Your comment (a) reminds me: computing the spectral gap in the neighbourhood of the equilateral triangles, I recall observing that there was a striking difference (in terms of symmetries) between the variation of the second (ie, first non-zero) eigenvalue with triangle angles, and the variation of the third eigenvalue angles. Perhaps this is a well-known fact in spectral analysis? Here are (coarse grid) figures of what I mean:

Figure of the second Neumann eigenvalue as a function of triangle angles:

http://people.math.sfu.ca/~nigam/polymath-figures/zoomed-in/Lambda1big_20.jpg

Figure of the third Neumann eigenvalue as a function of triangle angles (sorry, figure title has a typo.)

http://people.math.sfu.ca/~nigam/polymath-figures/zoomed-in/Lambda2_20.jpg

I wondered if this observation could somehow be used in trying to establish (a)? I used this informally to help me decide how to refine grids in parameter space.

]]>(a) now that narrow triangles have been dealt with, the other troublesome region of the configuration space is the nearly equilateral triangles (the region near the point H in the above diagram), mainly due to the fact that the second and third Neumann eigenvalue are very close to each other here (so that our previous numerical strategies would break down). Abstractly we know that there is an open region around H where the hot spots conjecture holds, but it may now be worthwhile to try to see how explicitly large of a region we can make here. One nice thing here is that for the perfectly equilateral triangle, all the spectral statistics (e.g. the Neumann eigenvalues, the Dirichlet eigenvalues, the mixed Neumann-Dirichlet eigenvalues when some sides are Neumann and others are Dirichlet) are all explicitly computable through reflection arguments and Fourier analysis, and through Rayleigh quotient arguments one can then get approximate control on these same statistics for nearly equilateral triangles.

(b) Miyamoto’s method proceeds by starting with the second eigenfunction with a critical point at , and subtracting off of it some multiple of an explicit solution to the eigenfunction equation with a critical point at with a definite sign on the normal derivative, to obtain a new solution with a degenerate critical point and a definite sign on the normal derivative, which can then be used to lead to a contradiction. But it might be possible to get additional information on by finding other ways to line up and to create a degenerate critical point. For instance, instead of getting the critical points of and to line up, one could ask instead for the nodal lines of and to be tangent to each other, as this would also mean that some linear combination of have a degenerate critical point (cf. the Lagrange multiplier method). Similarly if has a critical point at some point other than the nodal line of . By playing around with various choices for (e.g. radial examples, whose nodal line is a circle, or cosine examples, whose nodal line is a line) this may provide some geometric constraints on the nodal line which could potentially be useful. (For instance we had a conjecture that the nodal line was convex, which might potentially be attackable by this sort of method.) In conjunction with point (b), we might try all this first for the equilateral triangle, where the nodal lines are explicitly computable and the geometry is all explicit and symmetric.

]]>Unfortunately, it’s a bit tricky to figure out which critical points are local extrema and which are saddles; the former occurs when is positive and the latter when it is negative, and the eigenfunction equation isn’t of much use in simplifying this constraint. (But on an edge, any positive local minimum or negative local maximum is necessarily a half-saddle point.)

]]>Consider an acute triangle ABC, and let be the inward normal from AB towards C. Then the directional derivative vanishes at AB and at C, as well as at any other critical point of , and solves the eigenfunction equation. Let be a nodal domain of , and consider the quantity

(*)

This is non-negative for all but at most one of the nodal domains (Lemma 1). On F, this quantity is equal to

(Corollary 3). The set is the union of various intervals in AB, BC, AC, whose endpoints are boundary critical points of . The portion on AB vanishes because vanishes here. Now consider an interval PQ on AC (with P closer to A than Q) of this set:

.

We may write , where is the tangential derivative in the direction from A to C, and is the angle . Using the Neumann condition (which implies ), the above integral becomes

.

On the other hand, from the eigenfunction equation we have , so this integrates to

As are critical points of , vanishes there, and so

We thus have a relatively simple formula for the quantity (*):

where are the intervals of AC on the boundary of F, and are the intervals of BC on the boundary of F.

If we consider a nodal domain on which is positive, then we have and . Unfortunately this doesn’t quite give a sign for (*) because can be both positive and negative. But perhaps we can control how the nodal curve of interacts with the nodal curve of using this sort of analysis.

Note also that if u changes sign on, say, AC, then it must have an even number of critical points on the interior of AC, since it is a local maximum on A and a local minimum on C or vice versa. So if the hypotheses of Corollary 6 break down, then there are quite a few critical points on the boundary and so there should be quite a few nodal domains for , which one can hopefully use to one’s advantage.

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