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Title: Mean value property over squares? Post by Amadeus on Apr 3rd, 2006, 3:23am 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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Title: Re: Mean value property over squares? Post by Michael_Dagg on Apr 3rd, 2006, 2:19pm It is about characterization. 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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Title: Re: Mean value property over squares? Post by JP05 on Apr 4th, 2006, 12:09pm 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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Title: Re: Mean value property over squares? Post by Michael_Dagg on Apr 7th, 2006, 6:45pm Your remark about contour integration is correct. 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. 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. |
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Title: Re: Mean value property over squares? Post by JP05 on Apr 8th, 2006, 11:44am 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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Title: Re: Mean value property over squares? Post by Michael_Dagg on Apr 10th, 2006, 4:56pm 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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Title: Re: Mean value property over squares? Post by JP05 on Apr 11th, 2006, 11:57am References? |
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Title: Re: Mean value property over squares? Post by JP05 on Apr 11th, 2006, 3:13pm 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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Title: Re: Mean value property over squares? Post by Icarus on Apr 11th, 2006, 4:47pm 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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Title: Re: Mean value property over squares? Post by Amadeus on Apr 12th, 2006, 3:18am 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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Title: Re: Mean value property over squares? Post by Michael_Dagg on Apr 12th, 2006, 12:07pm 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 mean-value property. Give me a few days and I will provide you with references for your question regarding the n-gons. |
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Title: Re: Mean value property over squares? Post by Michael_Dagg on Apr 13th, 2006, 11:33am A pretty result regarding characterization (and also contains a result about harmonic conjugates) is given in Goldstein & Ow.: http://links.jstor.org/sici?sici=0002-9939%28197107%2929%3A2%3C341%3AOTMPOH%3E2.0.CO%3B2-1&size=LARGE#abstract Another is: http://links.jstor.org/sici?sici=0002-9890%28199701%29104%3A1%3C52%3ATSAHF%3E2.0.CO%3B2-A&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 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. http://www.facesofsantafe.com/wu/br.jpg |
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Title: Re: Mean value property over squares? Post by JP05 on Apr 14th, 2006, 7:48pm 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. |
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Title: Re: Mean value property over squares? Post by Icarus on Apr 14th, 2006, 8:46pm 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. (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1100790578) 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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