mm-519 === Subject: Re: measurable fields of Hilbert spaces Content-Length: 355 Originator: rusin@vesuvius This may be what you want: take the Haussdorf-Besikovitch measure of a Hilbert or Banach space or a subset, with respect to a uniform directed divergence. A general method of finding such a uniform directed divergence is not known, but specific problems no doubt do yield to attack. A. Hunger Buses are bosons. === Subject: Re: Axiom of Choice & Lebesgue Measure Problem Content-Length: 446 Originator: rusin@vesuvius > You can check Stan Wagon's book on the Banach-Tarski paradox, ISBN > 0-521-45704-1. In chapter 13 you will find the known consistency > results involving the axiom of choice and Lebesgue measure. Yes, still available in paperback, last chapter The Role of the Axiom Mike Carroll === Subject: Integer analogs of log, exp series Epigone-thread: troiplerlwhou Content-Length: 2249 Originator: rusin@vesuvius The idea below must be well-known but after searches failed (what are good key words?) Id like to ask here for references to related work. Consider functions f, g from an abelian group (A,*) to a commutative ring R[[x]] of power series. When the ring R contains Q (rationals), log and exp are often used to give bijections between multiplicative (f(a*b)=f(a).f(b)) and additive (g(a*b)=g(a)+g(b)) homomorphisms, provided in the image the constant terms in the power series are 1 (for f) and 0 (for g). Now for integral (integer coefficients) versions of such bijections. In what follows, write series as f = f_0 + f_1x +f_2x^2 + ..., with constant terms f_0=1, g_0=0, and let C(n,k) denote the binomial n choose k, where in fact n is a function A -> Z. The binomials below arise more or less as they do in formulas for integer-valued polynomial functions p (one-variable case p(x) is an often-quoted exercise in Polya-Szego; multivariate case is just as easy but Ive never seen a reference). In the case A -> Z[[x]], a bijection f <-> g can be constructed by a recursive formula with terms indexed by partitions of integers: $$f_n = sum{ prod C(g_i,j_i) : sum i j_i = n } (n>0; j_i>0)$$ i.e., f_0 = 1, g_0 = 0, f_1 = g_1, f_2 = g_2 + C(g_1,2), f_3 = g_3 + g_2g_1 + C(g_1,3), f_4 = g_4 + g_3g_1 + C(g_2,2) + g_2C(g_1,2) + C(g_1,4), etc. Thus the multiplicative homomorphisms f can be understood. This generalizes to power series over Z in a set of commuting variables. Its not clear (or useful) to me if much more might be possible. books/series what I would most like is a _full_ reference (like Ch, Sec, formula number) to quote, as I dont have easy access to books - the nearest adequate mathematics library is over 700km away. The above idea, for Z[[x,y]], suffices for my purposes (expressing the information contained in certain knot polynomials in the form of primitive integral Vassiliev invariants). However, I should add that non-commutative (say quantum) versions of the bijections, say with (A,*) more general, would presumably be of some interest. Peter Johnson Universidade Federal de Bahia Salvador, Bahia, Brazil === Subject: Determinant maximization with trace constraints Epigone-thread: khimclarquah Content-Length: 765 Originator: rusin@vesuvius Hi all, I am considering a determinant maximization problem with trace constrains, which is MAX det(A+F'*B*F) with respect to F s.t. trace(F'F)=M where M is a constant value and A is diagonal matrix with positive entries and B is diagonal matrix with non-negative entries. Without loss of generality, we can arrange the diagonal entries of A and B in opposite direction, e.g., diagonal entries of A are arranged in non-decreasing order, while diagonal entries of B are arranged in non-increasing order. My conjecture is that, under such condition, one solution is the diagonal matrix F with entries determined by water-filling algorithm. But I cannot prove it. Could anyone please give me some ideas of how to solve this optimization problem. === Subject: Re: Higher dimensional Klein/Pluecker terminology Content-Length: 919 Originator: rusin@vesuvius > Does anyone know of standard names for the higher dimensional > analogues of Pluecker coordinates, the Klein correspondence, or the > Klein quadric? In particular, if you map a pair of n-space vectors, > defining a plane through the origin and corresponding projective line, > to the wedge product, it lies on a projective variety. Does it have a > name? I would understand Pluecker coordinates as n-dimensional, if the context called for that. As Stein Arild Strmme said (if I understand him) the Klein quadric generalizes to Grassmann variety. I am not sure what the Klein correspoondence is, unless it is the bijection between the set of lines (or k-spaces) and the points of the Grassmann variety. -- Chris Henrich The total lack of evidence is the surest sign that the conspiracy is working. === Subject: Re: Higher dimensional Klein/Pluecker terminology Content-Length: 678 Originator: rusin@vesuvius [Gene Ward Smith] | Does anyone know of standard names for the higher dimensional | analogues of Pluecker coordinates, the Klein correspondence, or the | Klein quadric? In particular, if you map a pair of n-space vectors, | defining a plane through the origin and corresponding projective line, | to the wedge product, it lies on a projective variety. Does it have a | name? It is called the Grassmann variety (of lines). SA -- Stein Arild Strmme +47 55584825, +47 95801887 Universitetet i Bergen Fax: +47 55589672 Matematisk institutt www.mi.uib.no/stromme/ Johs Brunsg 12, N-5008 BERGEN stromme@mi.uib.no === Subject: Galois group of a given quartic equation Epigone-thread: twomprendlex Content-Length: 484 Originator: rusin@vesuvius Given a, b and c integers, let the Galois group of Q(x) = x^4 - 2 c x^3 + (c^2 - a^2 - b^2) x^2 + 2 a^2 c x - a^2 c^2 = 0 be G. For a and b different, can anybody either show that G cannot be A4 nor Z4 or give examples with these groups? All other cases of G is possible (ie. S4, S3, A3, D4, V = Z2 x Z2, Z2, {e}). If a = b, a complete anaylsis of the case can be given. We can prove that the only possibles groups then are V = Z2 x Z2, Z2 and D4. Kent Holing Norway === Subject: Re: Galois group of a given quartic equation Content-Length: 2269 Originator: rusin@vesuvius >Given a, b and c integers, let the Galois group of >Q(x) = x^4 - 2 c x^3 + (c^2 - a^2 - b^2) x^2 + 2 a^2 c x - a^2 c^2 = 0 >be G. >For a and b different, can anybody either show that G cannot be A4 nor >Z4 or give examples with these groups? A4 can be achieved. The Galois group is contained in A4 iff the discriminant of the polynomial is a rational square. A brute-force check of small combinations of integral a,b,c shows that the discriminant is not often a square, except when a = 0 or when b = a. But one additional example comes up: a=1, b=41, c=57. That example yields a quartic with Galois group Z2 x Z2 . However, it suggests there may be additional ways to keep the discriminant square. After failing to find any interesting generalizations of the obvious types, I tried looking for other examples having a = 1 and c = b + 16. (I had noticed that linear combinations of variables and perfect squares showed up a lot in other tricks I had pulled with this polynomial.) With these two constraints, the discriminant is - 27 b^4 + 31904 b^3 + 776448 b^2 + 6242400 b + 16581375. So now the question is, can we find rational numbers b which make this be a perfect square, knowing that it's square when b = 41 ? This is a question of elliptic curves: using this known point on the curve we can reduce it to canonical form (Y^2 = X^3 - 8847468 X - 118751785924608), compute torsion (1) and rank (1, I think). But the point is that there will be infinitely many values of b which make this quartic be square. One of them --- apparently the simplest after b = 41 --- is b = -4867108924096299900875640871/2078873513798342722876563129 and using a = 1 and c = 16 + b , this leads to the quartic 4321715086372268251406123617870969227589757195122270641*x^4 - 118058675101761891412526445783260584727107165316384569794*x^3 + 778258024430517321989049362739330264143147045915729561967*x^2 + 118058675101761891412526445783260584727107165316384569794*x - 806268488795907432230329054223036357236129975845817471249 whose galois group Maple reports to be Alt(4). I don't recall the algebraic characterization of quartics with Galois group contained in Z4 . dave PS -- We can more generally express the problem in terms of a, b, and d = c - b ; the Galois group is in Alt(4) iff a certain quartic in b is a square. If there is a value of b with this condition, then the curve is birationally equivalent to its Jacobian, which I make out to be Y^2 = X^3 - (135 d^4 + 108) X - 54 d^2 ( 2d^8 + 5d^4 + 20 ) These curves all have _integer_ points for d = 4, 5, 6, 7, 8 and have rank 1 through d = 10. I don't know if there is a generic point (i.e. a point on the curve over Q(d) ). === Subject: Re: Galois group of a given quartic equation charset=Windows-1252 Content-Length: 1050 Originator: rusin@vesuvius Dave Rusin showed that the Galois group of Q(x)=x^4 - 2 c x^3 + (c^2 - a^2 - b^2) x^2 + 2 a^2 c x - a^2 c^2 can be A4, where a,b,c are rational numbers. His example involved large numbers. Searching for (a,b,c) making the discriminant a square, as he suggests, I found the following triples of small integers for which the Galois group of Q(x) is A4: (5,333,365), (7,207,241), (11,171,277), (19,256,333), (61,469,873), (69,1029,1597), (93,672,971), (139,189,923). These were found with ubasic and the Galois groups verified as A4 using Maple 7. Also, to my surprise, Q(x) can be completely reducible when the discriminant is a square, for example, with (a,b,c) = (168,660,1105) the discriminant is (19*23*31*83*151)^2. and Q(x)=(x+195)*(x-357)*(x-280)*(x-1768). I don't know if Z4 can be achieved as a Galois group of Q(x). Jim Buddenhagen === Subject: Re: Galois group of a given quartic equation Content-Length: 874 Originator: rusin@vesuvius > This is a question of elliptic curves: using this known point on > the curve we can reduce it to canonical form (Y^2 = X^3 - 8847468 X > - 118751785924608), compute torsion (1) and rank (1, I think). Just a side point: this curve has analytic rank 3, and the following three points are independent (so the rank is at least 3). (53968, 6160760) (224641/4, 60862751/8) (18288149248/56169, -2468812272295904/13312053) (Results courtesy of Mark Watkins and Magma.) Geoff. ------------------------------------------------------------- --------------- - Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@maths.usyd.edu.au | Gameplayer by vocation. ------------------------------------------------------------- --------------- - === Subject: School Announcement Content-Length: 2093 Originator: rusin@vesuvius Summer School on Mathematical and Theoretical Crystallography Nancy, France, 20-24 June 2005 This is the second official activity of the MaThCryst workgroup, after The program of the School can be seen at our website http://www.lcm3b.uhp-nancy.fr/mathcryst/nancy2005.htm The School is addressed to young scientists as well as to all those who are interested in fundamental aspects of crystallography seldom covered by basic university courses, because of time limitations. Each session will include a lecture and guided exercises and / or examples / demonstrations. Poster presentations by the participants are welcome, and the corresponding abstracts will be made available electronically from our website. Full papers for the proceedings are also welcome, and will undergo a normal peer-review process. Financial support is provided by the International Union of Crystallography and by the European Crystallographic Association, to cover - totally or partially - travel and accommodation expenses by young scientists (graduate students and post-docs). Applications for financial support should include a curriculum vitae, a list of publication and a certificate of current status (copy of student card or letter by the supervisor). Inquiries should be sent to mathcryst.satellite@lcm3b.uhp-nancy.fr We kindly ask readers of this newsgroup to forward this information to colleagues who may be interested in the School. We would also be grateful to get links to our website in your institutions, associations and personal pages. ///////////////////////////////////////////////////////////// Dr. Massimo Nespolo Associate Professor of Mineralogy and Crystallography Laboratoire de Cristallographie et de Modelisation des Materiaux Mineraux et Biologiques (LCM3B) UMR - CNRS 7036 Universite' Henri Poincare' Nancy1 BP 239 F54506 Vandoeuvre-les-Nancy cedex France http://www.lcm3b.uhp-nancy.fr/ ///////////////////////////////////////////////////////////// === Subject: Nonsmooth Analysis Content-Length: 313 Originator: rusin@vesuvius let f be locally Lipschitz continuous and Bouligand-differentiable (= directionally differentiable). What is the link between the B-derivative of f at x, which is a mapping assigning each direction a scalar, and the B-subdifferential or its convex hull, which is a set of subgradients? bjj === Subject: Paper published by Algebraic and Geometric Topology Content-Length: 1555 Originator: rusin@vesuvius The following paper has been published: Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-30.abs.html Title: The Z-graded symplectic Floer cohomology of monotone Lagrangian sub-manifolds Author(s): Weiping Li Abstract: We define an integer graded symplectic Floer cohomology and a Fintushel-Stern type spectral sequence which are new invariants for monotone Lagrangian sub-manifolds and exact isotopes. The Z-graded symplectic Floer cohomology is an integral lifting of the usual Z_Sigma(L)-graded Floer-Oh cohomology. We prove the Kunneth formula for the spectral sequence and an ring structure on it. The ring structure on the Z_Sigma(L)-graded Floer cohomology is induced from the ring structure of the cohomology of the Lagrangian sub-manifold via the spectral sequence. Using the Z-graded symplectic Floer cohomology, we show some intertwining relations among the Hofer energy e_H(L) of the embedded Lagrangian, the minimal symplectic action sigma(L), the minimal Maslov index Sigma(L)$ and the smallest integer k(L, phi$ of the converging spectral sequence of the Lagrangian L. Secondary: 53D12, 70H05 Keywords: Monotone Lagrangian sub-manifold, Maslov index, Floer cohomology, spectral sequence Received: 3 December 2002 Author(s) address(es): Department of Mathematics, Oklahoma State University Stillwater, Oklahoma 74078-0613 Email: wli@math.okstate.edu === Subject: Inonu-wigner contraction Content-Length: 211 Originator: rusin@vesuvius contraction. There was no reply to my post in sci.physics.research. Can anybody here help? References and pointers would be helpful too. === Subject: Asymptotics and generating function Epigone-thread: skarjoafrem Content-Length: 352 Originator: rusin@vesuvius Let a sequence f(n) having the generating function : 1/(1-z^p)^q*exp(A(z)) p and q are integers and A is analytic for IzI<=1 . Is there a simple asymptotic formula for f(n) in terms of A(1)? If I'm not wrong it is easy when A is analytic for IzI1. But here A is analytic for IzI<=1 only. XD. === Subject: Re: Asymptotics and generating function Content-Length: 997 Originator: rusin@vesuvius >Let a sequence f(n) having the generating function : >1/(1-z^p)^q*exp(A(z)) >p and q are integers and A is analytic for IzI<=1 . Your IzI, I guess, is what most people write as |z|. >Is there a simple asymptotic formula for f(n) in terms of A(1)? If I'm >not wrong it is easy when A is analytic for IzI1. But here A >is analytic for IzI<=1 only. If by analytic for |z| <= 1 you mean analytic at every point with |z| <= 1 (and thus in a neighbourhood of every such point), this implies that there is r > 1 such that it is analytic for |z| < r. But I really don't believe your statement that for r > 1 it should be in terms of A(1). Perhaps in terms of the values of A at all the p'th roots of 1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Asymptotics and generating function Epigone-thread: skarjoafrem Content-Length: 933 Originator: rusin@vesuvius >p and q are integers and A is analytic for IzI<=1 . >Your IzI, I guess, is what most people write as |z|. Exactly. Sorry for this notation. >Is there a simple asymptotic formula for f(n) in terms of A(1)? If I'm >not wrong it is easy when A is analytic for IzI1. But here A >is analytic for IzI<=1 only. >If by analytic for |z| <= 1 you mean analytic at every point with >|z| <= 1 (and thus in a neighbourhood of every such point), this implies >that there is r > 1 such that it is analytic for |z| < r. Let me give an exemple interesting me : A(z)=sum(i=0,oo, z^(2*i+1)/(i+1/2))-z^(2*i+2)/(i+1)) If I understand, A is not analytic for |z| <= 1 but for |z| < 1 I'm interested with F(z)=1/(1-z^2)*exp(A(z)) and how getting the asymptotic behaviour of f(n) in F(z)=sum (f(k)*z^k). Here A(1)=Psi(1/2)... Xavier Debez. === Subject: Re: Asymptotics and generating function Epigone-thread: skarjoafrem Content-Length: 2307 Originator: rusin@vesuvius As a sample, let's consider A(z) = Sum(z^(r*i+1)/(i+1/r),i=0..infinity), and investigate the asymptotic properties of the Taylor series for f(z) = exp(A(z)). It turns out that A(z) = -sum(omega^(-j)*log(1-omega^(j)*z),j=0..4), where omega is a primitive fifth root of unity, omega = 0.3090169944 + 0.9510565163 i, approximately. Now f(z) = exp(A(z)) has singularities at the fifth roots of unity, z = omega^j . First, at z = 1, f(z) is 1/(1-z) times an analytic function, so f(z) is asymptotically C/(1-z) where C = exp(-ln(1-omega)/omega-ln(1-omega^2)/omega^2-ln(1-omega^3)/ omega^3-ln(1-omeg a^4)/omega^4) = 22.25098593367932 . At z=omega^(-1), f(z) is (1-z*omega)^(-omega^(-1)) times an analytic function, and similarly for the others. Now we need to apply the version of Darboux's formula as in Szego, Orthogonal Polynomials, Theorem 8.4. You can't tell from Wilf, but Szego allows complex exponent, as we have here. Since the exponents are all real part > -1 for the terms other than the z=1 term, I expect the z=1 term is the most important one (but to see for sure you would investigate the asymptotics of the binomial coefficients for complex exponent). If this is true, then the asymptotics should be the same as for the single term C/(1-z), which has all terms = C. And it looks like that is true: Maple says f(z) has series 1.+5.*z+12.50000000*z^2+20.83333333*z^3+26.04166667*z^4+ 26.04166667*z^5+ 22.53472222*z^6+19.66765873*z^7+20.10478671*z^8+22.74340002*z ^9+24.39253 334*z^10+23.37918182*z^11+21.21412669*z^12+20.53145593*z^13+ 21.95341999* z^14+23.57949556*z^15+23.44177987*z^16+21.86931880*z^17+ 20.90308450*z^18 +21.67701591*z^19+23.10387565*z^20+23.36132648*z^21+ 22.20841145*z^22+21. 18685123*z^23+21.57489827*z^24+22.79808017*z^25+23.25300988*z ^26+22.4018 5561*z^27+21.40381630*z^28+21.54540138*z^29+22.58930037*z^30+ 23.14595996 *z^31+22.51812191*z^32+21.57265048*z^33+21.54940860*z^34+ 22.44074441*z^3 5+23.04801336*z^36+22.58972886*z^37+21.70647659*z^38+ 21.56954552*z^39+22 .33184455*z^40+22.96067570*z^41+22.63388920*z^42+21.81431526* z^43+21.597 36026*z^44+22.25023518*z^45+22.88345390*z^46+22.66043864*z^47 +21.9024531 1*z^48+21.62849219*z^49+... and it looks like the coefficients could be converging to C = 22.25... The coefficient of z^999 is 22.32 and of z^1000 is 22.24 . === Subject: Re: Asymptotics and generating function Content-Length: 708 Originator: rusin@vesuvius >A(z)=sum(i=0,oo, z^(2*i+1)/(i+1/2))-z^(2*i+2)/(i+1)) A(z) = 2 (z - z^2/2 + z^3/3 - z^4/4 + ...) = 2 ln(1+z) >If I understand, A is not analytic for |z| <= 1 but for |z| < 1 Correct. It has a logarithmic singularity at z=-1. >I'm interested with F(z)=1/(1-z^2)*exp(A(z)) and how getting the >asymptotic behaviour of f(n) in F(z)=sum (f(k)*z^k). Here >A(1)=Psi(1/2)... F(z) = (1+z)/(1-z) = 2/(1-z) - 1 so f(n) = 2 for n >= 1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Asymptotics and generating function Content-Length: 1607 Originator: rusin@vesuvius >If by analytic for |z| <= 1 you mean analytic at every point with >|z| <= 1 (and thus in a neighbourhood of every such point), this > implies >that there is r > 1 such that it is analytic for |z| < r. > Let me give an exemple interesting me : > A(z)=sum(i=0,oo, z^(2*i+1)/(i+1/2))-z^(2*i+2)/(i+1)) Unbalanced parentheses... Assuming it is: A(z)=sum(i=0,oo, z^(2*i+1)/(i+(1/2))-z^(2*i+2)/(i+1)) I get A(z) = 2*ln(1+z), a complex logarithm, so exp(A(z)) = (1+z)^2 is analytic. That is all we need to do what Prof. Israel said. > If I understand, A is not analytic for |z| <= 1 but for |z| < 1 > I'm interested with F(z)=1/(1-z^2)*exp(A(z)) and how getting the > asymptotic behaviour of f(n) in F(z)=sum (f(k)*z^k). Here > A(1)=Psi(1/2)... ... here I get A(1) = 2*ln(2) You can find this explained in Wilf's book generatingfunctionology in section 5.2. If the generating function has at worst poles on the boundary of the disk of convergence, we can get good asymptotics for the coefficients. There is a more delicate theory (in section 5.3) where the function has algebraic singularities on the boundary of the disk of convergence. Maybe exp(A(x)) = sqrt(1-z) . You could easily modify your example above so that A(z) = (1/2)*ln(1-z), and then we have this situation. As I recall, Wilf's book only does the case of one algebraic singularity on the circle, but the references he gives handle the case of finitely many. -- G. A. Edgar edgar at math.ohio-state.edu === Subject: Re: Asymptotics and generating function Epigone-thread: skarjoafrem Content-Length: 545 Originator: rusin@vesuvius My example was too simple to help me. One of the general cases I consider is : F(z)=exp(sum(i=0,oo, A(z)) where : A(z)=sum(i=0,oo, a*z^(r*i+1)/(i+1/r)+z^(r*i+r)/(i+1)) and a, r are any integer>0. I tried to introduce B(z)=sum(i=0,oo, a*z^(r*i+r)/(i+1)) so that : F(z)=exp(A(z)-B(z)+B(z))=1/(1-z^r)^(a+1)*exp(A(z)-B(z)) A-B is now defined for |z|<=1. There are r poles for F : the r-th roots of unity but it seems I can't apply Wilf &5 directly here. Xavier Debez. === Subject: Re: Asymptotics and generating function Content-Length: 880 Originator: rusin@vesuvius > My example was too simple to help me. One of the general cases I > consider is : > F(z)=exp(sum(i=0,oo, A(z)) where : > A(z)=sum(i=0,oo, a*z^(r*i+1)/(i+1/r)+z^(r*i+r)/(i+1)) > and a, r are any integer>0. > I tried to introduce B(z)=sum(i=0,oo, a*z^(r*i+r)/(i+1)) so that : > F(z)=exp(A(z)-B(z)+B(z))=1/(1-z^r)^(a+1)*exp(A(z)-B(z)) > A-B is now defined for |z|<=1. There are r poles for F : the r-th > roots of unity but it seems I can't apply Wilf &5 directly here. > Xavier Debez. What is the simplest case you cannot do? Maybe that is the one for us to start with. -- G. A. Edgar edgar at math.ohio-state.edu === Subject: A solution to a quadratic problem Epigone-thread: slaqueltwi Content-Length: 241 Originator: rusin@vesuvius I want to find the solution to the meet following set of equations x^T A_i x=a_i, i=1,...,N where x is a M*1 vector M How would one solve the following system of equations? > (1) 17*u^2 - 144*u*v + 1904*v^2 = 43*U^2 - 60*U*V + 258*V^2 > (2) 162*u^2 - 8568*u*v + 18144*v^2 = 10*U^2 - 172*U*V + 60*V^2 > and u,v,U,V are in Z. > We seek solutions so that (u,v) and (U,V) have the same value in either (1) or > (2) (coincidental solutions?) > This is from trying to factor an homogenous quartic space from the elliptic > curve [0,6800,0,11559996,0] > Randall Provided the discriminants of each equation satisfy suitable mutual conditions (equal, or have the same square-free factor?) I think it's possible to find a linear transform of u, v, U, V, or maybe a birational transform of the dehomogenized pair, that reduces the original pair to the more canonical form: ax^2 + bxy + cy^2, dx^2 + exy + fy^2 = z^2, t^2 resp However, I'm moving house at the moment and all my notes are in boxes. [sci.math.research added - hope you don't mind, but I'd be interested in an expert answer to this myself!] John R Ramsden (jramsden@glasshouse.cam) com not cam === Subject: Re: Question: regularity of solutions to p-Laplacian problems Epigone-thread: ricliskang Content-Length: 1037 Originator: rusin@vesuvius >Let Omega be a smooth bounded domain in R^n, n ge 1 and consider the >problem >- Delta_p u = f(x) in Omega > u = 0 on the boundary of Omega >where Delta_p is the p-Laplacian operator, 1 < p < infty. When f is in >L^infty(Omega), it is known that this problem has a unique weak solution u >that is C^1 up to the boundary. I was wondering if anyone knows of any >results where one gets the same result, but only assuming that f is in >L^q(Omega) for an appropriate q < infty. Any references would be greatly >appreciated! The only references I know on the subject are the following ones: 1) Tolksdorf, Peter Regularity for a more general class of quasilinear elliptic equations. J. Differential Equations 51 (1984), no. 1, 126--150 2) Guedda, Mohammed; V.8eron, Laurent Local and global properties of solutions of quasilinear elliptic equations. J. Differential Equations 76 (1988), no. 1, 159--189. I hope this willk help you nicolas saintier === Subject: asymptotic of green function Epigone-thread: valgimblel Content-Length: 398 Originator: rusin@vesuvius Hello I'm looking for informations on the asymptotic behaviour of the Green's function G_epsilon for an operator $L_epsilon $ of the type $$ L_epsilon (u) = -Delta u + epsilon^{-2} h u $$ where $hin C^infty(Omega)$, $Omega subset R^n$ smooth bounded and $epsilon$ is a positive parameter which tends to 0. Any informations or references will be welcomed ! Nicolas Saintier === Subject: Contracting operator in a Banach space Epigone-thread: spoxstoghung Content-Length: 403 Originator: rusin@vesuvius I have this little pb that I can't figure out: Take E, a Banach space of real functions. || || is the associated norm on E. Suppose a you have a linear operator P on E so that: 1/ P(Id) = Id where Id is the identity function 2/ P preserves the positivity. i.e: f>=0 => P(f)>=0 How can we show that 1 + 2 imply that P is a contracting operator ? ie how can we show that: || Pf || <= ||f|| === Subject: Re: Contracting operator in a Banach space Originator: grubb@lola Content-Length: 450 Originator: rusin@vesuvius >I have this little pb that I can't figure out: >Take E, a Banach space of real functions. >|| || is the associated norm on E. >Suppose a you have a linear operator P on E so that: >1/ P(Id) = Id where Id is the identity function >2/ P preserves the positivity. i.e: f>=0 => P(f)>=0 >How can we show that 1 + 2 imply that P is a contracting operator ? >ie how can we show that: || Pf || <= ||f|| Use -||f|| Id <=f<=||f|| Id --Dan Grubb === Subject: Paper published by Geometry and Topology Originator: israel@math.ubc.ca (Robert Israel) The following paper has been published: URL: http://www.maths.warwick.ac.uk/gt/GTVol8/paper31.abs.html Title: Unimodal generalized pseudo-Anosov maps Author(s): Andre de Carvalho, Toby Hall Abstract: An infinite family of generalized pseudo-Anosov homeomorphisms of the sphere S is constructed, and their invariant foliations and singular orbits are described explicitly by means of generalized train tracks. The complex strucure induced by the invariant foliations is described, and is shown to make S into a complex sphere. The generalized pseudo-Anosovs thus become quasiconformal automorphisms of the Riemann sphere, providing a complexification of the unimodal family which differs from that of the Fatou/Julia theory. Secondary: 57M50 Keywords: Pseudo-Anosov homeomorphisms, train tracks, unimodal maps, horseshoe Proposed: Joan Birman Seconded: David Gabai, Yasha Eliashberg Author(s) address(es): Departamento de Matematica Aplicada, IME-USP Rua do Matao 1010, Cidade Universitaria 05508-090 Sao Paulo, SP, Brazil and Department of Mathematical Sciences, University of Liverpool Liverpool L69 7ZL, UK Email: andre@ime.usp.br, T.Hall@liv.ac.uk === Subject: Analytic Interpolation Originator: israel@math.ubc.ca (Robert Israel) I have an infinite family of sequences a_{m,n}, m, n in N. Three, (a_{m,5},a_{m,6} and a_{m,7}, m in N), are shown here: http://users.forthnet.gr/ath/jgal/math/figs/seq.gif All the sequences eventually become constant (to the right). I need to interpolate for m in N, using real analytic functions in the region between the first term and the first constant term (that's the S shaped region). I also have a recursive relationship for all the terms. (A recursive relationship for ALL terms of ALL sequences). Questions: 1) Do there exist real analytic functions that can interpolate them? (In the sequences only the initial terms are different. All the rest are given by a recursion). (The sequences seem to beg for some sort of variant of the exp function, but all my numerical analysis books are away from home). I know that polynomials can be used, but I am rather fond of the exp function. 2) If anyone can provide some online links and/or references, I'd be greatful. -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Anyone familiar with applications of the skip-fib seq 1 2 5 13 34 89 233 610, ... Originator: israel@math.ubc.ca (Robert Israel) ...particularly with reference to phyllotaxis-related applications (or other bio-related apps) is earnestly entreated to contact David Halitsky of Cumulative Inquiry (CI) at dhalitsky@cumulativeinquiry.com Please respond also if you are familiar with applications of this sequence as instatianted by ordered trees over n+1 vertices with max depth 3. CI-sponsored research has found a new case of this sequence which arises from a consideration of what dim2 series-parallel posets can be built from strings over a 4-letter alphabet of the form {++,-+.--.+-}, e.g. the alphabet defined (in three different ways) from the four DNA or mRNA bases. === Subject: Re: Fractional Integral Equation Epigone-thread: plucanskend Originator: israel@math.ubc.ca (Robert Israel) I can refer you to my paper Power Series Solutions of Fractional Differential Equations which has been accepted in the International fractional differential equations with non-constant coefficients. M. B. >Does anyone the solution for the following fractional integral >equation: >A*D^{-b}f(x)+x^{-b}*f(x)=1