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The extreme sport of eigenvalue hunting. Evans Harrell Georgia Tech www.math.gatech.edu/~harrell Research Horizons Georgia Tech 1 March 2006.

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Presentation on theme: "The extreme sport of eigenvalue hunting. Evans Harrell Georgia Tech www.math.gatech.edu/~harrell Research Horizons Georgia Tech 1 March 2006."— Presentation transcript:

1 The extreme sport of eigenvalue hunting. Evans Harrell Georgia Tech www.math.gatech.edu/~harrell Research Horizons Georgia Tech 1 March 2006

2 Spectral geometry, or What do eigenvalues tell us about shapes? M. Kac, Can one hear the shape of a drum?, Amer. Math. Monthly, 1966.

3 Spectral geometry, or What do eigenvalues tell us about shapes? M. Kac, Can one hear the shape of a drum?, Amer. Math. Monthly, 1966. Already in 1946, G. Borg considered whether you could hear the density of a guitar string.

4 Spectral geometry, or What do eigenvalues tell us about shapes? M. Kac, Can one hear the shape of a drum?, Amer. Math. Monthly, 1966. Already in 1946, G. Borg considered whether you could hear the density of a guitar string, but he failed to think of such a colorful title.

5 Inverse spectral theory Asking both questions at the same time, would mean: If we look for eigenvalues (normal modes) of the differential operator -  + V(x) acts on functions on a region (or manifold or surface) M, What do we know about V(x) or M?

6 Schrödinger operators -  + V(x)

7 Nanoelectronics Nanoscale = 10-1000 X width of atom Foreseen by Feynman in 1960s Laboratories by 1990.

8 Nanoelectronics Quantum wires Quantum waveguides Designer potentials Semi- and non-conducting “threads” Simplified mathematical models

9 Can you hear the shape of a drum, or the density of a string, or the strength of an interaction? Can you determine the domain Ω &/or the potential V(x) from the eigenvalues of the Laplace or Schrödinger operator?

10 So, can you hear the density of a string? I.e., find V(x) given the eigenvalues of -d 2 /dx 2 + V(x) on an interval with some reasonable boundary conditions (Dirichlet, Neumann, periodic)?

11 Can you even hear the density of a string? No! In 1946 Borg showed there is usually an infinite-dimensional set of “isospectral” V(x).

12 Well, can you hear the shape of a drum?

13 So, can you hear the shape of a drum? Gordon, Webb, and Wolpert, 1991

14 Can you hear the interaction in quantum mechanics from scattering experiments? No! Bargmann exhibited two different potentials with the same scattering data in 1949.

15 Can you hear the interaction in quantum mechanics from scattering experiments? No! Bargmann exhibited two different potentials with the same scattering data in 1949, thereby destroying the careers of whole tribes of chemists and causing bad blood between the disciplines ever since!

16 Some things are “audible” You can hear the area of the drum, by the Weyl asymptotics: For the drum problem k ~ C n (Vol(  )/k) 2/n. (A mathematician’s drum can be n- dimensional, and even be a curved manifold.)

17 Some things are “audible” The Schrödinger equation also exhibits Weyl asymptotics, which determine both –the volume of the region, and –the average of V(x).

18 To extremists, things tend to look simple…

19 Classic extreme spectral theorem Rayleigh conjectured, and Faber and Krahn proved, that if you fix the area of a drum, the lowest eigenvalue is minimized uniquely by the disk. This requires Dirichlet boundary conditions - the displacement is 0 at the edge.

20 Classic extreme spectral theorem Rayleigh conjectured, and Faber and Krahn proved, that if you fix the area of a drum, the lowest eigenvalue is minimized uniquely by the disk. Seemingly, rounder  deeper tone.

21 Classic extreme spectral theorem Rayleigh conjectured, and Faber and Krahn proved, that if you fix the area of a drum, the lowest eigenvalue is minimized uniquely by the disk. Seemingly, rounder  deeper tone However, if your drum is annular (fixing edge length and width), circular geometry maximizes 1.

22 Some loopy nano-problems s = arclength,  = curvature, and g = a “coupling constant”

23 Isoperimetric theorems for - d 2 /ds 2 + g  2

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25 Minimality when g ≤ 1/4. If 0 < g ≤ 1/4, the unique curve minimizing 1 is the circle

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27 A non linear functional

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31 Minimizer therefore exists. Its Euler equation is

32 1 1 1

33 Many open questions: Is the bifurcation value g=1? What is the minimizer for g ≥ 1? Equivalent to open questions of Lieb-Thirring inequalities, Fourier series.

34 An electron near a charged thread Recent article with Exner and Loss in Lett Math. Phys. Fix the length of the thread. What shape binds the electron the least tightly? Conjectured for about 3 years that answer is circle.

35 Reduction to an isoperimetric problem of classical type.

36 Science is full of amazing coincidences! Mohammad Ghomi and collaborators had considered and proved similar inequalities in a study of knot energies, A. Abrams, J. Cantarella, J. Fu, M. Ghomi, and R. Howard, Topology, 42 (2003) 381-394!

37 A family of isoperimetric conjectures for p > 0: Right side corresponds to circle.

38 A family of isoperimetric conjectures for p > 0: Right side corresponds to circle. The case C -1 arises in an electromagnetic problem: minimize the electrostatic energy of a charged nonconducting thread.

39 Proposition. 2.1. First part follows from convexity of x  x a for a > 1:

40 Proof when p = 2

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43 Inequality equivalent to

44 Inductive argument based on

45 What about p > 2? Funny you should ask….

46 What about p > 2? Funny you should ask…. The conjecture is false for p = . The family of maximizing curves for ||  (s+u) -  (s)||  consists of all curves that contain a line segment of length > s.

47 What about p > 2? Funny you should ask…. The conjecture is false for p = . The family of maximizing curves for ||  (s+u) -  (s)||  consists of all curves that contain a line segment of length > s. At what critical value of p does the circle stop being the maximizer?

48 What about p > 2? At what critical value of p does the circle stop being the maximizer? This problem is open. We calculated ||  (s+u) -  (s)|| p for some examples: Two straight line segments of length π: ||  (s+u) -  (s)|| p p = 2 p+2 (π/2) p+1 /(p+1). Better than the circle for p > 3.15296…

49 What about p > 2? Examples that are more like the circle are not better than the circle until higher p: Stadium, small straight segments p > 4.27898…

50 What about p > 2? Examples that are more like the circle are not better than the circle until higher p: Stadium, small straight segments p > 4.27898… Polygon with many sides, p > 6

51 What about p > 2? Examples that are more like the circle are not better than the circle until higher p: Stadium, small straight segments p > 4.27898… Polygon with many sides, p > 6 Polygon with rounded edges, similar.

52 Circle is local maximizer for all p < 

53 Another theorem implying extreme cases are simple Let A be a positive semidefinite linear operator on L 2 (X, dµ) with the property that A 1 = 0.

54 Another theorem implying extreme cases are simple Let A be a positive semidefinite linear operator on L 2 (X, dµ) with the property that A 1 = 0. (I am thinking of A = - ∆ on a manifold without boundary, but it could be of the form - div T(x) grad, for example.)

55 Another theorem implying extreme cases are simple Let A be a positive semidefinite linear operator on L 2 (X, dµ) with the property that A 1 = 0. Consider H = A + V(x) for some real-valued function V. If we fix the integral of V, then the lowest eigenvalue 1 is maximized when V is constant.

56 Proof Recall the Rayleigh-Ritz inequality, 1 ≤ And choose u = 1. We see that 1 ≤ Ave(V). V = cst is a case of equality. To see that it is the unique such case, assume that A+V- 1 is a positive operator and use the square root lemma to define B≥0 such that B 2 = A+V- 1. Calculate ||B 1|| = = = 0. Therefore B 1 = 0 so 0 = B 2 1 = V(x) - 1. QED.

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