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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
Spectral geometry, or What do
• M. Kac, Can one hear the shape of a drum?,
Amer. Math. Monthly, 1966.
Spectral geometry, or What do
• 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.
Spectral geometry, or What do
• 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.
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?
Schrödinger operators
-  + V(x)
Nanoelectronics
• Nanoscale = 10-1000 X width of atom
• Foreseen by Feynman in 1960s
• Laboratories by 1990.
Nanoelectronics
•
•
•
•
Quantum wires
Quantum waveguides
Designer potentials
Simplified mathematical models
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?
So, can you hear the density of a
string?
• I.e., find V(x) given the eigenvalues of
-d2/dx2 + V(x)
on an interval with some reasonable
boundary conditions (Dirichlet, Neumann,
periodic)?
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).
Well, can you hear
the shape of a drum?
So, can you hear
the shape of a drum?
Gordon,
Webb, and
Wolpert,
1991
Can you hear the interaction in
quantum mechanics from
scattering experiments?
• No! Bargmann exhibited two different
potentials with the same scattering data in
1949.
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!
Some things are “audible”
• You can hear the area of the drum, by the
Weyl asymptotics:
• For the drum problem
k ~ Cn (Vol()/k)2/n.
(A mathematician’s drum can be ndimensional, and even be a curved
manifold.)
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).
To extremists,
things tend to look simple…
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.
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.
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.
Some loopy nano-problems
s = arclength, k = curvature, and g = a “coupling constant”
Isoperimetric theorems for
- d2/ds2 + g k2
Minimality when g ≤ 1/4.
If 0 < g ≤ 1/4, the unique curve minimizing 1 is the
circle
A non linear functional
A non linear functional
Minimizer therefore exists.
Its Euler equation is
1
1
1
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.
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
Reduction to an isoperimetric
problem of classical type.
Reduction to an isoperimetric
problem of classical type.
Science is full of amazing coincidences!
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!
A family of isoperimetric
conjectures for p > 0:
Right side corresponds to circle.
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.
Proposition. 2.1.
QuickTime™ and a
TIFF (Uncompr essed) decompressor
are needed to see this picture.
First part follows from convexity of x  xa for a > 1:
Proof when p = 2
Inequality equivalent to
Inductive argument based on
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.
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?
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)||pp = 2p+2(π/2)p+1/(p+1) .
Better than the circle for p > 3.15296…
Examples that are more like the circle are not better than
the circle until higher p:
Stadium, small straight segments p > 4.27898…
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
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.
Circle is local maximizer
for all p < 
QuickTime™ and a
Photo - JPEG decompressor
are needed to see this picture.
Another theorem implying
extreme cases are simple
Let A be a positive semidefinite linear
operator on L2(X, dµ) with the property
that A 1 = 0.
Another theorem implying
extreme cases are simple
Let A be a positive semidefinite linear
operator on L2(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.)
Another theorem implying
extreme cases are simple
Let A be a positive semidefinite linear
operator on L2(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.
Proof
Recall the Rayleigh-Ritz inequality,
1 <u, u> ≤
<u, (A + V) u>
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 B2 = A+V-
1.
Calculate
||B 1|| = <B 1, B 1> = <1, (A+V-
1)1> = 0.
Therefore B 1 = 0 so 0 = B2 1 = V(x) - 1.
QED.
```
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