The hot spots conjecture

The hotspots conjecture can be expressed in simple English as:

Suppose a flat piece of metal, represented by a two-dimensional bounded connected domain, is given an initial heat distribution which then flows throughout the metal. Assuming the metal is insulated (i.e. no heat escapes from the piece of metal), then given enough time, the hottest point on the metal will lie on its boundary.

In mathematical terms, we consider a two-dimensional bounded connected domain D and let u(x,t) (the heat at point x at time t) satisfy the heat equation with Neumann boundary conditions. We then conjecture that

For sufficiently large t > 0, u(x,t) achieves its maximum on the boundary of D

This conjecture has been proven for some domains and proven to be false for others. In particular it has been proven to be true for obtuse and right triangles, but the case of an acute triangle remains open. The proposal is that we prove the Hot Spots conjecture for acute triangles!

Note: strictly speaking, the conjecture is only believed to hold for generic solutions to the heat equation. As such, the conjecture is then equivalent to the assertion that the generic eigenvectors of the second eigenvalue of the Laplacian attain their maximum on the boundary.

Threads

• Initial web page for proposal
• A more mathematical description of the problem
• Current research thread (June 3, 2012)
• Current discussion thread (June 9, 2012)

Possible approaches

Combinatorial approach

• A writeup of a combinatorial approach

Sturm comparison approach

How about approximating the eigenfunction by polynomials? The Neumann boundary conditions already imply that the corners are critical points. Perhaps starting with that observation, for sufficiently low degree, the geometry of the triangle implies that one of them must be a max/min. In an ideal world, a 2d version of the Sturm comparison theorem (if one exists) could then show that this feature of the polynomial approximation remains true for the actual eigenfunction.

In my understanding, Sturm-type results allow you to establish some qualitative property for solutions of an ODE (a 1d elliptic problem) provided that the desired property holds for a nearby solution of a nearby ODE. The property usually considered is the presence of a zero (node) of an eigenfunction in some interval. I think a similar result holds for critical points of eigenfunctions.

Say you could establish that the second eigenfunction of the Laplacian on a right angled triangle satisfies the desired maximum principle by showing that its only critical points are at the vertices (so that whatever the maximum is, it would have to be at one of these points). Then appealing to a comparison theorem might be able to show that the same property holds for almost-right acute triangles. Mind you, this is all speculation at this point…

Ah, so maybe a Sturm-type result can make rigorous the statement: “the second eigenfunction depends continuously on the domain”? By physical considerations, this statement ought to be true.

But then it seems this would only work to prove the conjecture for triangles close to right triangles?

As a side note, if such a continuity statement were to hold then the hot spots conjecture for acute triangles must be true by the following *non-math* proof: If it weren’t true then by continuity there would be an open set of angles where it failed to hold. But after simulating many triplets of angles the conjecture always holds. C’mon, What are the odds we missed that open set? :-)

It appears that Sturm’s classical work has far reaching generalizations, as described for instance in this monograph: Kurt Kreith, Oscillation Theory LNM-324, (Springer, 1973). In particular, Chapter 3 features some comparison theorems for solutions to elliptic equations.

Subdivision

Maybe ideas based on self-similarity of the whole triangle to its 4 pieces can help (i.e. modeling the whole triangle as its scaled copies + the heat contact). Then without going into the graph approximation (which looks fruitful anyway), one can see some properties.

One advantage in the graph case is that after dividing and dividing the triangles you get to the graph G_1 which is simply a tree of four nodes and there I think the theorem shouldn’t be too hard to prove. And from there there might be an argument by induction. In the continuous case, no matter how many times you subdivide the triangle, after zooming it it is still the same triangle.

On the other hand, in the continuous case, each of the sub-triangles is truly the same as the larger triangle. Does anyone know of good examples where self-similarity techniques are used to solve a problem?

In either case, there is the issue that while the biggest triangle has Neumann boundaries, the interior subtriangles have non-Neumann boundaries…

Something that is true and uses that the whole triangle is the union of the 4 congruent pieces is the following. From each eigenfunction of the triangle with eigenvalue lambda we can build another eigenfunction with eigenvalue 4 lambda by scaling the triangle to half length, and then using even reflection through the edges. Note that this eigenfunction will have an interior maximum since every point on the boundary has a corresponding point inside (on the boundary of the triangle in the middle). I’m skeptical this observation could be of any use.

Hmm… but, apart from the case of an equilateral triangle, when you divide a triangle into four sub-triangles, I don’t believe that the sub-triangles are reflections of one another.

Brownian motion

Suppose I start off a Brownian motion at time 0 at some point [math]\displaystyle{ x }[/math] in the triangle, and the Brownian motion reflects off the sides of the triangles. At time $latex t$ the probability density for the Brownian motion to be at location [math]\displaystyle{ y }[/math] is given by the solution [math]\displaystyle{ v(t,y) }[/math] of the heat equation with [math]\displaystyle{ v(0,y)=\delta(x-y) }[/math] and Neumann boundary conditions. So if I can show that the Brownian motion is eventually more likely to be in a corner (per unit area) than anywhere else, that will gives us the result.

Here's a way to approach it: Suppose I fix [math]\displaystyle{ x }[/math]. I ask how many ways are there for Brownian motion to get from [math]\displaystyle{ x }[/math] to [math]\displaystyle{ y }[/math] in time [math]\displaystyle{ t }[/math]. For [math]\displaystyle{ t }[/math] large, If [math]\displaystyle{ y }[/math] is in a corner I might imagine that there are more ways of Brownian motion to get to [math]\displaystyle{ y }[/math] than otherwise, because the Brownian motion has all the ways it would in free space, but also a ton of new ways that involve bouncing off the walls. Perhaps you can using some argument showing that there are more paths into some corners than anywhere else in the triangle. (Perhaps, using the Strong Markov Property and Coupling?)

This approach is also discussed at http://www.math.missouri.edu/~evanslc/Polymath/Polymath.pdf, and a similar approach was pursued in [BB1999].

Computational strategies

Consider the region in the plane [math]\displaystyle{ \mathcal{R} :=\{ (\alpha, \beta) \vert \frac{\pi}{2}\leq \alpha+\beta \lt \pi, 0\lt \alpha, \beta \leq \frac{\pi}{2} \} }[/math] Acute triangles will be characterized by angles [math]\displaystyle{ (\alpha, \beta, \pi-\alpha -\beta) }[/math], where [math]\displaystyle{ (\alpha, \beta) \in \mathcal{R}. }[/math]

So we can identify acute triangles with points in [math]\displaystyle{ \mathcal{R} }[/math] and ask: for each such triangle, where do the extrema of the 2nd Neumann eigenfunction of the Laplacian go?

Note that the line [math]\displaystyle{ \alpha = \beta }[/math] in [math]\displaystyle{ \mathcal{R} }[/math] gives us the isoceles triangles. Triangles corresponding to [math]\displaystyle{ (\alpha, \beta) }[/math] will have the same (upto similarity) eigenfunction as [math]\displaystyle{ (\beta, \alpha) }[/math] So we need to examine triangles corresponding to the region

[math]\displaystyle{ \mathcal{R}_1 :=\{ (\alpha, \beta) \vert \frac{\pi}{2}\leq \alpha+\beta \lt \pi, 0\lt \alpha\leq \beta \leq \frac{\pi}{2} \} }[/math]

The points [math]\displaystyle{ (\frac{\pi-\epsilon}{2}, \frac{\pi-\epsilon}{2}) }[/math] for [math]\displaystyle{ \epsilon\gt 0 }[/math] small have been examined in the section on nearly degenerate triangles.

If we can numerically demonstrate for a dense subset [math]\displaystyle{ \mathcal{J} }[/math] of [math]\displaystyle{ \mathcal{R}_1 }[/math] that the conjecture holds, then the suggestion is to use continuity arguments to conclude the result for all triangles in the region. More precisely, if on [math]\displaystyle{ \mathcal{J} }[/math] the extrema of the eigenfunctions are well away from other stationary points, and if the second derivatives are bounded away from zero, then the location of the extrema cannot shift dramatically as the angles in the triangle shift.

For each point in [math]\displaystyle{ \mathcal\lt J\gt , }[/math] we would construct an approximation [math]\displaystyle{ u_J }[/math] to the desired eigenfunction to high accuracy, and locate the extrema.

Special cases

Isosceles triangles

Say the triangle has vertices [math]\displaystyle{ (a,0) }[/math], [math]\displaystyle{ (-a,0) }[/math] and [math]\displaystyle{ (0,h) }[/math]. If [math]\displaystyle{ u(x,y) }[/math] is the second eigenfunction, then also is [math]\displaystyle{ u(-x,y) }[/math] and their sum [math]\displaystyle{ v(x,y)=u(x,y)+u(-x,y) }[/math]. The function v is even in x, so it is also a Neumann eigenvalue of the right triangle with vertices [math]\displaystyle{ (0,0) }[/math], [math]\displaystyle{ (0,h) }[/math] and [math]\displaystyle{ (a,0) }[/math]. According to the description of the problem the case of the right triangle has already been worked out, so this would reduce to that case unless v is identically zero.

If v is identically zero, it means that the second eigenfunction u was odd in x. In particular [math]\displaystyle{ u(0,y)=0 }[/math] for all y. The function u would be the first eigenfunction on the same right triangle as above but now with Dirichlet boundary condition on x=0 and Neumann in the other two edges. This function cannot change sign (otherwise |v| has less energy). I would expect its maximum to take place at (a,0) but I don't know how to prove it (maybe some rearrangement along lines?).

But how do you know that it is still the second eigenfunction(as opposed to some other eigen function)?

Note by symmetry that the space of second eigenfunctions splits into the symmetric eigenfunctions and the anti-symmetric ones. The symmetric eigenfunctions are also second eigenfunctions of the right-angled triangles on either side of the axis of symmetry, so in this case the conjecture follows from the right-angled case.

Equilateral triangles

It is convenient to work in the plane [math]\displaystyle{ \{ (x,y,z): x+y+z=0\} }[/math], and with the triangle with vertices (0,0,0), (1,0,-1), (1,-1,0). A function on this triangle with Neumann data can be extended symmetrically to the entire plane, in such a way that it becomes periodic with respect to translation by (2,-1,-1), (-1,2,-1), and (-1,-1,2). In particular, it has a Fourier series arising from functions of the form [math]\displaystyle{ \exp(2 \pi i (ax+by+cz)/3 ) }[/math] for integers a,b,c summing to 0; also, the Fourier coefficients need to be symmetric with respect to the reflections [math]\displaystyle{ (a,b,c) \mapsto (-a,-c,-b), (-c,-b,-a), (-b,-a,-c) }[/math]. To minimise the Rayleigh quotient, we see that (a,b,c) should only live among the six frequencies (1,0,-1), (1,-1,0), (0,1,-1), (-1,1,0), (0,-1,1), (-1,0,1), leading to the complex solution

[math]\displaystyle{ \exp( 2\pi i (y-z)/3 ) + \exp( 2\pi i (z-x)/3 ) + \exp( 2\pi i (x-y)/3 ) }[/math]

and its complex conjugate as basis vectors for the two-dimensional eigenspace. (See [McC2011], [McC2002] for more discussion.) All real second eigenfunctions are projections of this complex eigenfunction.

This complex eigenfunction appears to map the equilateral triangle to a slightly concave version of that triangle, so that the extreme points are always on the corners.

Has anyone done numerical work yet to estimate how sensitive the eigenvalue degeneracy lifting is to perturbations of a starting equilateral triangle? Any proof will have to address the issue that the third eigenvalue may be arbitrarily close to the second and has an eigenfunction with extremes at a different pair of corners.

Here is some preliminary data. As I am not sure how to simulate the true eigenfunctions of the triangle, the following is for the graph whose eigenvectors should roughly approximate the true eigenfunctions:

For the case that a=b=c=1 (Equilateral Triangle): HotSpotsAny(64,1,1,1,2) yields the corresponding eigenvalue -0.001070825047269 HotSpotsAny(64,1,1,1,3) yields the corresponding eigenvalue -0.001070825047269 HotSpotsAny(64,1,1,1,4) yields the corresponding eigenvalue -0.003211901603854

We see that indeed the eigenvalue -0.001070825047269 has multiplicity two.

Perturbing a slightly we have that for a=1.1, b = 1, c = 1 (Isosceles Triangle where the odd angle is larger than the other two): HotSpotsAny(64,1.1,1,1,2) yields the corresponding eigenvalue -0.001078552707489 HotSpotsAny(64,1.1,1,1,3) yields the corresponding eigenvalue -0.001131412869938

Whereas for a=.9, b = 1, c = 1 (Isosceles Triangle where the odd angle is smaller than the other two): HotSpotsAny(64,.9,1,1,2) yields the corresponding eigenvalue -0.001004876221957 HotSpotsAny(64,.9,1,1,3) yields the corresponding eigenvalue -0.001062028119964

In either case we see that the third eigenvalue is perturbed away from the second.

What strikes me as interesting is how different the outcome is in increasing a by 0.1 versus decreasing a by 0.1. In the former case, the second eigenvalue barely changes whereas in the latter case the second eigenvalue changes quite a bit. I imagine this ties into the heuristic “sharp corners insulate heat” — reducing a to .9 produces a sharper corner leading to more heat insulation and a much smaller (in absolute value) second eigenvalue, whereas increasing a to 1.1 makes that corner less sharp but the other two corners are still relatively sharp so the heat insulation isn’t affected as much. Just a guess though…

It looks like the degeneracy lifting might be linear. It’s the ratio of the second and third eigenvalues that matters more, which is close to the same in either case, and approximately half as far from 1 as the perturbation. Using (9/10, 1, 1) should be the same as using (1, 10/9, 10/9) and then scaling all eigenvalues by 90%.

Obtuse triangles

The obtuse case was settled in [BB1999] using a reflected Brownian motion approach. The “hot” and “cold” spots are located at the most distant vertices.

Right-angled triangles

The right-angled case of the conjecture is known, for instance it follows from the more general results in [AB2002] on lip domains (domains between two Lipschitz graphs of constant at most 1).

Thin sectors

Consider the sector [math]\displaystyle{ \Omega_0:=\{ (r \cos \theta, r \sin \theta): 0 \leq r \leq R; 0 \leq \theta \leq \alpha \} }[/math] with [math]\displaystyle{ \alpha }[/math] small. By separation of variables, one can use an eigenbasis of the form [math]\displaystyle{ u(r,\theta) = u(r) \cos \frac{\pi k \theta}{\alpha} }[/math] for [math]\displaystyle{ k=0,1,2,\ldots }[/math]. For any non-zero k, the smallest eigenvalue of that angular wave number is the minimiser of the Rayleigh quotient [math]\displaystyle{ \int_0^R (u_r^2 + \frac{\pi^2 k^2}{\alpha^2} u^2)\ r dr / \int_0^R u^2\ r dr }[/math]; the [math]\displaystyle{ k=0 }[/math] case is similar but with the additional constraint [math]\displaystyle{ \int_0^R u\ r dr = 0 }[/math]. This already reveals that the second eigenvalue will occur only at either k=0 or k=1, and [math]\displaystyle{ \alpha }[/math] small enough it can only occur [math]\displaystyle{ k=0 }[/math] (because the [math]\displaystyle{ k=1 }[/math] least eigenvalue blows up like [math]\displaystyle{ 1/\alpha^2 }[/math] as [math]\displaystyle{ \alpha \to 0 }[/math], while the [math]\displaystyle{ k=0 }[/math] eigenvalue is constant).

Once [math]\displaystyle{ \alpha }[/math] is small enough that we are in the k=0 regime (i.e. the second eigenfunction is radial), the role of [math]\displaystyle{ \alpha }[/math] is no longer relevant, and the eigenfunction equation becomes the Bessel equation [math]\displaystyle{ u_{rr} + \frac{1}{r} u_r + \lambda u = 0 }[/math], which has solutions [math]\displaystyle{ J_0(\sqrt{\lambda} r) }[/math] and [math]\displaystyle{ Y_0(\sqrt{\lambda} r) }[/math]. The [math]\displaystyle{ Y_0 }[/math] solution is too singular at the origin to actually be in the domain of the Neumann Laplacian, so the eigenfunctions are just [math]\displaystyle{ J_0(\sqrt{\lambda} r) }[/math], with [math]\displaystyle{ \sqrt{\lambda} R }[/math] being a zero of [math]\displaystyle{ J'_0 }[/math]. The second eigenfunction then occurs when [math]\displaystyle{ \sqrt{\lambda} R }[/math] is the first non-trivial zero of [math]\displaystyle{ J'_0 }[/math] ($latex 3.8317\ldots$, according to Wolfram alpha). This is a function with a maximum at the origin and a minimum at the circular arc, consistent with the hot spots conjecture.

Thin not-quite-sectors

From here, I was thinking along the following lines. The wedge can be turned into a nearly degenerate triangle [math]\displaystyle{ \Omega_1 }[/math] by joining the two rays by the chord connecting the points [math]\displaystyle{ (R,0) }[/math] and [math]\displaystyle{ (R,\epsilon) }[/math].

Now suppose there is a map F such that the thin side of the triangle can be written parametrically as [math]\displaystyle{ (R+F(\theta),\theta) }[/math]. We know precisely the 2nd Neumann eigenfunction and eigenvalue on the thin wedge [math]\displaystyle{ \Omega_0 }[/math]. We hypothesize this eigenfunction on the thin triangle is maximized at the same corner. We consider the situation on intermediate domains [math]\displaystyle{ \Omega_\delta }[/math] which are described as [math]\displaystyle{ \{(r(\theta),\theta)\vert 0\lt r\lt R+\delta F(\theta), 0\lt \theta\lt \epsilon\} }[/math] We can expand the Neumann eigenfunction in this intermediate region in terms of a series in powers of [math]\displaystyle{ \delta }[/math] as [math]\displaystyle{ u_\delta(r,\theta) = \sum_{n=0}^\infty \delta^n v_n(r,\theta) }[/math]. The first term in the sequence is just in terms of [math]\displaystyle{ J_0 }[/math]. I think one can compute the higher order terms easily, and that these will satisfy a related, but not identical, problem on the fixed wedge [math]\displaystyle{ \Omega_0 }[/math].

If the map [math]\displaystyle{ F }[/math] is smooth (analytic?) I think such a series must be in fact analytic; and that while [math]\displaystyle{ \delta=1 }[/math] may be outside the disk of analyticity, it is still in the domain of analyticity. This will enable us to get the desired lower bounds on the eigenvalue. In fact, I think the analytic continuation literature may have some pointers. I'll try to add references tomorrow.

However, if the mapping is not smooth this approach may fail. In this situation, one may yet be able to get a good lower bound on the true eigenvalue on the triangle. If [math]\displaystyle{ F }[/math] is not smooth, there will be a smooth map [math]\displaystyle{ G }[/math] which will not parametrize the chord, but some closeby curvilinear segment. The behaviour of the eigenfunction on the resulting domain can be obtained by the process described above.

I think it is important is to get away from the wedge (where the tangents to the arc are perpendicular to the sides of the sector, rendering it 'morally right-angled') in some smooth fashion to a geometry where the tangent to the curve [math]\displaystyle{ (R+\delta F(\theta), \theta) }[/math] is not perpendicular to the sides.

Nearly degenerate isosceles triangles

I think the case of a thin isosceles triangle [math]\displaystyle{ T_\varepsilon }[/math] with vertices [math]\displaystyle{ (0,0),(1,+\varepsilon), (1,-\varepsilon) }[/math] is also going to be OK. Let [math]\displaystyle{ u_\varepsilon }[/math] be a [math]\displaystyle{ L^2 }[/math] normalised second eigenfunction on [math]\displaystyle{ T_\varepsilon }[/math] with eigenvalue [math]\displaystyle{ \lambda_\varepsilon }[/math], thus [math]\displaystyle{ \int_{T_\varepsilon} u_\varepsilon^2 = 1 }[/math], and by integration by parts [math]\displaystyle{ \int_{T_\varepsilon} |\nabla u_\varepsilon|^2 = \lambda_\varepsilon }[/math] and [math]\displaystyle{ \int_{T_\varepsilon} |\nabla^2 u_\varepsilon|^2 = \lambda_\varepsilon^2 }[/math]. Ideally I would like to continue integrating by parts but I am having a bit of trouble dealing with the boundary terms; still the theory of regularity of Neumann eigenfunctions on a domain as nice as a triangle is presumably very well developed, so let me assume that we can get higher regularity bounds also.

Using a Poincare inequality, I think one can show that [math]\displaystyle{ \lambda_\varepsilon }[/math] is bounded above uniformly in the limit [math]\displaystyle{ \varepsilon \to 0 }[/math], and by perturbing the Bessel function example from the sector case I think one can also get a bound from below. So (after passing to a subsequence if necessary) we can assume that [math]\displaystyle{ \lambda_\varepsilon }[/math] converges to a limit [math]\displaystyle{ \lambda }[/math] as [math]\displaystyle{ \varepsilon \to 0 }[/math].

Now we look at the rescaled functions [math]\displaystyle{ v_\varepsilon(x,y) := \varepsilon^{1/2} u_\varepsilon(x,\varepsilon y) }[/math] on the triangle [math]\displaystyle{ T_1 }[/math]. These functions have unit norm [math]\displaystyle{ L^2 }[/math], and obey the [math]\displaystyle{ H^1 }[/math] bounds

[math]\displaystyle{ \int_{T_1} |\partial_x v_\varepsilon|^2 + \varepsilon^{-2} |\partial_y v_\varepsilon|^2 = O(1) }[/math]

and the [math]\displaystyle{ H^2 }[/math] bounds

[math]\displaystyle{ \int_{T_1} |\partial_{xx} v_\varepsilon|^2 + \varepsilon^{-2} |\partial_{xy} v_\varepsilon|^2 + \varepsilon^{-4} |\partial_y v_\varepsilon|^2 = O(1) }[/math].

This gives enough compactness to let [math]\displaystyle{ v_\varepsilon }[/math] converge strongly in H^1 and weakly in H^2 (again after passing to a subsequence) to some limit [math]\displaystyle{ v }[/math], which will be constant in the y direction, thus [math]\displaystyle{ v(x,y) = v(x) }[/math].

Now, the functions [math]\displaystyle{ v_\varepsilon }[/math] obey the PDE

[math]\displaystyle{ \partial_{xx} v_\varepsilon + \varepsilon^{-2} \partial_{yy} v_\varepsilon + \lambda_\varepsilon v_\varepsilon = 0 }[/math]

and the boundary conditions

[math]\displaystyle{ \partial_y v_\varepsilon = \pm \varepsilon^2 \partial_x v_\varepsilon }[/math]

when [math]\displaystyle{ y = \pm x }[/math]. We rewrite the PDE as

[math]\displaystyle{ \partial_{yy} v_\varepsilon = \varepsilon^2 (-\partial_{xx} v_\varepsilon - \lambda_\varepsilon v_\varepsilon). }[/math]

The RHS is [math]\displaystyle{ \varepsilon^2 (- v''(x) - \lambda v(x)) }[/math] plus errors which are [math]\displaystyle{ o(\varepsilon^2) }[/math] in some suitable function norm. Integrating this in y (from -x to x) and comparing with the boundary condition, we get that

[math]\displaystyle{ 2 \varepsilon^2 v'(x) + o(\varepsilon^2) = 2x \varepsilon^2 ( -v''(x) - \lambda v(x) ) + o(\varepsilon^2) }[/math]

and thus on taking limits we see that v obeys the Bessel equation

[math]\displaystyle{ v'' + v'/x + \lambda v = 0 }[/math].

The same sort of analysis as in the sector case then tells us that v has to be (a constant multiple of) [math]\displaystyle{ J_0(\sqrt{\lambda} x) }[/math] with [math]\displaystyle{ \sqrt{\lambda} }[/math] being the first root of J'_0, so it has a non-degenerate maximum at x=0 and non-degenerate minimum at x=1. Hopefully, there is enough uniform regularity on the [math]\displaystyle{ v_\varepsilon }[/math] to then conclude that [math]\displaystyle{ \partial_{xx} v_\varepsilon }[/math] is bounded away from zero near x=0 and x=1 for [math]\displaystyle{ \varepsilon }[/math] small enough, which (in conjunction with the convergence properties of [math]\displaystyle{ v_\varepsilon }[/math] to [math]\displaystyle{ v }[/math], and the Neumann boundary conditions) should ensure that [math]\displaystyle{ v_\varepsilon }[/math] still has a non-degenerate maximum at x=0 and a non-degenerate minimum at x=1. This would resolve the hot spots conjecture for all sufficiently degenerate isosceles triangles.

General domains

A counterexample to the hot spots conjecture for a domain with two holes was established in [BW1999]. A further counterexample in [BB2000] shows that both the maximum and minimum may be attained in the interior. On the other hand, the conjecture was established for convex domains with two axes of symmetry in [JN2000].

Regularity and stability theory

In [CZ1994] some bounds for the Neumann heat kernels [math]\displaystyle{ P_t(x,y) }[/math] on a general domain are given; in particular, one can bound this kernel by a multiple of the Euclidean heat kernel for t small enough. Using this bound, [BP2008] showed uniform stability of the second eigenfunction (in the uniform topology) and eigenvalue with respect to uniform perturbations of the domain.

In a sector [math]\displaystyle{ \{ (r \cos \theta, r \sin \theta): r \gt 0; 0 \leq \theta \leq \alpha \} }[/math], solutions to the eigenfunction equation [math]\displaystyle{ \Delta u = -\lambda u }[/math] with Neumann data can be computed using separation of variables in polar coordinates as

[math]\displaystyle{ u(r,\theta) = \sum_{k=0}^\infty u^{(k)}(r,\theta) }[/math]

where

[math]\displaystyle{ u^{(k)}(r,\theta) = c_k J_{\pi k / \alpha}(\sqrt{\lambda} r) \cos( \frac{\pi k}{\alpha} \theta ), }[/math]

[math]\displaystyle{ c_k }[/math] are real coefficients, and [math]\displaystyle{ J_\beta }[/math] are the usual Bessel functions

[math]\displaystyle{ J_\beta(r) = \frac{1}{2^{\beta-1} \Gamma(\beta+\frac{1}{2}) \sqrt{\pi}} r^\beta \int_0^1 (1-t^2)^{\beta-1/2} \cos(rt)\ dt }[/math].

One consequence of this particular representation of the Bessel functions is that we observe the pointwise bounds

[math]\displaystyle{ c_\beta \leq J_\beta(r)/r^\beta \leq C c_\beta }[/math]

for all [math]\displaystyle{ 0\lt r\lt 1 }[/math] (say) and [math]\displaystyle{ \beta \geq 0 }[/math], and some constants [math]\displaystyle{ c_\beta, C }[/math]. Among other things, this gives pointwise bounds of the form

[math]\displaystyle{ u^{(k)}(r,\theta) = O( (r/r_0)^{\pi k/\alpha} (\frac{1}{\alpha} \int_0^\alpha |u(r_0,\theta)|^2\ d\theta)^{1/2} ) }[/math]

whenever [math]\displaystyle{ \sqrt{\lambda}r, \sqrt{\lambda} r_0 \lt 1 }[/math]. If u is locally in L^2, this gives bounds near the corner of the sector of the form

[math]\displaystyle{ u^{(k)}(r,\theta) = O( (r/r_0)^{\pi k/\alpha} ) }[/math]

uniformly in k. Among other things, this implies for acute sectors [math]\displaystyle{ \alpha\lt \pi/2 }[/math] that [math]\displaystyle{ \nabla^2 u }[/math] is bounded (and even Holder continuous) all the way up to the corner. In particular, Neumann eigenfunctions on acute triangles are [math]\displaystyle{ C^{2,\varepsilon} }[/math] all the way up to the boundary.

For the thin triangle case, it would be good to get bounds of this type that remain uniform (perhaps after suitable normalisation) as the triangle degenerates.

Possible counterexamples

If we could find an heat distribution on a right triangle such that the hot spot is never on the vertices or the hypotenuse or the short side then could we combine two such heat distributions symmetrically in an isosceles acute triangle and get an isosceles acute triangle such that the hot spots are always on the interior?

An issue is that if you reflect a right triangle and its eigenfunction the eigenfunction you get for the larger triangle might not be the *second* eigenfunction.

Miscellaneous remarks

If the sides of the triangle have lengths A, B, and C, currently you are using edge weights a=A, b=B, and c=C. This solves a different heat equation, when you pass to the limit, than the one you want. The assumption that each small triangle is at a uniform temperature gives an extra boost to the overall heat conduction rate in those directions along which the small triangles are longer, so in the limit you are modeling heat conduction in a medium with anisotropic thermal properties. It is just as though you started with a material of extremely high thermal conductivity and then sliced it in three different directions, with three different spacings, to insert thin strips of insulating material of a constant thickness.

Based on some edge cases, I suspect that the correct formulas for edge weights to model an isotropic material in the limit are as follows: [math]\displaystyle{ a = 1/(-A^2+B^2+C^2) }[/math], [math]\displaystyle{ b=1/(A^2-B^2+C^2) }[/math], [math]\displaystyle{ c=1/(A^2+B^2-C^2) }[/math]. Interestingly enough, these formulas are only defined (with positive weights) in the case of acute triangles, which suggests that this approach, if it works, may not provide an independent proof of the known cases.

Courant's nodal line theorem tells us that the nodal curve [math]\displaystyle{ \{u=0\} }[/math] of a second eigenfunction can cut the domain into at most two components (otherwise, with three or more components, one could reduce the Rayleigh quotient while still having mean zero). On the other hand, the eigenfunction must change sign on the boundary (if it were positive on the boundary, the eigenfunction equation would be impossible to satisfy at the minimum on the boundary), and so the nodal curve cannot be closed, but must instead cut across the domain.

Nadirashvili [Na1986] has shown that the multiplicity of the second eigenvalue is at most 2; a simpler proof was given in [BB1999].

Software

• A MATLAB program used to generate and display the eigenvectors of the graphs G_n. The input format is HotSpotsAny(n,a,b,c,e), where a,b,c are the edge weights, n is the number of rows in the graph, and e determines which eigenvector of the graph Laplacian is displayed. (Note that in the proposal write up the graph G_n has 2^n rows so if you wanted to simulate the Fiedler vector for G_6 with edge weights a=1,b=2,c=3 then you would type HotSpotsAny(64,1,2,3,2)). For large n, the eigenvectors of the graph Laplacian should approximate the eigenfunctions of the Neumann Laplacian of the corresponding triangle… so the program can be used to roughly simulate the true eigenfunctions as well.

Numerics

Using a finite element method, here are the eigenfunctions of the Neumann Laplacian for a couple of acute triangles:

• http://people.math.sfu.ca/~nigam/polymath-figures/isoceles1.jpg
• http://people.math.sfu.ca/~nigam/polymath-figures/scaleneeig.jpg

The domain is subdivided into a finite number N of smaller triangles. On each, I assumed the eigenfunction can be represented by a linear polynomial. Across interfaces, continuity is enforced. I used Matlab, and implemented a first order conforming finite element method. In other words, I obtained an approximation from a finite-dimensional subspace consisting of piece-wise linear polynomials. The approximation is found by considering the eigenvalue problem recast in variational form. This strategy reduces the eigenvalue question to that of finding the second eigenfunction of a finite-dimensional matrix, which in turn is done using an iterative method. As N becomes large, my approximations should converge to the true desired eigenfunction. This follows from standard arguments in numerical analysis.

Additionally, I've computed a couple of approximate solutions to the heat equation in acute triangles. The initial condition is chosen to have an interior 'bump', and I wanted to see where this bump moved. Again, I've plotted the contour lines of the solutions as well, and one can see the bump both smoothing out, and migrating to the sharper corners:

• http://people.math.sfu.ca/~nigam/polymath-figures/isoceles.avi
• http://people.math.sfu.ca/~nigam/polymath-figures/isoceles2.avi
• http://people.math.sfu.ca/~nigam/polymath-figures/scalene.avi

I think in the neighbourhood of the corner with interior angle [math]\displaystyle{ \frac{\pi}{\alpha} }[/math], the asymptotic behaviour of the (nonconstant) eigenfunctions should be of the form [math]\displaystyle{ r^\alpha \cos(\alpha \theta) + o(r^{\alpha}) }[/math], where r is the radial distance from the corner.

Next, if one had an unbounded wedge of interior angle [math]\displaystyle{ \frac{\pi}{\alpha} }[/math], the (non-constant) eigenfunctions of the Neumann Laplacian would be given in terms of [math]\displaystyle{ P_n(r,\theta) = J_{\alpha, n} (\sqrt(\lambda) r) cos(\alpha n \theta), n=0,1,2\ldots }[/math]. The [math]\displaystyle{ J_{\alpha,n} }[/math] are Bessel functions of the first kind. The spectrum for this problem is continuous. When we consider the Neumann problem on a bounded wedge, the spectrum becomes discrete.

This makes me think approximating the second eigenfunction of the Neumann Laplacian in terms of a linear combination of such [math]\displaystyle{ P_n(r,\theta) }[/math] could be illuminating.

This idea is not new, and a numerical strategy along these lines for the Dirichlet problem was described in [FHM1967]. Betcke and Trefethen have a nice recent paper [BT2005] on an improved variant of this 'method of particular solutions'.

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