I put explicit details (including code, output, and explanation) at

http://www.hbeLabs.com/lshape/index.html

that demonstrates clearly how to obtain the 150 digits or more. (I’m still making small edits, but it seems close

to what I want to say.)

Perhaps someone can reproduce it in matlab? Have a lot of fun!

–Bob Jones ]]>

You may be surprised to learn that I am applying your original point-matching method, but on steroids.

Include enough digits in the intermediate calculations and it is numerically very well behaved, i.e., no spurious solutions or ill-conditioning appears in this shape (and probably many others that exhibit exponential convergence). I use fractional order Bessel functions and properly symmetrized sinusoids centered on the re-entrant angle, my matching points are uniformly distributed as 3i/(N+1) for i=1..N along the length 3 matching edge (in the half-L), number of matching points equals number of terms in expansion, and no interior points per GSVD, nor extra points on the edges, for example. Calculate the determinant and be clever about root finding (working with dL/L=1E-150 and determinant values of 1E-200000 may scare people at first). Make sure Bessel functions (and sinusoid, etc) are good to such precision. As I increment N (by 3 for the L-shape, so same proportion of points stays in each segment), the approximate eigenvalue alternates above and below an asymptote while smoothly approaching it. As one increments N, the previous results can be used to help bracket the root. If no mistakes, it works perfectly from N=20 to 700, so far. (Change the distribution even a little and it may fail, but still works to some extent.) Note, exponential convergence rate is as expected, about 4 to 5 basis functions per digit, so nothing new there. I think the trade-off in numerics pays off as compared with other popular methods like GSVD, especially if one pursues many digits. It works for any (non-degenerate, non-closed-form) eigenvalue as long as you can get enough expansion terms. (Closed form eigenvalues converge many orders of magnitude faster, but the same pattern.)

Anyone can do it.

To get the 150+ digits below, I use up to N=708 and 711 matching points and perhaps 750 digits in the intermediate calculations. (Keep the precision well above the number of terms in the expansion.) I was greatly affected by your now famous paper in the early 1990s when I observed something similar (but not exactly, I wasn’t expanding about the correct angle), but it only made it into my thesis. Only now, I decided to pick up again, as a hobby. I am quite surprised that no one else saw that nice numerical behavior and ease with which one can bound eigenvalues. It was only last week that I first tried the L-shape (among others) and it worked like a charm. I had to share it with you.

9.6397238440219410527114592623648231562672895258219

064561095797005640356478633703907228731650087967888

3115756686104335651595260970541019403854362864964220

Lower bound, append 2007, upper bound, append 0851

Happy New Year!

Bob Jones

9.63972384402194105271145926236482315626728952582190645610957970056403564786337039072287316500879678883 ]]>

9.63972384402194105271145926236482315626728952582190

64561095797005640356478633703907228731650087967888

lower bound, append 2157; upper bound, append 3349

Convergence rate: about 4.51 basis functions per digit.

For the last year or so, I have been working on the Helmholtz problem as a hobby. I also did it while working on my physics PhD in the early 1990s. Anyway, yesterday, I applied my new eigenvalue bounding method to your famous L-shape. With a modest effort (15 minutes of CPU), I coaxed my computer to calculate the lowest eigenvalue to 82 correct digits (did I count them right?):

9.639723844021941052711459262364823156267289525821906456109579700564035647863370390

I understand that you have a particular fondness of this number. Based on the convergence rate, I can probably get about 250 digits with about 2 or 3 days of cpu time on my laptop. With a bigger computer, one can probably get a thousand digits with relative ease.

Most of my efforts have been on the regular pentagon. For example, with 2 days of cpu, I calculate the lowest Dirichlet eigenvalue within the unit-edged regular pentagon to 502 digits (at twice the convergence rate of the L-shape).

10.996427084559806648376217352434650641833359995632

37894580825875576367724334246467686955218722701415

76003245842094863020175628274679805263604950882189

07917111669971729458438577840231176346790319533010

11923430107693443760577351037965994015994422766013

66034288347043392716545024815696192116422461745439

59163018349746617500442725050308489108663578447310

92819681411540178947571787915793369719045227221102

00171725391320867384069050750319502551467771360357

70227635725742015466458835434076721938562837383080

92

I am fairly confident that these are all correct digits, but surprisingly, there are no other comparable results so there is no way to be certain yet. My hobby is actually getting a little out of control as I am also in the process of calculating the lowest 8000+ pentagon Dirichlet eigenvalues to at least 60 digits (an arbitrary, but now seemingly small number). I am doing one eigenvalue every five minutes on average. I can also do Neumann eigenvalues with equal difficulty and convergence rates.

I am a little embarrassed to say that I don’t use matlab for this, but I am sure matlab can be coaxed into reproducing these results with a good multi-precision toolbox and trustworthy Bessel functions. I actually used the pari pg calculator with the gnu mp library.

Of course, I am preparing a little paper to share the technique, but I thought I would like to share these results with you now. The technique is based on an observation I made in 1993 but never did anything with it until now, and a (now obviously clear) suggestion by Dr Barnett to use the non-analytic vertex to get exponential convergence.

Enjoy, and Happy New Year!

–

Bob Jones rsjones7 at yahoo dot com

I like your stories very much. Therefore please don’t be angry with my naive remarks. I am amazed by your very delicate suggestion set(gca…). It would be much more difficult to find without this hint, in comparison with the following small correction, which is obvious without any computer. Thus, I should have started with

“If n>1 is odd”

instead of

“If n is odd”,

because 1 is also an odd number. Please excuse me.

Remaining your sincere fan forever,

yours Vassili.