be enough. Consider the two-dimensional system: $$ (\partial+1)\phi(z,z')=0 $$ $$ (\partial+\partial')\phi(z,z')=0 $$ and, $$ (\partial+z)\psi(z,z')=0 $$ $$ (\partial+\partial')\psi(z,z')=0. $$ Both have the same principal symbols, namely, $p$ and $p+p'$, however, the first has the solution $\phi(z,z')=e^{z'-z}$, while the second has no solution (other than $\psi(z,z')=0$).

It seems to be suggested in the first link that the second has a solution on an integral curve, perhaps, but it is not clear to me if that is the right context and if so, how does one construct such a curve?

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3 Answers 3

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(This really should be a comment, but it was too long.)

Part of the problem is that when you're dealing with a system of (linear) PDE's, the principal symbol is not the "right" object. Instead, you want to look at the characteristic ideal. It's defined as follows:

Let $D$ be the ring of linear partial differential operators in $z_1,\ldots,z_n$. A system of PDE's may be viewed as an $r\times s$ matrix $\mathsf P$ with entries in $D$. Let $N$ be the left column space of $\mathsf P$, i.e. the left $D$-module generated by the columns of $\mathsf{P}$. Let $\operatorname{gr}(D)=\{\sigma_P\mid P\in D\}$ be the (commutative) ring of principal symbols of differential operators, and let $\operatorname{gr}(N)=\{\sigma_V\mid V\in N\}$ be the $\operatorname{gr}(D)$-module of principal symbols of $N$. The characteristic ideal of the system of PDE's is (the radical of) the ideal $$ \operatorname{Ann}_{\operatorname{gr}(D)}\bigl(\operatorname{gr}(D)^r/\operatorname{gr}(N)\bigr).$$ The reason we need to take the radical is because the un-radicalized ideal depends on choice of coordinates while its radical does not.

There's a coordinate-free definition, but it's a bit more involved to state.

Edit: In your case: System 1: Have $r=1$ and $N=D\{\partial+1, \partial +\partial'\} $. One can show using Gröbner basis theory that $\operatorname{gr}(N)=(\partial, \partial') $.

System 2: Have $r=1$ and $N=D\{\partial+z, \partial +\partial'\} $. You might guess that $\operatorname{gr}(N)$ would be the same, but you'd be wrong! In fact, notice that the commutator $[ \partial +\partial', \partial+z] $ is $1$, so $N=D $, and in particular $\operatorname{gr}(N)= \operatorname{gr}(D) $.

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  • $\begingroup$ Would you mind elaborating on this? How would it look in the current problem? I am a physicist, so I'm afraid I may need it spelled out a bit more. Specifically, I don't understand what the Ann function represents. $\endgroup$
    – alphanzo
    Commented Oct 17, 2015 at 6:52
  • $\begingroup$ @alphanzo Ann is called the annihilator. It consists of all the elements of the ring which kill the entire module. The radical of an ideal $I$ is the set of ring elements $f$ such that $f^k\in I$ for some $k\geq1$. $\endgroup$
    – Avi Steiner
    Commented Oct 17, 2015 at 15:26
  • $\begingroup$ Thank you for this insight! This is very interesting and I'd like to read more if you can provide a reference. In this case, is it true that $r\times s=1\times 2$, so that the Characteristic ideal is simply $(p,\ p+p')=(0,0)$, which is itself isomorphic the Characteristic variety, $Ch(M)\subset T^*X$ for the module $M=D_X/I$, for $I=D_X(\partial+\beta(x))+D_X(\partial+\partial')$...? $\endgroup$
    – alphanzo
    Commented Oct 18, 2015 at 18:36
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    $\begingroup$ A good reference for the case of diffops on $\Bbb C^n$ with polynomial coefficients is Gröbner Deformations of Hypergeometric Differential Equations by Sato, Sturmfels, and Takayama. As for more general situations, any basic text on D-modules will cover this, probably under a section on characteristic varieties (the variety in the cotangent space cut out by the characteristic ideal) of coherent D-modules. For these, though, you'll want to be comfortable with sheaf language. $\endgroup$
    – Avi Steiner
    Commented Oct 18, 2015 at 18:45
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    $\begingroup$ @alphanzo You're correct that a holonomic system (or more precisely it's characteristic variety) must be involutive. That's actually true of all coherent D-modules. A nonzero holonomic guy is in addition Lagrangian. We need to allow $0$ to also be holonomic (even though its characteristic variety is empty) so that the category of holonomic D-modules is abelian. That's why the second system is still considered holonomic. Incidentally, your second system actually no non-trivial solutions. $\endgroup$
    – Avi Steiner
    Commented Oct 25, 2015 at 16:06
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The Wikipedia page that you cite contains the mistake that it does not distinguish between determined and overdetermined systems (as yours are), or, rather, more importantly, it does not distinguish between involutive systems (as your first one is) and noninvolutive ones (as your second one is). In overdetermined or symbol-degenerate cases (i.e., systems in which every co-direction is characteristic), the principal symbol alone does not contain enough information to determine whether the system is involutive or not, and this has an even greater effect on the nature of the solutions than does the principal symbol. The author of that page probably only had the determined, nondegenerate case in mind while writing it.

As usual with information found on the Internet, caveat lector.

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  • $\begingroup$ Thank you for the comment. That is indeed the correct way to go about it in the Frobenius picture when the equations are linear. However, in the second link the general case is considered micro locally. It seems that the second system does have a solution if only in the micro local sense of the term. I'm wondering if it implies there exists a hypersurface M⊂C2 on which a solution exists for restricted values (z,z′)∈M? $\endgroup$
    – alphanzo
    Commented Oct 14, 2015 at 3:27
  • $\begingroup$ First, involutivity has nothing to do with linearity; they are independent notions. Second, I don't know what it means to have a solution 'in the micro-local sense', so I can't help you there. What's certainly true is that there are solution curves in both cases (i.e., 1-dimensional integral manifolds in the space of $1$-jets), but the only $2$-dimensional integral manifold in the second case is the one given by the (unique) solution $\psi = 0$. $\endgroup$
    – Robert Bryant
    Commented Oct 14, 2015 at 4:40
  • $\begingroup$ @RobertBryant can you please elaborate on what involutivity means in this sense? $\endgroup$
    – Bran
    Commented May 14, 2022 at 15:52
  • $\begingroup$ @Bran: For the case in hand, a solution of the first system is a $2$-dimensional integral manifold of the $1$-form $\theta_1 = \mathrm{d}\phi+\phi\,\mathrm{d}z-\phi\,\mathrm{d}z'$ while a solution of the second system is a $2$-dimensional integral manifold of the $1$-form $\theta_2 = \mathrm{d}\phi+z\phi\,\mathrm{d}z-z\phi\,\mathrm{d}z'$. Both $1$-forms have integral curves, but $\theta_1=0$ is `involutive' while $\theta_2=0$ is not 'involutive' in the sense of $2$-plane fields. Does that help? $\endgroup$
    – Robert Bryant
    Commented May 15, 2022 at 14:02
  • $\begingroup$ @RobertBryant yes, clear. Thank you! $\endgroup$
    – Bran
    Commented May 15, 2022 at 21:29
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Let me elaborate a little about what Robert wrote. When analyzing a system of PDE's, there are two separate issues that need to be considered. Below, I'll assume that you want to study the space of smooth solutions to a homogeneous linear system of PDE's.

Before you even look at the characteristic variety (i.e., study the system microlocally), you have to look at the formal properties of the system. In other words, are there any formal power series solutions (without considering convergence) to the system? If there aren't any, then there's no point in going any further. If there are, the next important question is how "many" formal solutions are there?

Figuring this out is not always easy, especially with an overdetermined system like yours. Note that each of your systems has 2 unknown real-valued functions but 4 real-valued equations. As Robert says, the two systems have completely different formal properties. To analyze them properly, you often have to do what's known as prolongation which involves adding more unknown functions and equations obtained by differentiating the original equations. To determine whether you have formal solutions and "how many", you have to keep doing this until the system satisfies a condition known as involutivity.

Only after you have an involutive system does the characteristic variety (or ideal or sheaf) of the involutive system (and not the original system) tell you anything meaningful about the system of PDE's.

In your examples, the symbol and therefore the characteristic (variety, ideal, sheaf) are the same for both systems. However, since the second one is not involutive, the characteristic object tells you nothing about the second system. However, if you prolong the second system until it becomes involutive, then the characteristic object of the prolonged system will tell you something about the second system of PDE's and its solutions. In particular, they will be different from those of the first system.

The theory for this is quite involved. I'll let others provide more specific references for all of this.

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