# Talk:Differential operator

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Field: Analysis

## Missing ?

Why things like ${\displaystyle \partial _{x}}$, ${\displaystyle \partial _{xy}}$ and perhaps also things like ${\displaystyle \partial _{x}\partial _{x}=\partial _{x}^{2}}$, ${\displaystyle \partial _{x}\partial _{y}=\partial _{xy}}$ or ${\displaystyle \nabla \nabla =\nabla ^{2}=\Delta }$ are not mentioned. I'm physicist and the german Wikipedia is waiting for me ;-). Wolfgangbeyer 08:23, 21 Feb 2004 (UTC)

I would believe that ${\displaystyle \partial _{x}}$ is written D_x with the "big-D" notation as well. They're just symbolisms. But, even though you probably didn't mean the discrete difference operator thing too, with the nablas, that reminded me to say something about it too :) Dysprosia 08:31, 21 Feb 2004 (UTC)

There's material about f(D) (operator notation), and some much more advanced theories. They could all be here; but there are some distinctions worth maintaining. I don't think the Laplacian should ideally come ahead of one-variable theory.

Charles Matthews 09:05, 23 Feb 2004 (UTC)

I agree. I'd like to shove it down below, but don't know what to say to introduce it. Dysprosia 10:17, 23 Feb 2004 (UTC)

There's a technical wrinkle about formally self-adjoint DOs not being quite the same thing as Hilbert space self-adjoint operators (because unbounded, basically). Charles Matthews 18:03, 9 Nov 2004 (UTC)

I've unlinked to Self-adjoint operator and made mention of them being formal (oops). Should that be enough? Dysprosia 22:22, 9 Nov 2004 (UTC)

It might be useful to point out explicitly that the meaning here is

∫fLg = ∫gLf

as can be proved by integration by parts; since this draws attention to what is the essential problem, namely that the bit in the integration, evaluated at the end points, that you want to go to zero at infinity, does so only under some hypotheses. That might be enough for this page.

Charles Matthews 22:50, 9 Nov 2004 (UTC)

Ok. I've tried to make mention of this, please check that what I've said is what you mean. Dysprosia 23:56, 9 Nov 2004 (UTC)

## Vote for new external link

Here's my website of example problems with differential operators. Someone please put it in the external links section if you think it's helpful!

http://www.exampleproblems.com/wiki/index.php/Differential_Operators

## Asterisks *

I was briefly confused by the use of the asterisk in the definition of the adjoint operator and its use to define an inner product in a function space.

Could the overbar notation be used instead? The article links to the page on inner product spaces, which uses this notation. (Or perhaps some words to indicate what the notation means.)

I was looking for some connection between the two asterisks.

Fair enough. Now changed use an overbar. -- Jitse Niesen (talk) 13:51, 24 May 2006 (UTC)

## Wording : linear, mapping, geometry

I don't agree with the first sentence of the article "differential operator is a linear operator". There are nonlinear operatos as well. I'm also not very happy with the formulation "function of the differentiation operator". I would suggest somethink like "Differential operator is a mapping that depends only on the function and its derivatives." Furthwer, I would add to the article, at least as a remark, that in differential geometry, differential operators act not always on function, but sometimes on vector fields or sections of some vector bundle, see invariant differential operators. Maybe someone could merge this two articles into one. Franp9am 07:32, 11 September 2006 (UTC)

The discussion of the adjoint of an operator comes completely out of the blue. Why is it included? —Ben FrantzDale 01:13, 9 April 2007 (UTC)

I have the exact same question. Why can't we just link to the page Adjoint_of_an_operator? Sunbeam44 (talk) 20:08, 9 November 2008 (UTC)

## squaring = composition?

I'm not particularly familiar with this operator, and am trying to understand the following expression in a textbook: ${\displaystyle (xD)^{2}f(x)}$.

I'm guessing that I should interpret the first term not as ${\displaystyle x^{2}D^{2}}$ but as ${\displaystyle (xD)(xD)}$, so that the whole becomes: ${\displaystyle xD(xD(f(x))=xD(xf'(x))}$.

Is that correct? Should I have been able to determine that from the article? The section on "Properties of differential operators" looks like it is starting to go in that direction, but I wonder if it needs expanding into a more general "Algebra of ...". Hv (talk) 13:33, 6 January 2008 (UTC)

In all cases I can think of, ${\displaystyle (xD)^{2}f(x)}$ means ${\displaystyle xD(xDf(x))}$. I.e., squaring does mean composition. Silly rabbit (talk) 14:28, 6 January 2008 (UTC)

Correct. Some details:

${\displaystyle (xD)^{2}f(x)=(xD)(xD)f(x)=(xD)(xf'(x))=x(xf''(x)+f'(x))=x^{2}f''(x)+xf'(x).\,}$

${\displaystyle x^{2}D^{2}f(x)=x^{2}f''(x).\,}$

Michael Hardy (talk) 00:47, 7 January 2008 (UTC)

## Linear vs nonlinear

This article should either be renamed "Linear differential operator", or it should be expanded to treat both linear and non-linear cases. A similar comment applies to other articles on Wikipedia, such as the "Schwarzian derivative", linked from the lead, which (out of deference to this one?) fail to call differential operators differential operators. Arcfrk (talk) 03:52, 15 February 2008 (UTC)

Yes! - Lead still says "This article considers mainly linear operators, which are the most common type. However, non-linear differential operators, such as the Schwarzian derivative also exist."
but the rest of the article does not explain what makes a DO linear or non-linear, or how their properties differ.
( Schwarzian derivative still doesn't say it is a non-linear DO.) - Rod57 (talk) 11:13, 30 October 2016 (UTC)

## Order of operator

The article doesn't define the order of a differential operator. Could a second-order operator involve first-order derivatives, or does it use only second-order derivatives? Probably the first, but it should be explicit. If it used only second-order derivatives, would it be "homogeneous"? Laura Puffysphere (talk) 16:36, 11 April 2009 (UTC)