# Questions tagged [differential-operators]

In mathematics, a differential operator is an operator defined as a function of the differentiation operator.

399 questions
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### Applications of Pseudodifferential Operators

I am very interested in just about anything that has to do with PDE's, and inevitably pseudodifferential operators comes up. Its obvious that such a novel way of looking at PDE's would be important, ...
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### “Universal” differential identities

I now asked this at MO. Let $f:\mathbb{R}^d \to \mathbb{R}$ be smooth. The mixed derivatives commute: $f_{xy}=f_{yx}$. This identity is "universal" in the sense that it holds for any smooth map. ...
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### What are the eigenfunctions of the D'Alembert operator on pseudo-Riemannian manifolds?

Consider the operator $\Box=g^{\mu\nu}\nabla_\mu\nabla_\nu$ acting on a function space $\mathbf{F}(M)$, given by the set of functions $\phi:M\to\mathbb{R}$ whose values go to zero at infinity (at the ...
1answer
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### How to prove $(0,1)$ form is not $\overline\partial$-exact

On a complex manifold, if we are dealing with the $d$ operator, there's a pretty easy way of showing some form is not $d$-exact, simply by integrating in a closed loop. If you can find a loop that is ...
1answer
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### How to show that differential operator can be defined in terms of certain commutator operators

Let $U$ be any open subset of $\mathbb{R}^n$ (or, more general, of some smooth manifold). Define $\mathcal{D}_{-1}(U):=\{0\}$. For any two linear operators $A$ and $B$, the commutator operator $[A,B]$ ...
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### Elliptic and Fredholm partial differential operators

As I learn from the comments to this question a non elliptic operator on a compact manifold can not be a fredholm operator. However, I learn also from the conversations in the same post that the ...
1answer
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### Linear transformations in infinite dimensional vector spaces

If we look at an $n$ - dimensional vector space $V$ and a linear transformation $$T : V \to V, \quad x \mapsto Tx \quad \forall \, x \in V$$ then given a choice of basis ...
1answer
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