Regular Polygonal Vortex Filament Evolution and Exponential Sums

Abstract

When taking a regular planar polygon of M\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$M$\\end{document} sides and length 2π\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$2\\pi $\\end{document} as the initial datum of the vortex filament equation, Xt=Xs∧Xss\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$\\mathbf{X}_{t}= \\mathbf{X}_{s}\\wedge \\mathbf{X}_{ss}$\\end{document}, the solution becomes polygonal at times of the form tpq=(p/q)(2π/M2)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$t_{pq} = (p/q)(2\\pi /M^{2})$\\end{document}, with gcd(p,q)=1\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$\\gcd (p,q)=1$\\end{document}, and the corresponding polygon has Mq\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$Mq$\\end{document} sides, if q\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$q$\\end{document} is odd, and Mq/2\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$Mq/2$\\end{document} sides, if q\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$q$\\end{document} is even. Moreover, that polygon is skew (except when q=1\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$q = 1$\\end{document} or q=2\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$q = 2$\\end{document}, where the initial shape is recovered), and the angle ρ\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$\\rho $\\end{document} between two adjacent sides is a constant. In this paper, we give a rigorous proof of the conjecture that states that, at a time tpq\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$t_{pq}$\\end{document}, cosq(ρ/2)=cos(π/M)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$\\cos ^{q}(\\rho /2) = \\cos (\\pi /M)$\\end{document}, if q\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$q$\\end{document} is odd, and cosq(ρ/2)=cos2(π/M)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$\\cos ^{q}(\\rho /2) = \\cos ^{2}(\\pi /M)$\\end{document}, if q\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$q$\\end{document} is even. Since the transition of one side of the polygon to the next one is given by a rotation in R3\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$\\mathbb{R}^{3}$\\end{document} determined by a generalized Gauss sum, the idea of the proof consists in showing that a certain product of those rotations is a rotation of angle 2π/M\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$2\\pi /M$\\end{document}, which is equivalent to proving that some exponential sums with arithmetic content are purely imaginary.