Author 
Topic: Mean value property over squares? (Read 6511 times) 

Amadeus
Newbie
Posts: 2


Mean value property over squares?
« on: Apr 3^{rd}, 2006, 3:23am » 
Quote Modify

Hi, Harmonic functions have the mean value property, which involves integrating over a circle. Why does the curve need to be a circle, why not e.g. a square or a triangle? I found this teaser problem in a book, and haven't come up with a satisfying answer. Can you help me?


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: Mean value property over squares?
« Reply #1 on: Apr 3^{rd}, 2006, 2:19pm » 
Quote Modify

It is about characterization. The meanvalue property over circles _characterizes_ harmonic functions. If you use squares (or trigangles), say, instead of circles, you get a different class of functions. You can get the Poisson kernel P for the square by letting F be a conformal mapping from the interior of the circle to the interior of the square, and then take P composed with F.


IP Logged 
Regards, Michael Dagg



JP05
Guest


Re: Mean value property over squares?
« Reply #2 on: Apr 4^{th}, 2006, 12:09pm » 
Quote Modify
Remove

This is the first time I have ever heard such a thing. It is true that the region over which contour integration occurs can be any closed region. What is the name of this other function class?


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: Mean value property over squares?
« Reply #3 on: Apr 7^{th}, 2006, 6:45pm » 
Quote Modify

Your remark about contour integration is correct. The name of the class is harmonic polynomials. It can be shown that if the meanvalue theorem holds for all integrable harmonic functions on a planar domain, then it is both necessary and sufficient that the domain be a disk. Similarly, it can be shown that the meanvalue property over a regular ngon (n>=3) does not characterize all integrable harmonic functions but instead a class of harmonic polynomials determined by n. Actually, there is a whole theory about these ideas that is called quadrature domains. You'll should be able to Google up papers about quadrature domains and will even find references from potential theory. In fact, you'll find references from real analysis (and to my knowledge the most detailed) since these ideas clearly hold for harmonic functions in R^2. But, I don't know if anyone has wrote about classes based specifically on the square or triangle but it is not difficult to do. Your statement that this topic is a teaser problem in a book is interesting; either I missed it because I know the answer or that I don't know the book. In exchange for the title of the book, I would be willing to provide written generalizations for both the square and the triangle in trade.

« Last Edit: Jun 12^{th}, 2007, 1:26pm by Michael Dagg » 
IP Logged 
Regards, Michael Dagg



JP05
Guest


Re: Mean value property over squares?
« Reply #4 on: Apr 8^{th}, 2006, 11:44am » 
Quote Modify
Remove

My mistake for connecting this with integration. It has been pretty drilled into my head from what complex variables I know is the shape of the region of integration can be any closed region. What you say seems to suggest that when n > infinity which cause the npolygons to make a circle that the classes of these harmonic polys converge to the complete class of harmonic functions. I am not sure if I said that right but the idea make sense?


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: Mean value property over squares?
« Reply #5 on: Apr 10^{th}, 2006, 4:56pm » 
Quote Modify

Nice idea. I think you said it right but you mean the number sides of the polygons and not the number of polygons, as n goes to infinity. Explore your idea further and see what you come up with. It might surprise you.


IP Logged 
Regards, Michael Dagg



JP05
Guest


Re: Mean value property over squares?
« Reply #6 on: Apr 11^{th}, 2006, 11:57am » 
Quote Modify
Remove

References?


IP Logged 



JP05
Guest


Re: Mean value property over squares?
« Reply #7 on: Apr 11^{th}, 2006, 3:13pm » 
Quote Modify
Remove

Having thought more about this leads me to say that something mysterious happens at infinity (or this is all false) because since for each ngon the class of harmonic functions is polynomials and there is no k > n such that the class contains nonpolynomial harmonic functions or even contains one. Since not all harmonic functions are polynomials and that the meanvalue property taken over polygons results in classes that are polynomials (so you say) this implies that there is no finite k that even gives up a class of nonpolynomials harmonic functions. So, what is the magic at infinity? I look at it from the viewpoint for example that I can see a series get closer to a sum as I take n to infinity and like a limit getting closer to its value. But, along the way we can see the results. There are many other examples. What I don't see here is a nonpolynomial harmonic function along the way. I am clear?


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: Mean value property over squares?
« Reply #8 on: Apr 11^{th}, 2006, 4:47pm » 
Quote Modify

JP05, you seem to be thinking that polynomials are isolated functions, like the integers in the set of all reals. They are not. Rather, they are like the rational numbers in the set of Reals, everywhere dense. You can get uniformly close to anything analytic with a polynomial, at least within a limited area: If D is a compact domain in R^{N}, and f : D > R is analytic, and if h>0, then there is a polynomial P : D > R such that sup  f  P  < h. So it is not necessary for any ngon to have nonpolynomial harmonic functions in its class, because as n gets higher, the polynomials are able to more closely approximate the harmonic functions of the limiting circle.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



Amadeus
Newbie
Posts: 2


Re: Mean value property over squares?
« Reply #9 on: Apr 12^{th}, 2006, 3:18am » 
Quote Modify

Wow, I'm pleasantly surprised at the amount of replies this topic has inspired. Thank you! Michael_Dagg, the book is Greene&Krantz's complex analysis book. I probably used the word 'teaser problem' in a wrong context as English is not my first language. Sorry for the confusion. I'd still very much appreciate any further information on the topic.


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: Mean value property over squares?
« Reply #10 on: Apr 12^{th}, 2006, 12:07pm » 
Quote Modify

The topic being a teaser puzzled me since I have never seen it discussed this way. The discussion about this topic in that book is fairly standard though. About JP05's questions, here is a link to a paper that talks about quadrature domains: http://www.math.kth.se/math/forskningsrapporter/gustavsson.shapiro.pdf This paper contains some interesting ideas and as you will see it starts with the meanvalue property. Give me a few days and I will provide you with references for your question regarding the ngons.

« Last Edit: Apr 12^{th}, 2006, 5:07pm by Michael Dagg » 
IP Logged 
Regards, Michael Dagg



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: Mean value property over squares?
« Reply #11 on: Apr 13^{th}, 2006, 11:33am » 
Quote Modify

A pretty result regarding characterization (and also contains a result about harmonic conjugates) is given in Goldstein & Ow.: http://links.jstor.org/sici?sici=00029939%28197107%2929%3A2%3C341%3AOTM POH%3E2.0.CO%3B21&size=LARGE#abstract Another is: http://links.jstor.org/sici?sici=00029890%28199701%29104%3A1%3C52%3ATSA HF%3E2.0.CO%3B2A&size=LARGE The Beckenback and Reade paper is the most accessible one I know of (requiring only knowledge of calculus in R^2) is shown below. Theorem 5 is the result. Here I give the consequences of Theorm 5 for the square and equilateral triangle. The harmonic polynomials shown are constructed from the series formulas in the paper. See (1.2) for reference to the ngons. When I say phi = 0, I mean that the side of the polygon to the right of its center is vertical: SQUARE RESULT: Let f be continuous on the simply connected open set D. Then its value at each point in D equals its average value on the boundary of every square in D , with phi = 0, centered at that point if and only if f is a harmonic function of this form: f(x,y) = B_0 + A_1 x + B_1 y + A_2 (x^2  y^2) + B_2 xy + A_3 (x^3  3 x y^2) + B_3 (3 x^2 y  y^3) + B_4 (4 x^3 y  4 x y^3) . TRIANGLE RESULT: Let f be continuous on the simply connected open set D Then its value at each point in D equals its average value on the boundary of every equilateral triangle in D , with phi = 0, centered at that point if and only if f is a harmonic function of this form: f(x,y) = B_0 + A_1 x + B_1 y + A_2 (x^2  y^2) + B_2 xy + B_3 (3 x^2 y  y^3) . +++++++++++++++++ In each, if you omit the requirement phi = 0, thus requiring the averaging to work for all squares and equilateral triangles no matter how they're turned, then the conclusion still holds if you delete the last summand. Also, the reference to J.L.Walsh at the bottom of p.232 is also an interesting paper that exploits similar ideas.

« Last Edit: Apr 13^{th}, 2006, 1:34pm by Michael Dagg » 
IP Logged 
Regards, Michael Dagg



JP05
Guest


Re: Mean value property over squares?
« Reply #12 on: Apr 14^{th}, 2006, 7:48pm » 
Quote Modify
Remove

Your contribution is highly appreciated and I trust that others will agree. I have to say that is the "longest" image I have ever seen. I apologize for my previous post where it may looked as if I "got overly excited" about the class harmonic polynomials "along the way" to infintity but I did not fully understand the situation and my behavior got the best of me.

« Last Edit: Apr 14^{th}, 2006, 8:15pm by JP05 » 
IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: Mean value property over squares?
« Reply #13 on: Apr 14^{th}, 2006, 8:46pm » 
Quote Modify

We've all been there. I "proved" once that a wellknown and accepted set theory was contradictory. When I looked at my proof a few months later, I realized that I had totally misunderstood the concepts. At least I didn't go nearly as far in my bogus reasonings as these people did. This is why discourse like this is so important in mathematics (and all other disciplines). By bringing your concerns forward, others were able to point out what you were missing and get you heading in the right direction now, instead of you building upon your misconception to the point that it is difficult to find where you went wrong. This is a correction that everyone needs from time to time. In other words, you have nothing to apologize for.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



