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If I have a partial differential operator $p(D)$, where $p$ is a polynomial with constant coefficients and $D$ is the derivative in Euclidean space. Its inverse is easily described in Fourier space: $\mathcal{F}[p(D)^{-1}] = 1/p(k)$ with $k$ being the Fourier variable (of course up to specification of boundary conditions). Conversely, If I have a translation invariant integral operator with kernel $G(x-y)$, it is the inverse of a partial differential operator with constant coefficients whenever its Fourier transform is of the form $\mathcal{F}[G(x)] = 1/p(k)$ for some polynomial $p(k)$ with constant coefficients. I think this converse does a nice job of characterizing integral operators $G(x-y)$ that are inverse to a partial differential operator with constant coefficients.

Question: When translation invariance is dropped and one moves from Euclidean space to a general manifold, is it possible to characterize integral operators, say given by an integral kernel $G(x,y)$, which are inverse to some partial differential operator $p(x,D)$?

I have a feeling that the answer, if one exists, should have something to do with the wave-front set of $G(x,y)$. Unfortunately, my understanding of this subject is rather poor, so references to some general literature on this kind of problem are also appreciated.

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  • $\begingroup$ You have to be more precise when you speak about the "inverse" of a partial differential operator. Take for example the Laplacian on a Riemann manifold. It has a nontrivial kernel consisting of constant functions. When the manifold is the Euclidean spane $\mathbb{R}^n$, do you consider this operator invertible? $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 3, 2012 at 8:41
  • $\begingroup$ Certainly. However, I left the issue of boundary conditions and other restrictions on the domain of the operators (which are needed to define an inverse in the presence of such "zero modes") undiscussed because I think it might actually distract from the main thrust of the question. However, if fixing such conditions is crucial, I'm happy to restrict to the case of hyperbolic operators with retarded or advanced boundary conditions, or equally well to elliptic operators without zero modes. $\endgroup$
    – Igor Khavkine
    Commented Apr 3, 2012 at 9:53
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    $\begingroup$ It's reasonable to talk about a right inverse of a differential operator without specifying any boundary conditions. As for characterizing such an operator, I don't see any easy way to do so. First, the inverse operator is not even necessarily a pseudodifferential operator. If it is, then a necessary condition is that the principal symbol is the reciprocal of a polynomial (i.e., the symbol of a differential operator). But things get more complicated with the right inverse of a hyperbolic PDE, which is a Fourier integral and not a pseudodifferential operator. $\endgroup$
    – Deane Yang
    Commented Apr 10, 2012 at 12:36
  • $\begingroup$ Doing some reading, I've found that there are characterizations of some classes of PDOs due to Beals and also Bony. I don't know if there is a similar characterization of FIOs; it would be interesting to learn just that. Now, supposing I knew that $G(x,y)$ is a PDO, how do I recover the principal symbol? Is it enough to write $G(x,y)=H(x,x-y)$ and the the Fourier transform of $H(x,y)$ in the second argument? How would I get the symbol if I knew that $G(x,y)$ is a FIO? After the principal symbol is known, what about lower order terms? $\endgroup$
    – Igor Khavkine
    Commented Apr 10, 2012 at 13:08
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    $\begingroup$ The simplest example of a hyperbolic operator is $\partial_n$ and when you solve $\partial_n u = f$, if $f$ has a singularity at a point, then $u$ might have a singularity not just at that point but along the integral curve of $\partial_n$ through that point. Any hyperbolic or even PDO of real principal type has the same property. This is why the kernel of the inverse will have a singular set off the diagonal. $\endgroup$
    – Deane Yang
    Commented Apr 10, 2012 at 17:20

3 Answers 3

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In some sense, much or all of microlocal analysis is precisely about these sorts of questions. As the previous answers indicate, different types of differential operators require rather different types of pseudodifferential or Fourier integral operators for their parametrices. Pseudodifferential operators are enough to study elliptic operators (or even more general operators in regions of phase space where they are micro-elliptic). There is a nice geometric definition of the structure of the Schwartz kernels of pseudodifferential operators, namely they are (classical) conormal distributions on the product space with respect to the diagonal. For wave operators one requires the more complicated class of FIO's. Their Schwartz kernels are typically given as Lagrangian, or more generally, marked Lagrangian distributions. There is a nice coordinate invariant description of these too. The question you ask (in one of your followups), namely to find simple conditions on G(x,y) that it be the retarded Green function of a hyperbolic differential operator, is pretty hopeless as stated. Certainly G would need to be a marked Lagrangian distribution, but to determine precisely which such distributions are actually the fundamental solutions of differential operators is not likely to have an easy answer.

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The problem is in fact a division problem: let $p(\xi)$ be a polynomial in $n$ variables. There exists a tempered distribution $T$ such that $$ p(\xi) T(\xi)=T(\xi) p(\xi)=1. $$ This is a result due to Hörmander and Lojasiewicz. The convolution by $\hat T(-x)$ provides then a fundamental solution for the operator $p(D)$.

Bazin.

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  • $\begingroup$ Yes, this is a standard construction and I described it, though perhaps less precisely, in the first paragraph of my question. I'm interested in its analog when the differential operator has non-constant coefficients, and so cannot be directly diagonalized using the Fourier transform. Deane and Piero made some good comments above, but I've yet to make a substantial dent in the relevant literature. $\endgroup$
    – Igor Khavkine
    Commented Apr 22, 2012 at 15:50
  • $\begingroup$ Yes, the approach that works for constant coefficient operators work in the variable case only for elliptic (and maybe more generally for hypoelliptic operators). Maybe you should focus on those, first? $\endgroup$
    – Deane Yang
    Commented Apr 22, 2012 at 17:28
  • $\begingroup$ Deane, my original motivation was to see if I could find a simple set of conditions on a distribution $G(x,y)$ to be the retarded Green function of a hyperbolic differential operator. I can think of a few necessary conditions, but no sufficient ones. I'll keep in mind that the elliptic case might be easier, but I've not yet had the time to delve into the suggested literature to make progress on either version. $\endgroup$
    – Igor Khavkine
    Commented Apr 22, 2012 at 18:55
  • $\begingroup$ For an elliptic operator $P$ of order $m$, you can construct an approximate inverse in the following sense. There exists $Q$ a pseudo differential operator of order $-m$ such that $$ PQ=Id+R,\quad QP=Id +S $$ where $R,S$ are regularizing operators (that is with smooth kernels). That construction can be microlocalized for a (pseudo)differential operator which is elliptic at a point $(x_0,\xi_0)$ and is a way to prove that $$ WF u\subset\WF(Pu)\cup\char P. $$ $\endgroup$
    – Bazin
    Commented Apr 23, 2012 at 20:27
  • $\begingroup$ @Bazin Could you please help me with this! $\endgroup$
    – Mr. Proof
    Commented Mar 10, 2022 at 5:58
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It is straightforward to show that if a linear partial differential operator $P$ has a pseudodifferential right inverse (or even a pseudodifferential right parametrix) then it must be micro-hypoelliptic, i.e. $WF(Pu)=WF(u)$ for all $u\in\mathscr{D}'$, due to the fact that the wave front set of the Schwartz kernel of any pseudodifferential operator must be conormal to the diagonal. If $P$ has constant coefficients this amounts to saying that $P$ is hypoelliptic, i.e. $Pu\in C^\infty$ implies $u\in C^\infty$. For instance, the Laplace and heat operators are hypoelliptic, but the wave operator is not.

More generally, the existence of a right inverse to a linear partial differential operator $P$ entails that $P$ is locally solvable: for every compact subset $K$ and every $f$ in finite-codimensional subspace of $C^\infty$, there is $u\in\mathscr{D}'$ which satisfies $Pu=f$ in an open neighborhood of $K$. In that case, Hörmander has shown that the principal symbol $p$ of $P$ must satisfy the so-called condition $(\Psi)$: any compact subset $K$ has an open neighborhood $Y$ where there is no positively homogeneous, complex-valued $q\in C^\infty(T^*Y\smallsetminus 0)$ such that $\mathrm{Im}(pq)$ changes sign along any bicharacteristic strip $\gamma$ of $\mathrm{Re}(pq)$ over $Y$ such that $q\neq 0$ on $\gamma$.

In general condition $(\Psi)$ does not entail that the right inverse will be a FIO (i.e. a linear operator whose Schwartz kernel is a Lagrangian distribution associated with an homogeneous canonical relation): if $p$ is real valued, this is indeed true as Deane Yang mentioned in his last comment to the OP's question (at least away from the diagonal, as Rafe Mazzeo remarked in his answer), but otherwise the bicharacteristic relation of $p$ may not be a manifold.

Finally, it should be remarked that all of the above applies to scalar $P$. For systems, the situation gets even more complicated, particularly when $P$ has characteristics of varying multiplicity.

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