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Take a differential operator with elliptic symbol, consider just the principal part of the operator. Can one invert this principal part with some parametrix type construction (at least construct a left inverse)?

Background: the concern is on linear PDE systems $$\mathcal L(x, \frac{\partial}{\partial x})u=\mathcal S(x), \qquad \mathcal B(x,\frac{\partial}{\partial x})u|_{\partial\Omega}=\varphi(x)$$ whose symbol coincides with their principal symbol, $$ \mathcal L_0(x,i\xi)=\mathcal L(x,i\xi),\qquad \mathcal B_0(x,i\xi)=\mathcal B(x,i\xi),$$ so there are no terms of lower order. The operator is considered on a bounded domain $\Omega\in\mathbb{R}^n$. There can be more equations in the systems than unknown functions $u=(u_1,\ldots, u_J)$. The symbol is assumed to be elliptic/full rank in $\Omega$ for $|\xi|\neq 0$.

The concern is not about existence, but only about regularity estimates given existence of the solution. To this purpose, some general theory asserts the existence of a left regularizer to $\mathcal A$, where $\mathcal Au=(\mathcal L\times\mathcal B)u=\left(\begin{array}\mathcal S\\\varphi\end{array}\right)$, that is $$ \mathcal R\mathcal A=\mathcal{Id}-\mathcal T,$$ where $\mathcal T$ is compact.

The question is whether one can achieve that $\mathcal T=0$ because there are no lower order terms, as $\mathcal L_0(x,i\xi)=\mathcal L(x,i\xi)$. Is invertibility of the operator easy to analyze when the elliptic symbol has no lower order terms?

The aim is to get a regularity estimate for the solution $u$ just with $\mathcal S$ and $\varphi$ on the right hand side, without Lebesgue norm of $u$.

I recall that in the theory of $\Psi DO$ in $\mathbb{R}^n$, I saw a construction of a parametrix for an elliptic principal symbol which acted as inverse but could not find the source again.

A subquestion: is there literature on (redundant) PDE systems apart from that Russian article from the 70ies (link), which has its own peculariar notation and proofs which are at sometimes difficult to follow, relying on Russian articles not available in English. If the task is promising, I want to study the mentioned problem but lack other literature which construct regularizers with more perspicuous proofs.

Any comments are appreciated. Thanks.

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Isn't the Laplacian with Neumann conditions a counterexample? There are no lower order terms, yet there is no inverse.

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  • $\begingroup$ So this cannot be true, thank you for the reality check. The potential has to be gauged, yes. But this seems an easy case of non-uniqueness. Simpler to analyze than, say, the Helmholtz eqn. and eigenvalues of the Laplacian (which has a lower order term). When I was asking the question, I wanted to understand a particular remark in the literature that assured: analysis of invertibility is closely tied up with lower order terms. Do you know of a technique analyzing unicity of (linear) PDE by considering the terms in $\mathcal T$, where having just highest order is an advantage? $\endgroup$ – Tom Jul 27 '13 at 14:27
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The remark you mentioned has to do with the fact that (thinking of strongly elliptic operators on bounded domains) lower order perturbations can change the dimension of the kernel or the co-range, even though the Fredholm index remains constant. I would say that having no lower terms is not particularly important for invertibility, what is important is that there is always a choice of lower order (in fact 0-th order) term that makes the operator invertible. This fact also helps when you derive estimates because you can then neglect the lower order terms and just work with the principal part. A point of view that is convenient here is that one is studying a family of operators with the same principal part, and that the particular problem with a particular lower order term that you are trying to solve is in some sense one you picked by "accident" from the large family.

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