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 Mean value property over squares?   « on: Apr 3rd, 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?

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

The mean-value 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.
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Michael Dagg
JP05
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 Re: Mean value property over squares?   « Reply #2 on: Apr 4th, 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?
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Michael Dagg
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 Re: Mean value property over squares?   « Reply #3 on: Apr 7th, 2006, 6:45pm » Quote Modify

The name of the class is harmonic polynomials.  It can be shown that if the mean-value 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 mean-value
property over a regular n-gon (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.
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 12th, 2007, 1:26pm by Michael Dagg » IP Logged

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Michael Dagg
JP05
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 Re: Mean value property over squares?   « Reply #4 on: Apr 8th, 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 n-polygons 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?

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Michael Dagg
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 Re: Mean value property over squares?   « Reply #5 on: Apr 10th, 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.
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Michael Dagg
JP05
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 Re: Mean value property over squares?   « Reply #6 on: Apr 11th, 2006, 11:57am » Quote Modify Remove

References?
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JP05
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 Re: Mean value property over squares?   « Reply #7 on: Apr 11th, 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 n-gon the class of harmonic functions is polynomials and there is no k > n such that the class contains non-polynomial harmonic functions or even contains one.

Since not all harmonic functions are polynomials and that the mean-value 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 non-polynomials 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 non-polynomial harmonic function along the way.

I am clear?

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Icarus
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 Re: Mean value property over squares?   « Reply #8 on: Apr 11th, 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 RN, 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 n-gon to have non-polynomial 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.
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 Re: Mean value property over squares?   « Reply #9 on: Apr 12th, 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.
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Michael Dagg
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 Re: Mean value property over squares?   « Reply #10 on: Apr 12th, 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.

This paper contains some interesting ideas and as you will see it starts with the mean-value property.

Give me a few days and I will provide you with references for your question regarding the n-gons.
 « Last Edit: Apr 12th, 2006, 5:07pm by Michael Dagg » IP Logged

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

A pretty result regarding characterization (and also contains a result about
harmonic conjugates) is given in Goldstein & Ow.:

Another is:

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 n-gons.

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 13th, 2006, 1:34pm by Michael Dagg » IP Logged

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Michael Dagg
JP05
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 Re: Mean value property over squares?   « Reply #12 on: Apr 14th, 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 14th, 2006, 8:15pm by JP05 » IP Logged
Icarus
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 Re: Mean value property over squares?   « Reply #13 on: Apr 14th, 2006, 8:46pm » Quote Modify

We've all been there. I "proved" once that a well-known 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.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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