# The polymath blog

## June 15, 2012

### Polymath7 research threads 2: the Hot Spots Conjecture

Filed under: hot spots,research — Terence Tao @ 9:48 pm

The previous research thread for the Polymath7Hot Spots Conjecture” project has once again become quite full, so it is again time to roll it over and summarise the progress so far.

Firstly, we can update the map of parameter space from the previous thread to incorporate all the recent progress (including some that has not quite yet been completed):

This map reflects the following new progress:

1. We now have (or will soon have) a rigorous proof of the simplicity of the second Neumann eigenvalue for all non-equilateral acute triangles (this is the dotted region), thus finishing off the first part of the hot spots conjecture in all cases.  The main idea here is to combine upper bounds on the second eigenvalue $\lambda_2$ (obtained by carefully choosing trial functions for the Rayleigh quotient), with lower bounds on the sum $\lambda_2+\lambda_3$ of the second and third eigenvalues, obtained by using a variety of lower bounds coming from reference triangles such as the equilateral or isosceles right triangle.  This writeup contains a treatment of those triangles close to the equilateral triangle, and it is expected that the other cases can be handled similarly.
2. For super-equilateral triangles (the yellow edges) it is now known that the extreme points of the second eigenfunction occur at the vertices of the base, by cutting the triangle in half to obtain a mixed Dirichlet-Neumann eigenvalue problem, and then using the synchronous Brownian motion coupling method of Banuelos and Burdzy to show that certain monotonicity properties of solutions to the heat equation are preserved.  This fact can also be established via a vector-valued maximum principle.  Details are on the wiki.
3. Using stability of eigenfunctions and eigenvalues with respect to small perturbations (at least when there is a spectral gap), one can extend the known results for right-angled and non-equilateral triangles to small perturbations of these triangles (the orange region).  For instance, the stability results of Banuelos and Pang already give control of perturbed eigenfunctions in the uniform norm; since for right-angled triangles and non-equilateral triangles, the extrema only occur at vertices, and from Bessel expansion and uniform C^2 bounds we know that for any perturbed eigenfunction, the vertices will still be local extrema at least (with a uniform lower bound on the region in which they are extremisers), we conclude that the global extrema will still only occur at vertices for perturbations.
4. Some variant of this argument should also work for perturbations of the equilateral triangle (the dark blue region).  The idea here is that the second eigenfunction of a perturbed equilateral triangle should still be close (in, say, the uniform norm) to some second eigenfunction of the equilateral triangle.  Understanding the behaviour of eigenfunctions nearly equilateral triangles more precisely seems to be a useful short-term goal to pursue next in this project.

But there is also some progress that is not easily representable on the above map.  It appears that the nodal line $\{u=0\}$ of the second eigenfunction $u$ may play a key role.  By using reflection arguments and known comparison inequalities between Dirichlet and Neumann eigenvalues, it was shown that the nodal line cannot hit the same edge twice, and thus must straddle two distinct edges (or a vertex and an opposing edge).   (The argument is sketched on the wiki.) If we can show some convexity of the nodal line, this should imply that the vertex straddled by the nodal line is a global extremum by the coupled Brownian motion arguments, and the only extremum on this side of the nodal line, leaving only the other side of the nodal line (with two vertices rather than one) to consider.

We’re now also getting some numerical data on eigenvalues, eigenfunctions, and the spectral gap.  The spectral gap looks reasonably large once one is away from the degenerate triangles and the equilateral triangle, which bodes well for an attempt to resolve the conjecture for acute angled triangles by rigorous numerics and perturbation theory.  The eigenfunctions also look reassuringly monotone in various directions, which suggests perhaps some conjectures to make  in this regard (e.g. are eigenfunctions always monotone along the direction parallel to the longest side?).

This isn’t a complete summary of the discussion thus far – other participants are encouraged to summarise anything else that happened that bears repeating here.

1. […] an active discussion in the last week or so, with almost 200 comments across two threads (and a third thread freshly opened up just now).  While the problem is still not completely solved, I feel optimistic that it should […]

Pingback by Updates on the two polymath projects « What’s new — June 15, 2012 @ 10:22 pm

2. Thanks, this summary is very helpful!

I’ve been pondering two suggestions on the previous thread, which concerned some sort of perturbation argument and the heat kernels. This got me to thinking about the Hankel function of the first kind, which is the fundamental solution
of the Helmholtz equation with the outgoing wave condition at infinity.

At least from the numerical data, it seems the second Neumann eigenvalue $\lambda_{\epsilon}$ varies smoothly with perturbations $\epsilon$ in angle.

At an eigenvalue of the interior Neumann Laplacian, we can think of the eigenfunction as solving the Helmholtz equation $(\Delta + \lambda_\epsilon) u=0$ with zero Neumann data. One could presumably write this $u$ in terms of a layer potential using the fundamental solution of the Helmholtz equation with wave number $\sqrt{\lambda_{\epsilon}}$, or some clever combination of the Calderon operators. I think the regularity of such operators on Lipschitz domains is documented, eg. in the books by Nedelec or Hsiao/Wendland or MacLean. I don’t know if the behaviour of these integral operators as $\epsilon$ varies is also characterized, but it most likely is.

Using these integral operators, one can reformulate the boundary value problem in terms of the solution of an integral equation, $\mathcal{T}_\epsilon \phi=F$ where $\mathcal{T}_\epsilon.$ Suppose $\mathcal{V}_\epsilon \phi$ then is the desired Neumann Laplace eigenfunction.

The boundary trace of $u$ will be given in terms of this density, and we know this trace is at least continuous. Can one say something about the composition of the trace operator with $\mathcal{V}_\epsilon \circ \mathcal{T}_\epsilon^{-1}$, as epsilon varies?

Is there a clever reformulation of the Hot Spot conjecture in terms of these layer densities?

Comment by Nilima Nigam — June 15, 2012 @ 10:52 pm

3. Some numerically computed 2nd Neumann eigenfunctions are in this public directory. Titles on the graphs give two of the angles (scaled by pi).

http://www.math.sfu.ca/~nigam/polymath-figures/EigenFunctions/

Comment by Nilima Nigam — June 15, 2012 @ 11:22 pm

• These are nice. Is it correct that alpha corresponds to the lower left corner and beta corresponds to the lower right corner? And I notice the triangles vary in shape from picture to picture… are the triangles in the figure actually drawn such that they have the correct angles? If so that definitely gives the full picture!

Also, maybe these pictures can shed light on the conjecture “are eigenfunctions always monotone along the direction parallel to the longest side?” proposed in the summary. It is a bit hard for me to make out exactly, but I think in triangles of the type “one hot two cold corners” (well often there is only one globally coldest corner and the other is only locally coldest) the level curves of $u$ roughly form concentric circles at each of the verticies.

If this is the case I believe such triangles disprove the conjecture: If the conjecture were true then, letting $\xi$ be a vector pointing in the direction of the longest side, you could always head in/against the direction $\xi$ to achieve ever hotter/colder values. I think this precludes three local extrema at the corners.

(I am not explaining this very well but basically I have in mind the sort of argument used in Linear Programming to prove that if you have a linear function and you want to maximize it over a convex polygon the maximum occurs at a vertex of the polygon. The argument goes that you head in the direction of the gradient of the function and eventually you ram against the wall, keep going, and end up in a corner.)

Comment by Chris Evans — June 16, 2012 @ 8:25 am

• Hi Chris,

– I’ve put low-resolution thumbnails of these runs on one page for easier viewing, you should be able to click on any to get the detailed plot (again somewhat low-resolution, .png seems to compress awkwardly)

http://people.math.sfu.ca/~nigam/polymath-figures/eigenpics.html

– Each triangle here corresponds to a different choice of parameter angles.

– On each plot, I’ve marked the location of the max(|u|) by a green star. You’ll notice there is only one such point in each triangle plotted, and it occurs on the vertex.

– Yes. If the nodal line doesn’t go through the vertex, the level curves of u, if you zoom in enough to the corners, are nearly concentric circles. This goes all the way back to some comment in an older thread: locally near a corner, the eigenfunction looks like that of a wedge, ie, like a P_n(r,\theta) . Near a corner I think it’s a combination of Bessel functions in the distance from the corner, and the angular variable.

– I just noticed the titles on my figures (generated by outputting some run-time data) are not scaled correctly. I can re-run the code to change the titles.

Comment by Nilima Nigam — June 16, 2012 @ 2:37 pm

• Ah, you’re right, the conjecture is false. Around any acute vertex, if the eigenfunction takes the value $c$ at that vertex, then the eigenfunction has the asymptotic expansion $c J_0(\sqrt{\lambda} r) + o(r^2) = c - cr^2/4 + o(r^2)$ where r is the distance to that vertex, so the level sets locally are indeed close to circles. (This was concealed in the super-equilateral isosceles case because the non-extreme vertex had $c=0$.) That’s a pity – now I don’t know if there are any plausible monotonicity conjectures one can make that would be valid for all acute triangles.

Comment by Terence Tao — June 16, 2012 @ 5:15 pm

4. Hi Terry,

I discussed with Chris this afternoon and I’m still worried about the PDE proof of Corollary 4. Sorry that I have come back to this point many times.

The point again is about the reflection part. It is too good to be true. The solution of the heat equation is OK when you do even reflection. However, the gradient of the new solution in the reflected region may not stay in the convex sector ${\rm S}$ anymore.

Let me give a clear example. Assume that we have $u=u(x,y):\mathbb R \times [0,\infty) \to \mathbb R$ with $\dfrac{\partial u}{\partial y}(x,0)=0$. Let
$w$ be the even reflection of $u$, i.e. $w(x,y)=u(x,y),\ y\ge 0$ and $w(x,y)=u(x,-y),\ y <0$. Then $\nabla w(x,y)=(u_x,-u_y)(x,-y),\ y<0$. So $\nabla w(x,y)$ may not be in ${\rm S}$ for $y<0$ anymore.

Comment by Hung Tran — June 16, 2012 @ 12:41 am

• Yes, the gradient of $\nabla u$ may leave S in the reflected region, but this is not relevant to the argument, as one only needs the gradient to lie in S (or in $S_{\varepsilon(t+1)}$) in the original domain. (In particular, when one discusses things like “the first time the gradient touches the boundary of $S_{\varepsilon(t+1)}$, it is understood that one is working in the original domain.)

Comment by Terence Tao — June 16, 2012 @ 2:59 am

• Then how could you conclude that $\Delta \nabla w(x,0,t)$ points inward or tangentially where $(x,0,t)$ is the first point that $\nabla u(x,0,t)$ touches the boundary of $S_{(1+\epsilon)t}$? As now, in the neighborhood of $(x,0,t)$, $\nabla w$ takes values not only in $S$ or $S_{(1+\epsilon)t}$.

Comment by Hung Tran — June 16, 2012 @ 3:38 am

• Ah, right. If one is working on a boundary point $(x,0,t)$, then as you say, in a neighbourhood of that point, $\nabla w$ will live either in $S_{\varepsilon(t+1)}$ or its reflection, but both of these sets (assuming for sake of concreteness that S is oriented in the first quadrant, with one side on the x axis) lie to the right of where $\nabla w(x,0,t)$ will be, which in this case is $(-\varepsilon(t+1),0)$. So $\Delta \nabla w(x,0,t)$ will have a non-negative first component and so will point inwards of $S_{\varepsilon(t+1)}$ (or of its reflection).

Alternatively, one can argue entirely inside the original domain without reflection using directional derivatives (and the Neumann boundary condition) to get the inward/tangential nature of the Laplacian.

Comment by Terence Tao — June 16, 2012 @ 4:05 am

• Thank you. I will double-check this carefully. Just a quick comment that we want to have $\Delta \nabla w(x,0,t)$ points inwards or tangentially to $S_{\epsilon(t+1)}$ (NOT its reflection) in order to obtain the contradiction. So in this example, we need both of the components are non-negative.

Anyway, I’ll move on to consider other directions.

Comment by Hung Tran — June 16, 2012 @ 5:22 am

• At the point $\nabla w(x,0,t)$, the outward normal to $S_{\varepsilon(t+1)}$ is parallel to the boundary, so that $S_{\varepsilon(t+1)}$ and its reflection have the same normal (and thus the same notion of “inwards” and “tangentially”).

Comment by Terence Tao — June 16, 2012 @ 5:48 am

• This is an important point. In fact as far as I can see it is the only place in the proof where it is used that the axes of S are oriented exactly along the directions of DB and AB.

I tried reproducing the argument for a triangle with Neumann conditions in the three edges. Setting $S$ to be the sector corresponding to the exterior of one of the angles in a triangle, I tried to check whether if when the initial gradients are all in $S$, they would stay in $S$. Note that this would imply some monotonocity along the three edges.

The same argument goes through as long as these two edges form an obtuse or right angle. It is exactly the same proof, but it fails for acute angles.

The way it fails for acute angles is funny. It is exactly because the prolongation of the edges of $S$ do not intersect the boundary $S_{\varepsilon(t+1)}$ perpendicularly. It is necessary that these two intersect perpendicularly to make Terry’s reflection argument work.

I guess that in light of thread 3 we should not expect a monotonicity result like this one to hold for acute triangles.

Comment by Luis S — June 18, 2012 @ 9:12 pm

5. I’ve been reading the latest summary, and I’m not sure what point 3. above and the orange region say about the hot spots conjecture for almost 30-60-90 triangles, in a neighbourhood in parameter space of point $C$ above .
I followed the link in the summary to (one of) the Banuelos and Pang papers. The abstract of the 2008 paper I glanced at mentioned a “snowflake type boundary” ; so, I can’t say I understand how the Banuelos and Pang paper from 2008 relates to the stability of the second eigenvalue & eigenfunction for the Neumann Laplacian in the orange region of the parameter space. Thanks for another illuminating summary.

Comment by meditationatae — June 16, 2012 @ 2:03 am

• Theorem 1.3 of the Banuelos-Pang paper gives continuity of the second eigenvalue, and Theorem 1.5 gives stability of the second eigenfunction, under the assumption of Gaussian bounds on heat kernels (which are certainly true for acute angled triangles) together with simplicity of the second value. (Banuelos-Pang developed these results with the intention of applying them to snowflake domains, but the results are more general than this.)

Comment by Terence Tao — June 16, 2012 @ 3:01 am

• Ok. Many thanks. That clears up my questions about the Banuelos-Pang paper.
David Bernier

Comment by meditationatae — June 16, 2012 @ 3:17 am

6. Here is a basic way to get stability of the second eigenfunction. Consider an acute triangle $\Omega$ with second eigenvalue $\lambda_2$ and third eigenvalue $\lambda_3 > \lambda_2$, thus the Rayleigh quotient $\int_\Omega |\nabla u|^2 / \int_\Omega |u|^2$ for mean zero $u$ is lower bounded by $\lambda_2$, with equality when $u$ is a scalar multiple of the second eigenfunction $u_2$, and lower bounded by $\lambda_3$ when $u$ is orthogonal to $u_2$.

Now consider a perturbed triangle $\Omega' = B \Omega$ for some linear transformation $B = 1+O(\varepsilon)$ close to the identity. The second eigenvalue of the perturbed triangle is the minimiser of the perturbed Rayleigh quotient $\int_\Omega |\nabla u|^2 + \varepsilon \nabla u \cdot A \nabla u / \int_\Omega |u|^2$, where $A = O(1)$ is the self-adjoint matrix such that $B^{-1} (B^{-1})^T = 1 + \varepsilon A$.

Consider the second eigenfunction $u'_2$ of the perturbed problem (viewed on the reference triangle $\Omega$ rather than on $\Omega'$), normalised to have L^2 norm 1. We can split it as $u'_2 = \alpha u_2 + \beta v$, where $v$ has unit norm and is orthogonal to $u_2$ (and also mean zero), and $\alpha,\beta$ are scalars with $\alpha^2+\beta^2=1$. We will show that $\beta=O(\frac{\varepsilon \lambda^{3/2}_2}{\lambda_3^{1/2} (\lambda_3-\lambda_2)})$, which shows that $u'_2$ is within $O(\varepsilon)$ in $L^2$ norm of (a scalar multiple of) the original eigenfunction $u_2$ when there is a large spectral gap.

Since $u'_2$ minimises the perturbed Rayleigh quotient, one can compare it against the original eigenfunction $u_2$ to obtain the inequality

$\int_\Omega |\nabla u'_2|^2 + \varepsilon \nabla u'_2 \cdot A \nabla u'_2 \leq \int_\Omega |\nabla u_2|^2 + \varepsilon \nabla u_2 \cdot A \nabla u_2$.

Expanding out $u'_2 = \alpha u_2 + \beta v$, noting from integration by parts that $\nabla u_2$ is orthogonal to $\nabla v$, and using $\alpha^2 = 1 - \beta^2$, then after some algebra one ends up at

$\beta^2 \int_\Omega |\nabla v|^2 + 2 \varepsilon \alpha \beta \int_\Omega \nabla u_2 \cdot A\nabla v + \varepsilon \beta^2 \int_\Omega \nabla v \cdot A \nabla v \leq \beta^2 \lambda_2 + \beta^2 \varepsilon \int_\Omega \nabla u_2 \cdot A \nabla u_2.$

After some Cauchy-Schwarz (bounding $\alpha$ by 1 and recalling that $\|\nabla u_2\|^2 = \lambda_2$) this becomes

$\beta^2 \| \nabla v \|_2^2 + O( \varepsilon \beta \lambda_2^{1/2} \|\nabla v \|_2 ) + O( \varepsilon \beta^2 \|\nabla v \|_2^2 ) \leq \beta^2 \lambda_2 + O( \beta^2 \varepsilon \lambda_2 )$

Now, one can bound the $\lambda_2$ factors on the RHS by $\frac{\lambda_2}{\lambda_3} \|\nabla v \|_2^2$. If one does this and rearranges for a while, one eventually ends up with the bound

$\beta \|\nabla v\|_2 = O( \frac{\varepsilon \lambda_2^{3/2}}{\lambda_3-\lambda_2} )$

(assuming $\varepsilon$ small enough); since $\|\nabla v \|_2 \geq \lambda_3^{1/2}$, this gives the bound on $\beta$.

This type of perturbation analysis measures the perturbation in L^2 and in H^1 with quite explicit bounds. What one really wants though is uniform (or even better, C^1 or C^2) bounds, as this gives enough control to start tracking where the extrema go, but I think one may be able to get that from the L^2 and H^1 bounds by using the Bessel function expansion formalism (but the bounds get worse when doing so, unfortunately). Still, one may be able to work everything out explicitly for, say, perturbations of the 30-60-90 triangle, so that we can get an explicit neighbourhood of that triangle for which one can establish the conjecture.

Comment by Terence Tao — June 16, 2012 @ 3:53 am

• One thing I realized about the above analysis is that it also works when there is multiplicity of the second eigenvalue, and in particular when $\Omega$ is the equilateral triangle. In this case, $u_2$ becomes the projection of $u'_2$ to the second eigenspace (normalized to have norm one). So this should imply that for a perturbed equilateral triangle, the second eigenfunction is close to some second eigenfunction of the equilateral triangle in H^1 norm at least. Integration by parts also gives uniform H^3 bounds on this eigenfunction, so by some interpolation (Gagliardo-Nirenberg type inequalities) should show that the second eigenfunction of the perturbed eigenfunction is close in C^0 (and probably also C^1) to a second eigenfunction of the equilateral triangle. I think I can show that for all second eigenfunctions of the equilateral triangle, the extrema only occur at the vertices (and are uniformly bounded away from the extrema once one is a bounded distance away from the vertices), so this should indeed establish the claim stated previously that the extrema can only occur at the vertices for perturbed equilateral triangles.

Comment by Terence Tao — June 16, 2012 @ 5:01 pm

• This would be great, and I think one can get some numerical evidence for the desired uniform bounds. I can compute the L^2-projection of an eigenfunction onto the eigenfunction of a nearby triangle. In other words, I can compute $\beta v$ in your notation, and try to get a feel for the sup-norm bounds on derivatives, for perturbed eigenfunctions in the neighbourhood of a few different ‘reference’ triangles. I’ll fix one side of the triangles to be 1.

I had a quick question. In your argument, the asymptotic constant in the bound for $\beta$ presumably depends on the measure of the original acute triangle, $\Omega.$ Let’s call it $C_\Omega$
In your series of calculations above, do you recall whether the constant scaled as the area or the square root of the area? Once again, I ask because this will help me decide how small to choose the neighbourhoods of the reference triangles- we don’t want $C_\Omega \epsilon$ to get big.

[Fixed latex. Incidentally, there was an error in the latex instructions on this blog (now fixed) – one needs to enclose latex commands in “$latex” and “$” rather than “<em>” and “</em>”. -T.]

Comment by Nilima Nigam — June 16, 2012 @ 7:23 pm

• The implied constants in the above analysis are actually dimensionless – they don’t depend on the area of $\Omega$. Instead, they depend on the operator norm of the perturbing matrix $A$.

Comment by Terence Tao — June 16, 2012 @ 7:29 pm

• So this constant is uniform for all acute triangles? In other words, if I perturb *any* triangle $\Omega$ by an affine map which is close to the identity, the same bound $\beta(\int_\Omega \|\nabla u\|^2 )^\frac{1}{2} = O(\frac{\epsilon\lambda_{2,\Omega}^{3/2}}{\lambda_{3,\Omega}-\lambda_{2,\Omega}}) \approx C(\frac{\epsilon\lambda_{2,\Omega}^{3/2}}{\lambda_{3,\Omega}-\lambda_{2,\Omega}}) + \,\,l.o.t.$ should hold? This would be good news for purposes of coding.

Comment by Nilima Nigam — June 16, 2012 @ 9:22 pm

• I wrote a more careful computation at http://michaelnielsen.org/polymath1/index.php?title=Stability_of_eigenfunctions . The upshot is that if one takes a perturbation $B\Omega$ of the triangle $\Omega$, and considers a second eigenfunction of $B\Omega$ which (after rescaling back to $\Omega$) takes the form $u_2+v$ for some second eigenfunction $u_2$ of $\Omega$ of norm 1, and some v orthogonal to u_2, then

$\displaystyle \| \nabla v \|_2^2 \leq \frac{(\kappa^2-1) \lambda_{2,\Omega}^{1/2}}{1 - \kappa^2 \frac{\lambda_{2,\Omega}}{\lambda_{3,\Omega}}}$

provided that the denominator is positive, where $\kappa = \|B\|_{op} \|B^{-1}\|_{op}$ is the condition number of B, and $\lambda_{3,\Omega}$ is first Neumann eigenvalue that is strictly greater than $\lambda_{2,\Omega}$. (This is slightly different from what I said before; the factor of $\lambda_2^{3/2}$ in my previous post should have instead been a $\lambda_2^{1/2} \lambda_3$.)

Comment by Terence Tao — June 17, 2012 @ 12:02 am

7. Here is a Mathematica notebook with calculations for the proof of simplicity. This is not the full proof. It reduces the problem to solving a set linear inequalities. There are 2 cases. I am also attaching a plot with acute triangles as yellow area and red/blue as the cases. Clearly everything is covered. Now I just need to find the cleanest way to handle those inequalities.

http://pages.uoregon.edu/siudeja/TrigInt.m (this one is needed to run the notebook)
http://pages.uoregon.edu/siudeja/simple.nb (first line needs to be edited)
http://pages.uoregon.edu/siudeja/simple_plot.pdf

The first case uses 3 reference triangles and rotated symmetric eigenfunction of equilateral for upper bound. The second case uses 2 reference triangles and Cheng’s bound.

Comment by Bartlomiej Siudeja — June 16, 2012 @ 6:50 pm

8. I’ve been thinking about how one would try to use rigorous numerics to verify the hot spots conjecture for acute-angled triangles. In previous discussion, we proposed covering the parameter space with a mesh of reference triangles, getting rigorous bounds on the eigenfunctions and eigenvalues of these reference triangles, and then using stability analysis to then cover the remainder of parameter space. This would be tricky though, in part because we would need rigorous bounds for the finite element schemes for each of the reference triangles.

But another possibility is to only use for the reference triangles the triangles for which explicit formulae for the eigenfunctions and eigenvalues are available, namely the equilateral triangles, the isosceles right-angled triangles, the 30-60-90 triangles, and the thin sector (a proxy for the infinitely thin isosceles triangle). In other words, to use the red dots in the above diagram as references.

The trouble is that if one does a naive perturbation analysis, one could only hope to verify the hot spots conjecture in small neighborhoods of each red dot, which would not be enough to cover the entire parameter space. But I think, in principle at least, that one has a way to amplify the perturbation analysis to get much larger neighborhoods of each reference triangle, by using not just the lowest eigenvalue and eigenspace of the reference triangle, but the first k eigenvalues and first k eigenspaces for some medium-sized k (e.g. k = 100). This would necessitate the use of 100 x 100 linear algebra (e.g. minimizing a quadratic form on 100 variables) but this seems well within the capability of rigorous numerics. As long as the strength of one’s perturbation analysis grows at a reasonable rate with k, this may well be enough to rigorously cover the entire parameter space.

For instance, consider perturbing off of the isosceles right triangle $\Omega$ with vertices (0,0), (0,1), (1,1). The eigenfunctions here are quite explicit (because one can reflect the triangle indefinitely to fill out the plane periodically with periods (0,2), (2,0), and then use Fourier series): every pair of integers (k,l) gives rise to an eigenfunction

$\cos(\pi(kx+ly)) + \cos(\pi(kx-ly)) + \cos(\pi(lx+ky)) + \cos(\pi(lx-ky))$

with eigenvalue $\pi^2 (k^2+l^2)$. Thus, for instance, the second eigenfunction is (up to constants) $\cos(\pi x)+\cos(\pi y)$ with eigenvalue $\pi^2$, the next eigenfunction is $\cos(\pi x) \cos(\pi y)$ with eigenvalue $2 \pi^2$, the next eigenfunction is $\cos(2\pi x ) + \cos(2\pi y)$ with eigenvalue $4\pi^2$, and so forth (the next eigenvalue, for instance, is $5\pi^2$).

As discussed in the wiki, if one wants to find the second eigenfunction for a perturbed triangle $B\Omega$, this amounts to minimizing the quadratic form

$Q(u,u) := \int_\Omega M \nabla u \cdot \nabla u$

amongst mean zero functions of norm 1, where $M := (B^{-1}) (B^{-1})^T$. This is a quadratic minimization problem on an infinite-dimensional Hilbert space. However, one can restrict u to a finite-dimensional subspace, namely the space generated by the first k eigenvalues of the reference triangle. For instance, one could consider u which are a linear combination of $\cos(\pi x) + \cos(\pi y)$, $\cos(\pi x) \cos(\pi y)$, and $\cos(2\pi x) + \cos(2\pi y)$. This minimization problem could be computed with rigorous numerics without difficulty.

What I believe to be true is that there is some rigorous stability inequalities that relate the minimizers of the finite-dimensional variational problem with the minimizers of the infinite-dimensional problem, with the error bounds improving in k (roughly speaking, I expect the H^1 norm of the error to decay like $\lambda_k^{-1/2}$ or even $\lambda_k^{-1}$). By increasing k, one thus gets quite good rigorous control on where the minimizers are.

Unfortunately, the H^1 norm is not good enough to control where the extrema go: one would prefer to have control in a norm such as C^0 or (even better) C^1. But it should be possible to use Bessel function expansions and the eigenfunction equation to convert H^1 control of an eigenfunction to C^0 or C^1 control, though this could get a bit messy with regards to the explicit constants in the estimates. But in principle, this provides a completely rigorous way to control the eigenfunction in arbitrarily large neighborhoods of reference triangles in parameter space, and so for some sufficiently large but finite k (e.g. k=100) one may be able to cover everything rigorously.

Of course, it would be preferable to have a more conceptual way to prove hot spots than brute force numerical calculation (since, among other things, a conceptual argument would be easier to generalize to other domains than triangles), but I think the numerical route is certainly a viable option to pursue if we don’t come up with a promising conceptual attack on the problem.

Comment by Terence Tao — June 17, 2012 @ 10:26 pm

• I’ve been pondering something similar. Even for a conceptual approach, there’s a duality between the problem of locating the extrema of eigenfunctions of
$- \nabla(M\nabla w)= \lambda w, \,\,M\nabla w \cdot n=0$ on a right-angled isoceles triangle and that of the hot spot conjecture. I’m of the view that a successful conceptual attack will deal with all the cases (except, maybe, the equilateral one), in one go.

As far as numerics go, here was my thinking:

A) one can get very good approximations of both the initial part of the spectrum and eigenfunctions of a positive definite operator on a right-angled isoceles triangle, using non-polynomial functions in a spectral approach.

B) you had a perturbation analysis in thread 6 from one triangle to another, nearby one;

C) one can pull back both triangles in your analysis to a right-angled isoceles one, and do the perturbation analysis there. The constants include, explicitly, the information about the pullback. I am doing this to check if how big an \epsilon I was allowed to pick depended on the shape of the triangles. I’ll post a link to this (naive) write-up to the wiki soon.

For any numerical attack, I think it’s extremely important to record how the eigenvalues are being located, or how the minimizer is being found.

For (A) I use approximations $u_N(x,y) = \sum_{i=1}^N c_ip_i(x,y)$ where the basis functions are not piecewise polynomials. I’m writing up both a Fourier spectral method and a method based on Bessel functions. This is in the debugging stage, and I will wait for better results before I upload details.

Comment by Nilima Nigam — June 17, 2012 @ 11:07 pm

• what about a partition-of-unity approach, where one takes 4 patches (3 which include the corners, and one for the center)? we’ve got great control of what happens near the corners.

Comment by Nilima Nigam — June 17, 2012 @ 11:10 pm

• I apologize about not describing what I meant by a spectral approach. On a square, I know that $\phi_{mn}(x,y) = \exp(imx)\exp(iny)$ are orthogonal. So one writes the approximation as $u_N(x,y) := \sum_{m,n}^N c_{nm} \phi_{mn}(x,y)$, writes a discrete variant of the eigenproblem, and solves it for the unknown coefficients. On clearly has to take the basis functions with the correct symmetry, while working on the triangle. These basis functions don’t satisfy the prescribed boundary conditions, so the discrete formulation has to be carefully done.

Comment by Nilima Nigam — June 17, 2012 @ 11:21 pm

• One advantage of the quadratic form minimization approach is that one does not need to explicitly keep track of boundary conditions, as the minimizer to the quadratic form will automatically obey the required Neumann conditions. It’s true though that it isn’t absolutely necessary to use the reference eigenfunctions as the basis for the approximation, and that other bases could be better (for instance, in the spirit of what was done to show simplicity of eigenvalues, one could take an overdetermined basis consisting of eigenfunctions pulled back from multiple reference triangles). This would of course make the rigorous stability and perturbation analysis more complicated, though, so some there is some tradeoff here. I guess we could do some numerical experiments to probe the efficiency of various types of bases for approximate eigenfunctions.

Comment by Terence Tao — June 18, 2012 @ 12:04 am

• The most efficient set of approximations functions I’ve found (in terms of fewest non-zero coefficients) are: some Bessel functions centered at the three corners, and then I threw in some trigonometric polynomials. These aren’t orthogonal, and I’d have to show the discrete problem was consistent with the original one.

Comment by Nilima Nigam — June 18, 2012 @ 12:46 am

• All explicitly known eigenfunctions for triangles are trigonometric polynomials (equilateral, right isosceles and 30-60-90. So a basis made of linear perturbations of these is very reasonable. However, transplantation of known cases gives extremely ugly Rayleigh quotients on arbitrary triangles. Trig functions do not mix well. But, this can be done. That is how I am proving simplicity.

For nearly degenerate triangles Bessel functions must be best (Cheng’s bound is obtained this way), but linear functions, varying along long side, adjusted to have average 0 are really good too. After all sides are almost perpendicular and variation must happen along the long side. For simplicity of nearly degenerate triangles I am using linear function (it gives slightly better results for not-so-nearly-degenerate cases.

Comment by Bartlomiej Siudeja — June 18, 2012 @ 3:58 pm

• This reminds me of a small remark I wanted to make on the nearly degenerate case. Instead of approximating by a thin sector, one can instead rescale to the right-angled triangle $\Omega$ with corners (0,0), (1,0), (0,1), in which case one is trying to minimise a Rayleigh quotient the numerator of which looks something like $\int_\Omega |\partial_x u|^2 + A |\partial_y u|^2\ dx dy$ for some large A (this is the formula for a thin right-angled triangle, in general there will also be a mixed term involving $\partial_x u \partial_y u$ which I will ignore for this discussion). One can then map this to the unit square S with corners (0,0), (1,0), (0,1), (1,1) via the transformation $(x,y) \mapsto (x,y/x)$, in which case the numerator now becomes $\int_0^1 \int_0^1 x |\partial_x u|^2 + A x^{-1} |\partial_y u|^2\ dx dy$ and the denominator $\int_0^1 \int_0^1 x |u|^2\ dx dy$. Morally, when A is large the second term in this numerator forces u to be nearly indepent of y. On the other hand, to minimise $\int_0^1 x |\partial_x u|^2\ dx / \int_0^1 x |u|^2\ dx$ assuming mean zero one easily computes that the minimiser comes from our old friend, the Bessel function $J_0(x / j_1)$ with eigenvalue $j_1^2$. It is then tempting to analyse this quotient using a product basis consisting of tensor products of Bessel functions in the x variable and cosines in the y variable, i.e. $J_0(x / j_{1,k} ) \cos( \pi l y)$ for various integers k,l. Among other things this should give an alternate proof of the hot spots conjecture in the nearly degenerate case, though I didn’t work through the details.

Comment by Terence Tao — June 18, 2012 @ 5:25 pm

• I like this line of thinking.

Using a basis of the form $J_0(x/j_{1,k}) cos(\pi ly)$ is pretty much what I’m coding up, and is the heart of the method of particular solutions (Fox-Henrici-Moler, Betcke-Trefethen and later Barnett-Betcke).

Comment by Nilima Nigam — June 18, 2012 @ 6:05 pm

• After doing some theoretical calculations, I’m now coming around to the opinion that bases of trigonometric polynomials in a reference triangle are not as accurate as one might hope for, unless the triangle being analysed is very close to the reference triangle. The problem lies in the boundary conditions: an eigenfunction in a triangle is going to be smooth except at the boundary, but when one transplants it to the reference triangle (say, the 45-45-90 triangle for sake of argument) and then reflects it into a periodic function (in preparation for taking Fourier series) it will develop a sawtooth-like singularity at the edges of the reference triangle due to the slightly skew nature of the boundary conditions (distorted Neumann conditions). As a consequence, the Fourier coefficients decay somewhat slowly: a back-of-the-envelope calculation suggests to me that even if one uses all the trig polynomials of frequency up to some threshold K (so, one is using about K^2 different trig polynomials in the basis), the residual error in approximating the sawtooth-type eigenfunction in this basis is about $O(\varepsilon K^{-3/2})$ in L^2 norm, and $O(\varepsilon K^{-1/2})$ in H^1 norm, if the triangle is within $O(\varepsilon)$ of the reference triangle. This is pretty lousy accuracy, in practice it suggests that unless $\varepsilon$ is very small, one would need to take K in the thousands (i.e. work with a basis of a million or so plane waves) to get usable accuracy.

So it may be smarter after all to try to use a basis that matches the boundary conditions better, even if this makes the orthogonality or the explicit form of the basis more complicated.

Comment by Terence Tao — June 20, 2012 @ 8:18 pm

• yes, the trig functions are pretty inefficient here. Here’s what works (without proof right now, so I won’t post results): take some bessel functions around the corners, add in some trig functions. take a linear combination. evaluate the normal derivatives at points on the boundary, and enforce that the linear combination does not vanish at randomly selected points in the interior. set up an overdetermined system (ie, take more points than unknowns), find the QR decomposition of the matrix. this gives you orthogonal columns, which correspond to a discrete orthogonal basis. use this to solve for the eigenvalues and eigenfunctions. it’s a tweak of what Trefethen and Betcke did, and the results are not as spectacular. but I can get results with 30 unknowns that I can’t using a basis of 200 purely trig functions.

Comment by Nilima Nigam — June 20, 2012 @ 8:52 pm

• Hmm, I can see that this scheme would give good numerical results, but obtaining rigorous bounds on the accuracy of the numerical eigenfunctions obtained in this manner could be tricky.

One possibility is to compute the residual $\| -\Delta u - \lambda u \|_{L^2(\Omega)}^2$ of the numerically computed eigenvalue $\lambda$ and eigenfunction $u$. Assuming $u$ is smooth enough (e.g. $C^3$ except at the vertices, where the third derivative blows up slower than $1/|x|$) and obeys the Neumann boundary conditions exactly, we have the expansion

$\| -\Delta u - \lambda u \|_{L^2(\Omega)}^2 = \sum_k |\lambda_k - \lambda|^2 |\langle u, u_k \rangle|^2$

where $u_k$ are the true orthonormal eigenbasis of the Neumann Laplacian. If $\lambda < \lambda_3$, this implies in particular that

$\| -\Delta u - \lambda u \|_{L^2(\Omega)}^2 \geq \frac{(\lambda_3 - \lambda)^2}{\lambda_3^2} \sum_{k \geq 3} \lambda_k^2 |\langle u, u_k \rangle|^2$

$= \frac{(\lambda_3 - \lambda)^2}{\lambda_3^2} \| \nabla^2( u - c u_2 ) \|_{L^2(\Omega)}^2$

where $c = \langle u, u_2 \rangle$. Thus, an $L^2$ bound on the residual, together with a sufficiently good rigorous lower bound on the third eigenvalue (in particular producing a gap between this eigenvalue and the numerical second eigenvalue $\lambda$), gives an $\dot H^2$ bound on the error:

$\| \nabla^2(u - c u_2) \|_{L^2(\Omega)} \leq \frac{\lambda_3}{\lambda_3 - \lambda} \| -\Delta u -\lambda u \|_{L^2(\Omega)}$.

One can also get control on lower order terms from the Poincare inequality, thus

$\| \nabla^i(u - c u_2) \|_{L^2(\Omega)} \leq \frac{\lambda_3^{1-i/2}}{\lambda_3 - \lambda} \| -\Delta u -\lambda u \|_{L^2(\Omega)}$

for i=0,1,2.

This way, we don't have to do any analysis at all of the numerical scheme that comes up with the numerical eigenvalue and eigenfunction $\lambda, u$, so long as we can rigorously compute the residual. Note from Sobolev embedding that H^2 controls C^0, so in principle this is enough control on the error to start locating extrema…

Comment by Terence Tao — June 20, 2012 @ 11:11 pm

• I agree, the numerical analysis is gnarly, Betcke and Barnett did it for the Dirichlet case on analytic domains for the method of particular solutions http://arxiv.org/pdf/0708.3533.pdf

For the finite element calculations and the more recent attacks, I’ve been keeping track of the numerical residual $\int_\Omega |\nabla u|^2 -\lambda|u|^2$. I can calculate the I’ve also been keeping track of the (numerical) spectral gap.
The devil is in the detail: for the finite element calculations, I can compute the integrals analytically (piecewise polynomials on triangles). For the new bessel+trig calculations, I can’t compute the integrals analytically. So for the latter, the residuals are computed through high-accuracy quadrature, but not analytically.

Comment by Nilima Nigam — June 20, 2012 @ 11:57 pm

9. The second eigenfunction of the Neumann Laplacian for the
unit interval $[0, 1]$ is
$u(x) = \cos \left(2\pi x\right)$ .
I’ve been thinking of trying to prove that the
equivalent of the hot spots conjecture for this
eigenfunction holds, i.e. that the max and min
are attained on the boundary of $[0, 1]$ .
The method of approach I had involved assuming
the supremum of the absolute value of $u(x)$
is attained only at a point $p$ in the
interior of $[0, 1]$, and using a
non-constant stretching/compressing map
$\phi$ from $[0, 1]$ to itself
to construct $x \mapsto u\left(\phi \left( x \right) \right)$ ,
followed by compressing/stretching the values of $u(x)$
so as to preserve the $L^2$-norm of the resulting
function, while hopefully decreasing the Rayleigh quotient.

for the unit interval’s second eigenfunction, but triangles
are another matter altogether.

I’d be grateful if someone could tell me how to
access and use the LaTex sandbox area for Polymath
blogs.
David

[The sandbox is at https://polymathprojects.org/how-to-use-latex-in-comments/ – T.]

Comment by meditationatae — June 18, 2012 @ 5:19 am

• I am not sure that I follow…

For the interval $[0,1]$ with Neumann boundary, the second eigenfunction is (I am pretty sure), as you say, $u_2(x)=\cos(2\pi x)$. Knowing that, the fact that $u_2(x)$ attains its max and min at the boundary points of $[0,1]$ just follows from understanding the cosine function (e.g. sketching the graph of $u_2$). Maybe I am missing something (your edit below makes me think you instead wanted to suggest something about the third eigenfunction perhaps)?

But in any case, for the interval $[0,1]$ I believe that explicit expressions for the eigenfunctions are known (though of course coming up with an understanding of them that doesn’t rely on the explicit expressions might be useful in that it could be generalized to two dimensional triangles).

Comment by Chris Evans — June 18, 2012 @ 6:01 am

• I read some lecture notes online about Neumann boundary conditions, and as I recall, for the unit interval one is led to an ODE for the eigenvalues and eigenfunctions of the Neumann Laplacian. So, the solution via this route is easy. From Terry’s and others’ comments, explicit expressions for the first non-trivial eigenfunction of the Neumann Laplacian for triangular regions are only known for a tiny region of the parameter space, e.g. equilateral triangles, the 45-45-90 degrees triangle and quite a bit about isosceles triangles with a very pointy angle, using the known exact solutions for sectors. I haven’t used LaTex much recently, and that’s why I didn’t write all that much in my post. In any case, I can describe it more in words. For the unit interval, the $u(x)$ on the unit interval is sufficiently well-behaved candidate function for being the second non-trivial eigenfunction, and I also assume that $u$ has mean zero on the unit interval, as well as having derivative zero at $x=0$ and $x=1$. If needed, I also assume that $u$ has two distinct zeros in the unit interval, just as $x \mapsto \cos \left(2\pi x\right)$ does. Without loss of generality, we can assume that $u(x)$ is strictly positive near 0 and near 1, and that $u(x)$ is strictly negative in the open interval from $r_{1}$ to $r_2$, these last two real numbers denoting the first and second zeros of $u(x)$ on the unit interval. What I’m aiming at is a proof by contradiction of the analog of hot spots for the eigenfunction associated to the second non-trivial eigenvalue, where the domain of interest is simply the unit interval, rather than triangular regions. As you say, the solutions are explicitly known here and computing them is quite easy. So I’m working without assuming the knowledge of these explicit cosine solutions, relying more on physical intuition and the method/technique of Rayleigh quotients. In my previous post, I called $p$ the point in the unit interval where the maximum of the absolute value of the candidate function $u(x)$ is attained. So $r_{1} < p 1$ results in smaller eigenvalues for the domain $[0, K]$. That’s one of the key things I keep in mind, because it reflects how the Rayleigh quotient changes when the graph of any well-behaved function $u(x)$ is dilated or expanded by a factor of $K$ along the x-axis ; we can assume that nothing happens along the y-axis. Introducing $\phi(x)$ which maps the unit interval to itself is a means to an end. The immediate objective is to get something like a perturbation of $u(x)$, except that I usually think of pertubation as adding, say, $\epsilon g(x)$ to $u(x)$, for some well-chosen $g(x)$ and a small $\epsilon$. Suppose $q$ is used to denote the point in the unit interval such that $q \mapsto p$ or in other word, $p = \phi(q)$. Then surely we want to have $0 < \phi'(q) 1$ and $\phi'(1)>1$. Locally near 0 and near 1, the Rayleigh quotient will be increased. We can deduce the value of the squared second derivative of the composite function $x \mapsto u\left(\phi \left( x \right) \right)$ at 0 and 1 from the assumption that $u$ is an eigenfunction. The only counterfactual assumption used (apart from benign requirements such as $u$ having two zeros on the unit interval), is that there is a point $p$ in the interior of the unit interval such that $|u(p)| > |u(0)|$ and $|u(p)| > |u(1)|$ . I think I’d like to keep constant the $L^2$-norm of the composite function. At $(q, \phi(q))$, this requires decreasing further modification of the composite function by multiplying by $sqrt(\phi'(q))$, which while locally near $q$ will maintain the contribution to the $L^2$-norm of the modified composite of $\phi$ with $u$, will locally decrease the square of the second derivative of $u$. The hope is that, since by assumption $|u(p)|$ is greater than $|u(0)|$ and $|u(1)|$, in total, the integral of the square of the modified composite of $u$ with $\phi$ will decrease, contradicting the Rayleigh principle. As I recall, the Rayleigh quotient method here requires that a candidate function for the second non-trivial eigenfunction both have mean zero, and be orthogonal to the first non-trivial eigenfunction. Also, the various $\phi'(x)$ values as x ranges over $[0, 1]$ must have a mean of 1, by the fundamental theorem of calculus. The end result hoped for is that the modified function, while still satisfying the orthogonality requirements of the Rayleigh-Ritz method, have a strictly smaller Rayleigh quotient, where one proves this using a specially chosen $\phi$. Say u~ is the modified function obtained starting with u. Then u~ is in the admisible space, but has strictly lower Rayleigh quotient than u [Contradiction].
So, under the benign assumptions about two nodes for the second non-trivial eigenfunction, that eigenfunction attains its min and max at 0 or 1.

Comment by meditationatae — June 18, 2012 @ 8:21 am

• A couple of thoughts:

-Apriori I would guess that the second eigenfunction for the domain $[0,1]$ with Neumann boundary conditions has *one* zero not two. That is, I would guess that an analogue of the Nodal Line Theorem would hold and as such it would have two nodal domains. Indeed, the function $\cos(2\pi x)$ has one zero (at $\frac{\pi}{2}$) in the interval $[0,1]$ (Edit: Clearly I am wrong here :-). See my reply below).

-If I understand correctly, your idea is to argue by contradiction: If the second eigenfunction $u_2(x)$ achieves an extremum in the interior of $[0,1]$, then by considering an appropriately chosen $\phi:[0,1]\to[0,1]$, we can show that $u_2(\phi(x))$ in fact has a smaller Raleigh quotient than $u_2(\phi(x))$ (while still having mean zero), contradicting the fact that $u_2(x)$ minimizes this quotient. This is definitely an interesting idea… in principle it could work for a triangle, but we would need to find a function $\phi:D\to D$ with all these properties.

Comment by Chris Evans — June 18, 2012 @ 9:19 pm

• I’ve been reading the mathematical parts of a technical report by Moo K. Chung and others (on applied mathematics and medical imaging) called: “Hot Spots Conjecture and Its Application to Modeling Tubular Structures” . For eigenfunctions of the Neumann Laplacian, their numbering in English starts with “first eigenfunction”, followed by “second eigenfunction”, where these are denoted respectively by $\psi_{ 0}$ and $\psi_{1}$ .

It’s clear from the presentation in that report that $\psi_{ 0}$ is a non-zero constant function, whatever the domain. They state that the nodal set of the $ith$ eigenfunction $\psi_{ i-1}$ divides the region or domain into at most $i$ sign domains.

For the unit interval, I believe that $ith$ eigenfunction $\psi_{ i-1}$ is given by the formula $\psi_{i-1} \left( x\right) = \cos \left(\left(i-1\right)\pi x\right)$. I find that $\cos \left(2\pi x\right)$ has roots at ${ 1 \over 4}$ and ${ 3 \over 4}$ .

With respect to your second thought, a proof by contradiction is indeed what I have in mind. The laplacian of the composition of functions represented by $x \mapsto u_{2}\left(\phi \left( x \right) \right)$ seems to me to depend not only on $\phi \prime \left( x\right)$, but also on $\phi \prime \prime \left( x\right)$. So an affine transformation map $\phi$, where $\phi \prime$ is everywhere constant, and allowing mappings from a domain $D$ to a possible different domain $D \prime$ for illustration, transforms the laplacian in a simple way (noting that $\phi \prime \prime$ is then identically zero.) In case $\phi$ is not affine, I think the relation between the laplacian of $u_{2}$ composed with $\phi$ and the laplacian of $u_{2}$ isn’t easy to understand on an intuitive level …

Comment by meditationatae — June 19, 2012 @ 6:00 am

• Ah, you are correct… $\cos(2\pi x)$ does indeed have two zeros at the places you state (I was thinking of $\cos(\pi x)$ which has its zero at $\frac{1}{2}$… and not $\frac{\pi}{2}$ as I wrote). And I see I was also confused about the numbering… sorry about that!

So you are considering the more general hot-spots problem then (showing that all of the eigenfunctions attain their extrema on the boundary)? Also, if you plan to argue by Raleigh quotients, you should only need to worry about first derivatives, no?

I played around a bit trying to make such an argument (for the first/second eigenfunction… i.e. for the simpler hot spots conjecture) but didn’t have any luck. An issue, as I see it, is that any argument has to account for the fact that there are examples of (non simply connected) domains which have the extrema of the first/second eigenfunction in the interior.

So any construction of $\phi$ would have to fail for that counter example domain. Then again, as $\phi$ in some sense “moves points around”, it may somehow be tied to the topology of the space… so maybe the construction of $\phi$ would depend on the domain being simply connected? But this is just wild conjecture… I have nothing specific in mind.

One other idea/conjecture that came up when I was playing around with this: Perhaps it is possible to show that “there cannot be an interior maximum without an interior minimum”. This is a weaker statement than the full hot-spots conjecture, but if it were true and if we also knew that the nodal line, which straddles a corner, does so in a convex way then we would be done: In the convex sub-domain the extremum would lie at the corner, therefore in the other sub-domain the extremum must also lie on the boundary.

Comment by Chris Evans — June 19, 2012 @ 7:08 am

• When I thought of trying to argue for “hot spots” (for the unit interval) by contradiction, it just so happens that the first test case that came to mind was the eigenfunction for the Neumann laplacian $u_{2} \left( x \right) = \cos \left( 2 \pi x \right)$ , which has a minimum of $- 1$ at $x = 1/2$ and maxima of $1$ at $x = 0$ and $x = 1$ . We would have to speak about “maximum of the absolute value of a non-trivial eigenfunction” rather than “maximum of a non-trivial eigenfunction” and “minimum of a non-trivial eigenfunction” , because for $u_{2} \left( x \right) = \cos \left( 2 \pi x \right)$, the minimum of $u_{2}$ is attained at $x = 1/2$ only.
When arguing by Rayleigh quotients, including a domain transformation such as the $\phi$ in an earlier comment, if for simplicity’s sake we stick to $u_{2}$ as the name of a hypothetical counter-example of “hot spots” or a variant of “hot spots” which refers to “points which extremize the absolute value of a non-trivial eigenfunction”, the way I was thinking of it, we would need to compute the second derivative (“laplacian”) of $u_{2} \circ \phi$. From a mental computation using the “chain rule”, the first derivative of the composition of two functions is a product of two terms, one of which in our case would be $\phi '$ . Then differentiating again, the “product rule” will yield a term where $\phi ''$ appears. If that is the case, does it matter? I think it does, in the sense that it makes “tweaking” the numerator of the Rayleigh quotient harder to do (the numerator involves the laplacian of a test function) . I might be overlooking something, so that I might not need to worry about $\phi ''$ … You mention topology of the domain, for instance the property of “simple connectedness” of a domain , as possibly relevant/”decisive” for the “hot spots” conjecture to hold. I think it’s an interesting topic, how topology of the domain affects the truth or falsity of the “basic” conjecture, i.e. the part relating to extrema of $u_{2}$.

Comment by meditationatae — June 19, 2012 @ 4:39 pm

• Yes, the interval shows that the “generalized hot-spots conjecture” is not true for the interval. That is, after multiplying by $(-1)$ if necessary, the second/third eigenfunction will have its hottest point in the interior. The same is true for a circular region I believe. But, as you say, the conjecture could be modified to consider the absolute value of $u$.

Honestly, I haven’t much thought about the generalized hot-spots conjecture… the simple one (about the first/second Neumann eigenfunction) is hard enough!

Concerning the Raleigh quotient, I believe that for the interval it is given by $\mathcal{R}\left[u\right]=\frac{\int_0^1 (u_x)^2\ dx}{\int_0^1 u^2\ dx}$ and so only first derivatives are needed, no?

Comment by Chris Evans — June 19, 2012 @ 5:49 pm

• Ah, you’re right. I just read about the energy functional and variational (calculus of variations) characterization of the first/second Neumann eigenfunction, and the expression to be minimized is just what you wrote in the last paragraph. So, under a transformation of domains, it seems like we don’t need to worry about the second derivative of the $\phi$ map from the domain to itself …

Comment by meditationatae — June 20, 2012 @ 6:46 am

10. Correction to my latest comment: I wrote “second eigenfunction” , and following Terry Tao’s summary, “third eigenfunction” for the eigenfunction linked to $\lambda_{3}$ seems like a better term.

Comment by meditationatae — June 18, 2012 @ 5:32 am

11. I am slowly finishing the proof of simplicity. It may require a computer for some of the longest calculations. Especially certain large polynomial inequalities with 2 variables are not easy to handle. I have implemented an algorithm for proving these. I am attaching a Mathematica notebook and a pdf version of it. Algorithm can be found in Section 5 of http://www.ams.org/mathscinet-getitem?mr=MR2779073.

There are still 2 cases missing, but general idea should be clear. I have included comments in the notebook, so it should be relatively easy to read.

Comment by Bartlomiej Siudeja — June 18, 2012 @ 10:18 pm

• I have updated the files under above links. The proof is now complete. Eigenvalues are simple.

I have started working on the best possible lower bound for spectral gap for a given triangle. For now, I implemented an upper bound based on many eigenfunctions from known cases. There are lists of eigenfunctions of known cases sorted according to eigenvalues. There is also an upper bound based on a chosen number of eigenfunctions for each known case. There is certainly room for improvement, but the upper bound is already very accurate. There is however an issue of optimizing a linear combination of eigenfunctions. If I take 5 or 10 eigenfunctions per known case I get great results. For 15 per case, bound is actually worse, which means there are local minima.

Just for comparison. For a triangle with vertices (0,0), (1,0), (0.501,0.85) (slightly off from equilateral) I am getting 17.62110831277495 as the upper bound. Using second order FEM with 262144 triangles I am getting 17.62110794. Nilima, could you compare this with your results?

Comment by Bartlomiej Siudeja — June 19, 2012 @ 6:26 pm

• sure, I can run this case. Do you mean the 2nd computed eigenvalue is 17.62110794? or the spectral gap?

Comment by Nilima Nigam — June 19, 2012 @ 6:39 pm

• The second eigenvalue. The third is 18.1342622359. So the gap is about 0.5 (numerically).

Comment by Bartlomiej Siudeja — June 19, 2012 @ 6:41 pm

• I just ran it with P1 elements, 80358 nodes. Ev = 17.6212. If I use 320844 nodes, the ev = 17.62110632.
You’re using a higher-order element, so I think our results are comparable.

Comment by Nilima Nigam — June 19, 2012 @ 7:29 pm

• I have updated the files again. There is a lower bound for the gap at the end of the notebook. For any triangle I tried, it gives about 60% of the numerical gap. The problem is of course with the lower bound for the third eigenvalue. The upper bound for the second eigenvalue is very good, even with just 3 eigenfunctions per known case.

Comment by Bartlomiej Siudeja — June 20, 2012 @ 2:44 am

• 60% is the worst case. For known cases it gives exact values.

Comment by Bartlomiej Siudeja — June 20, 2012 @ 3:48 am

• Hi Bartolomiej,
this is good news! I had a question: how much sharper are your bounds on the second eigenvalue, compared with those you’d get from the Poincare inequality here (the functions have mean zero, so one can get a pretty explicit bound). Maybe this is already in your posted notes?

Comment by Nilima Nigam — June 20, 2012 @ 3:45 pm

• Poincare inequality gives lower bound for the second eigenvalue, instead of upper. Optimal Poincare inequality for triangles gives lower bound in terms of just second equilateral eigenvalue. Same techniques as for lower bound in my notes, just less involved, work for Poincare.

Comment by Bartlomiej Siudeja — June 20, 2012 @ 3:54 pm

• Yes, I was thinking of the lower bound. However, do the techniques in your notes give tighter lower bounds than those given by a Poincare inequality?

Comment by Nilima Nigam — June 20, 2012 @ 6:14 pm

• Together with Laugesen we used the same techniques to prove optimal Poincare inequality. However, here I had to push the method to its limits (more reference triangles). Bounds are also a bit different, since we are dealing with second+third eigenvalue.

Comment by Bartlomiej Siudeja — June 20, 2012 @ 7:23 pm

12. I had a thought about a somewhat different tactic, which combines some of the ones discussed earlier. Let us consider the case of triangles with rational angles. These are dense in the set of all triangles, so if we can prove the Hot Spots conjecture for this case, it seems to me that we can easily make continuity arguments to the general case.
Now, consider the universal cover of such a triangle under the reflection group. First off, is it the case that this is an N-sheeted manifold, for some N a factor of the gcd of the denominators?
Second, we can consider the Brownian motion formulation pulled back to this cover, where each preimage of the triangle has a point heat source. Can’t we argue that, since the corners are effectively exposed to “multiple” neighborhoods (N * angle / $\pi$), whereas every other point is (locally) exposed to a single neighborhood, the corners must be hot spots? (E.g. move the source towards a given corner and look at the long-term behavior.)
This argument will fail if N=1. But there are only 3 such cases (the equilateral, isoceles right, and 30-60-90 triangles), and all of them have the Hot Spots conjecture verified.

Comment by Craig H — June 19, 2012 @ 9:07 pm

• I think I may be off on my definition of N and my neighborhood counting — but if N is finite, it shouldn’t matter.

Comment by Craig H — June 19, 2012 @ 9:10 pm

• I don’t quite follow the argument that being exposed to multiple neighborhoods should make the corner points the hottest. For one thing, by continuity, the points slightly in from the corner will also be very hot despite only being locally exposed to its nearest neighbors.

The picture in my mind of this manifold is that it looks something like a spiral staircase near the corner. The center of this staircase might seem the hottest, but then when you “fold up the triangles” to convert the heat flow back on the manifold back to the heat flow on the triangle, every point in the triangle ultimately has many points that map to it.

That’s not to say there isn’t a good idea here… I just don’t have a good intuition for what the N-sheeted manifold looks like :-) And for example in the case of an equilateral triangle which tiles the plane (a 1-sheeted manifold?) I don’t see any intuitive reason why the hottest point should be at the boundary :-/

As a side point, the intuitive reason I see behind why the hot spots should lie in the corners of sharpest angles comes from the probabilistic interpretation: The heat flow is dual to reflected Brownian motion and reflected Brownian motion has a hard time getting inside sharp corners. This is because reflected Brownian motion reflects perpendicularly off of the boundary. Therefore, hitting the boundary gives the reflected Brownian motion a nudge away from the corner! (Yes, for sharper corners, the perpendicular does not point as much out from the corner… but this is offset by the fact that the reflected Brownian motion hits the boundary much more frequently in a narrow corner).

In fact, I suspect that if you take any Neumann boundary region $D$ and append a spike of angle $\epsilon$ and let $\epsilon\to 0$, then for sufficiently small $\epsilon$, one of the hot-spots will be at the tip of the spike… irrespective of the shape of the original domain $D$. (And I imagine if you have two such spikes, then their tips will be the the two hot-spots).

Comment by Chris Evans — June 20, 2012 @ 2:23 am

• Alright, let me make my intuition here a little more precise: Around any of the preimages of the corners, the area of the ball of radius $R$ goes as $\pi p R^2$, where $p$ is the numerator of the angle ($\alpha = \pi p/q$). Whereas around any non-corner point $P$, there is some $\epsilon$ such that $\forall r < \epsilon$, Area($B(P,r)$) = $\pi r^2$. And you are going to get that the heat is a smooth function of $\epsilon$, with a maximum as $\epsilon \rightarrow 0$ (in fact, we know just what it looks like for small $\epsilon$ — it is constant plus angularly-modulated Bessel function).

Comment by Craig H — June 20, 2012 @ 1:32 pm

• I see that there is more area on the manifold near a corner point than an interior point… but I don’t follow the intuition of why that would make the point hotter… couldn’t it be argued just as well that there is more area for heat to escape off to?

Comment by letmeitellyou — June 20, 2012 @ 5:39 pm

• Also, I said that this argument pretty explicitly fails in the three $(\pi/p, \pi/q, \pi/r)$ cases where the reflected triangles tile the plane — but we have separate arguments for each of those cases as well.

Comment by Craig H — June 20, 2012 @ 1:34 pm

• Just FYI, I realized that in general we do not have a finite number of sheets. One can take, for example, the 45-67.5-67.5 isoceles triangle. Reflecting about the two congruent edges gives an octagon, but if on reflects long edge-short edge-long edge, the two images of the short edge are at right angles to each other. Repeating this process produces a square tiling of the plane. If you assume that the short edge has length 1, you can get an infinite number of square tilings of the plane translated by 1 in a diagonal direction — that is, the preimage of the corners of the original triangle contains Z[\sqrt{2}]^2. Since this set is dense, we can’t have a finite number of sheets.

Comment by Craig H — June 25, 2012 @ 7:39 pm

13. Perhaps it might be a good idea to focus numerical work on a single “generic” triangle – e.g. a 40-60-80 triangle – as a test case, and see to what extent we can show rigorously that extrema for this triangle (and for nearby triangles) only occur at the vertices? This would give an indication of how fine a mesh of triangles we would need to cover the whole sample space.

The way I see it, one could rigorously show the hot spots conjecture for a given triangle $\Omega$ by some variant of the following scheme:

1. Find an approximation $u$ to the second eigenfunction which is provably accurate in $C^0$ norm with some error $\varepsilon$, thus $|u(x)-u'(x)| \leq \varepsilon$ for all x, where u’ is the true second eigenfunction.

2. As a consequence, any point x at which $u(x)$ differs by more than $2\varepsilon$ from the extrema of u, cannot be an extremum of u’. Assuming that u behaves as expected, with extrema only at vertices, this should place all extrema of u’ within $O(\sqrt{\varepsilon})$ of the vertices.

3. It should be possible to perform a Bessel function expansion around each vertex with upper and lower bounds on coefficients. With sufficiently good such bounds, this should show that in the $O(\sqrt{\varepsilon})$ neighbourhood of each extremal vertex, there are no critical points other than the vertex.

1, 2, and 3 give the hot spots conjecture for the specific triangle (and will also give the conjecture for some explicit neighbourhood of that triangle in parameter space).

The catch is that it may require some absurdly high level of accuracy, e.g. $\varepsilon = 10^{-6}$, in order to work, in which case we may be able to resolve a single triangle (such as the 40-60-80 triangle) but would not be able to cover all of parameter space without a ridiculous amount of computer time. The square rooting of the epsilon parameter is a little disconcerting. Note though that if one had C^1 control instead of C^0 control then one could exclude all critical points outside of an epsilon-neighbourhood of the vertices rather than a sqrt(epsilon) neighbourhood, and with C^2 control one could exclude critical points everywhere. But such control may be unreasonable…

Comment by Terence Tao — June 22, 2012 @ 1:37 am

• I think what you propose is do-able. Presumable you want to check a generic triangle to avoid any special symmetries which arise on the right-angled one? I’d much rather pull back the problem on the 40-60-80 triangle and compute on the right-angled one. Then there isn’t any fussing with mesh quality etc.

One would begin with the literature to check for both a priori and a posteriori sup-norm estimates for algorithms. Here’s a nice paper on FEM for eigenvalues, but the estimates are not in the sup norm. http://www.ing-mat.udec.cl/~rodolfo/Papers/DPR.pdf

In the finite-element world, one usually examines L^2 (or other Hilbert space) norms, but I think there’s enough on sup norm estimates as well. What we’ll hopefully get is a theorem which provides an estimate which relies only on the computed eigenfunction $u_h$, to get control on $sup|u(x)-u_h(x)| \leq C h^r$. There’s already lots in the literature on such estimates in the energy norm, but we’ll have to look for equivalent estimates in $C^1$.

Here $C>0$ is a constant which may depend on the triangle, $h$ is some parameter in the method which goes to zero as the approximating sequence $u_h$ approaches $u$. One could compute several approximation terms, and use an extrapolation technique (Richardson’s or something more sophisticated) to get an extremely refined estimate.

Until we see what’s in the literature on the sup-norm estimates on the gradients, we can’t really assess how hard it will be to get the accuracy. For what it’s worth, the measure of error I use to monitor code, $\|\nabla u_h\|_0^2 - \lambda_h\|u\|_0^2$, is down to $10^{-13}$ on these problems without trying hard.

Comment by Nilima Nigam — June 22, 2012 @ 4:20 am

• Yes, I picked the 40-60-80 triangle because it looked quite typical, so that one isn’t deceived by symmetry. I agree though that in practice we would pull it back to a nicer triangle such as the right-angled triangle.

I’m curious as to how you are computing the numerical eigenvalue $\lambda_h$; it doesn’t appear to simply be the Rayleigh quotient, since otherwise your error $\| \nabla u_h\|_0^2 - \lambda_h \|u\|_0^2$ would simply be zero. In any case, your accuracy of $10^{-13}$ is reassuring, it probably gives us a fair amount of room to play with.

Here is one simple way to get $C^0$ or $C^1$ estimates away from vertices. Let x be an interior point of the triangle, and suppose first that the ball B(x,R) of radius R centred at x lies completely inside the triangle. Let $-\Delta u = \lambda u$ be a true eigenfunction. The radial averages

$f(r) := \frac{1}{2\pi} \int_0^{2\pi} u(x + r e^{i\theta})\ d\theta$

(using complex notation) obey the Bessel ODE $f''(r) + \frac{1}{r} f'(r) + \lambda f(r) = 0$, and thus must be a scalar multiple of the Bessel function $J_0(\sqrt{\lambda} r)$. In particular, on integrating from r=0 to r=R we conclude a mean value theorem

$\displaystyle u(x) = \frac{ \int_{B(x,R)} u(y)\ dy}{ \int_{B(x,R)} J( \sqrt{\lambda} |y-x| )\ dy}$ (1)

which, among other things, can be used to convert L^2 type bounds on u to C^0 bounds on u in the interior region of the triangle. Since the eigenfunction equation commutes with derivatives in the interior, we can also convert H^1 bounds to C^1 bounds in a similar fashion:

$\displaystyle \nabla u(x) = \frac{ \int_{B(x,R)} \nabla u(y)\ dy}{ \int_{B(x,R)} J( \sqrt{\lambda} |y-x| )\ dy}$.

If one is near an edge of the triangle, but not near a vertex, one can achieve a similar formula after reflecting across the edge, so long as B(x,R) does not touch a vertex. Once one touches a vertex, things start getting a bit crazy and I don’t see a clean formula, but as I said in the previous comment we may be able to handle the area near the vertices by a different method.

If the approximate eigenfunction u_h also approximately obeys estimates such as (1), then the error $u-u_h$ will also approximately obey an equation like (1), which should in principle thus give good C^0 or C^1 estimates on the error away from vertices…

Comment by Terence Tao — June 22, 2012 @ 5:48 am

• To compute the numerical eigenvalue, I first formulate the eigenvalue problem in discrete form. This is done either (1) variationally (through finite elements) or (2) by the modified method of particular solutions (linear combinations of Bessel functions and trigonometric polynomials).

For (1), I end up with a large generalized eigenvalue problem $Ax = \lambda Bx$, where $A_{ij} = \int_\Omega \nabla \phi_i \nabla \phi_j$, and $B_ij = \int_\Omega \phi_i \phi_j$. This is solved using Arnoldi iterations to get the eigenvalues.

For (2), I’m still trying to tweak the method, but the eigenvalues are located by a numerical minimization, which in turn is done using the Nelder-Mead algorithm. I am nervous about using this for validated numerics unless I can prove the method is consistent in infinite precision arithmetic.

Comment by Nilima Nigam — June 22, 2012 @ 2:26 pm

• Here is a way to quantify the fact that there are no extrema near a vertex, other than at the vertex itself.

Using polar coordinates around a vertex of angle $\alpha$, with angular coordinate $\theta \in [0,\alpha]$, a Neumann eigenfunction $-\Delta u = \lambda u$ obeys the equation

$\partial_{rr} u + \frac{1}{r} \partial_r u + \frac{1}{r^2} \partial_{\theta\theta} u + \lambda u = 0$

and thus (by separation of variables, or equivalently by Fourier series in the angular variable) has the Bessel expansion

$u(r,\theta) = \sum_{k=0}^\infty c_k J_{k\pi/\alpha}(\sqrt{\lambda} r) \cos( k \pi \theta / \alpha )$ (1)

in any sector $S := \{ (r,\theta): 0 \leq r \leq R; 0 \leq \theta \leq \alpha \}$ inside the triangle. One can estimate the size of the coefficients $c_k$ by computing the $L^2$ norm on S using Plancherel’s theorem:

$\int_S |u|^2 = \frac{\alpha}{2} \sum_{k=0'}^\infty |c_k|^2 \int_0^R J_{k\pi/\alpha}(\sqrt{\lambda} r)^2\ r dr$ (2)

where the prime in the summation indicates that the $k=0$ term is counted twice. One could also compute the $\dot H^1$ norm using the identity $|\nabla u|^2 = |\partial_r u|^2 + \frac{1}{r^2} |\partial_\theta u|^2$ and the Plancherel identity:

$\int_S |\nabla u|^2 = \frac{\alpha}{2} \sum_{k=0'}^\infty |c_k|^2 \int_0^R |J'_{k\pi/\alpha}(\sqrt{\lambda} r)|^2 + \frac{k^2\pi^2}{\alpha^2 r^2} |J_{k\pi/\alpha}(\sqrt{\lambda} r)|^2\ r dr$. (3)

If we normalise u to have L^2 norm 1 on the triangle, then we have $\int_S |u|^2 \leq 1$ and $\int_S |\nabla u|^2 \leq \lambda$, which gives some l^2 type control on the coefficients $c_k$. Of course, to make this optimal, one should make the sector S as large as possible while still inside the triangle.

From (1) and the triangle inequality we have

$|u(0)| - |u(r,\theta)| \geq u(0) (1 - |J_0(\sqrt{\lambda} r)|) - \sum_{k=1}^\infty |c_k| |J_{k\pi/\alpha}(\sqrt{\lambda} r)|$
|
so if $|u(r,\theta)|$ is greater or equal to $|u(0)|$, we must have

$\sum_{k=1}^\infty |c_k| |J_{k\pi/\alpha}(\sqrt{\lambda} r)| \geq u(0) (1 - |J_0(\sqrt{\lambda} r)|).$ (4)

Note that for small r, $1 - |J_0(\sqrt{\lambda} r)|$ is comparable to $r^2$, and the higher order Bessel functions $J_{k\pi/\alpha}(\sqrt{\lambda} r)$ are comparable to $r^{k\pi/\alpha}$, so this already rules out any competitor to u(0) that is sufficiently close to the vertex. By using (2) or (3) and Cauchy-Schwarz with (4), one can get an explicit region of r for which competitors are ruled out.

Comment by Terence Tao — June 22, 2012 @ 4:55 pm

• Some results.
In http://www.math.sfu.ca/~nigam/polymath-figures/dump-data.odt I’ve documented the following:

Case 1: For a right-angled triangle (2 sides of 1), you’ll find values of:

# of degrees of freedom, $\lambda_{calculated}$, $max of u_h,$ $min of u_h,$ $u_h$ at the three corners, and the error $\lambda_h-\lambda$.

Case 2: For the 40-60-80 triangle, we don’t have the exact eigenvalue. I’ve tabulated values of

# of degrees of freedom, $\lambda_{calculated}$, $max of u_h,$ $min of u_h,$ $u_h$ at the three corners,

Method: I’ve computed these quantities using finite elements + shifted Arnoldi iteration for the eigenvalue. The finite element eigenfunction is then interpolated onto a very fine grid before we locate the max and min. For each triangle, I’ve computed using piecewise linears, then piecewise quadratic, and finally piecewise cubic functions. I’ve used the freeware FreeFem++, and am happy to pass along the code to anyone who’d like to play with it.

Observations:

1. First, the Arnoldi iteration will not favor an eigenfunction $u_h$ over $- u_h$. So one can see that max and min value of u may switch from iteration to iteration, but the absolute values don’t.

2. Next, for all these experiments, one of the corner values is precisely the maximum value of $abs(u_h)$. Using Terry’s argument above, there are no others in the neighbourhood.

I’m happy to assert we’ve got 7 digits accuracy for the maxima and minima, but would have to run this on a larger machine than my laptop to get better.

Comment by Nilima Nigam — June 22, 2012 @ 8:27 pm

• I was trying to recreate your numbers using my code, and I have a few questions for case 1. What vertices are you using? Right isosceles triangle with sides 1 and 1 should have eigenvalue pi^2. Also, what do you mean by degrees of freedom? Is this the size of the matrix in the eigensolver? Finally, I was using shift-and-invert transform to speed up eigenvalue convergence. This seems to work pretty well (speed-wise), but makes it impossible to get as accurate results as you are getting (10^-9 at best for eigenvalue). Have you considered using shift-and-invert, or is this just a bad idea?

In observation 1, is it possible to apply boundary condition, say 1 at a vertex, to force maximum at a specific vertex? Kind of like applying Dirichlet to the matrix in the eigensolver. You would also get a nicely scaled eigenfunction. I am not even sure this is possible.

Comment by Bartlomiej Siudeja — June 22, 2012 @ 11:43 pm

• For case 1, I used the vertices (0,0), (1,0), (1,1).
I thought the 2nd eigenfunction on this triangle should have form $\cos(\pi x) \cos (\pi y)$, giving $- \Delta 2\cos(\pi x) cos (\pi y)= 2\pi^2$. At any rate, I’m reporting the first eigenvalue corresponding to a non-constant eigenfunction. Turns out it’s very close to $2\pi^2$.

The number of degrees of freedom is the number of unknowns, and therefore the number of unknowns in the eigensolver. The matrices are sparse, so it’s also approximately (upto a factor of a constant) the number of entries in the matrix.

For a piece-wise linear finite element, this will correspond to the number of vertices in the mesh (counted correctly). For piecewise quadratic elements there will be more finite element basis functions per triangle, and so forth. These dofs scale linearly with $1/h$ where $h$ is some measure of mesh size.

How are you computing the eigenvalue? Using a shift-and-invert on top of an Arnoldi iteration is overkill, so I don’t do it. But if you are using an inverse Rayleigh iteration or something, that should be OK…. upto the expense of inversion.

Numerically, why would you apply a Dirichlet condition to a vertex? By pinning down that node, one will influence the discrete eigenvalue problem you’re trying to solve, and potentially in a nonlinear manner. Different finite element packages do different things to enforce Dirichlet conditions, so this would become another factor to consider in the analysis.

I wanted to use a conforming approximation strategy, and would therefore be reluctant to enforce any boundary conditions which are not natural. In principle, if we think $u_h\in H^1(\Omega)$, pinning it down at one node won’t matter IF we do the eigenvalue solve exactly, and IF we don’t care about point values of $u_1$. But we’re looking for the behavior of $u_h$ and derivatives at points,

Comment by Nilima Nigam — June 23, 2012 @ 12:33 am

• This one is actually the third eigenfunction. There is also cos(pi x)+cos(pi y) which gives eigenvalue pi^2.

I am using SLEPc eigensolver included in FEniCS project, and matrices A, M as you described here somewhere. If I use shift-and-invert I am getting at least 10x speed gain. Even without using any shift. I was also using Krylov-Schur instead of Arnoldi.

I did not mean to imply applying 0 boundary condition to a vertex. More like applying 1 to a vertex. I know this is not very natural, but it should force eigenfunction to have same sign at the vertex. And should make no difference for derivative calculations.

Comment by Bartlomiej Siudeja — June 23, 2012 @ 2:42 am

• I don’t get it. To keep calculations simple, let’s consider the triangle flipped over, so the corners are $(0,0), (1,0), (0,1)$. $u=cos(\pi x)+ cos(pi y)$ certainly has $-\Delta u = \pi^1$, and the normal derivatives vanish on the straight edges. Indeed, $u$ is the first non-constant eigenfunction on the square.

On the third side of the triangle, which is along the line $y=1-x$, the outward (unscaled) normal is $(1,1)$. So, the normal derivative of $u$ on this edge is simply $u_x+u_y= -\pi (\sin(\pi x) + \sin(\pi (1-x) ) = -\pi [\sin(\pi/2) cos( (\pi-2x)/2) ]$, which does not vanish. Am I missing something?

At any rate, imposing $u=1$ will have the same problem. Actually, a bit worse: you’ll end up with a non-symmetric formulation, since the computed $u$ will not be in a vector space: $U_h:= \{ u_h \in H^1(\Omega) \vert u_h(corner)=1\}$. To deal with this, one has to use test functions in a genuine vector space $V_h:= \{ v_h \in H^1(\Omega) \vert v_h(corner)=0\}.$ Since $U_h \not = V_h$, the resultant discrete formulation is not symmetric. That’s fine, but I’m not sure there’s much point.

Comment by Nilima Nigam — June 23, 2012 @ 4:22 am

• Second eigenvalue for a square has multiplicity 2, constant in x or in y. Now the sum and the difference of the two will be either constant on the diagonal, or have 0 normal derivative there. So one of the two newly created eigenfunctions will be the eigenfunction of the right isosceles triangles. On the flipped triangle that would be cos(pi x)-cos(pi y). This is actually an example of a subdomain with the same smallest eigenvalue as the whole domain.

Comment by Bartlomiej Siudeja — June 23, 2012 @ 4:55 am

• interesting…. but the normal derivative didn’t vanish on the diagonal, so what is going on?

Comment by Nilima Nigam — June 23, 2012 @ 4:59 am

• Try cos(\pi x)-cos(\pi y). This one should have 0 normal derivative on (1,0) — (0,1). For your original triangle (with (1,1) as vertex) the plus version should work.

Comment by Bartlomiej Siudeja — June 23, 2012 @ 5:08 am

• You’re right!
Thanks for your patience- this has been a helpful discussion for me. This error, and the fact that I also got the second eigenvalue on the 40-60-80, suggested I had a systematic bug in the code I wrote today. I found it, fixed it, and have replaced the data:

http://www.math.sfu.ca/~nigam/polymath-figures/dump-data.odt

The conclusions remain the same.

Comment by Nilima Nigam — June 23, 2012 @ 6:06 am

• I have also noticed that using too fine mesh gives worse results. I guess rounding errors kick in. With cubics I get the best results with DOF of 10000-20000 (10^-10 for eigenvalue of right isosceles). After that I loose a bit of accuracy.

Comment by Bartlomiej Siudeja — June 23, 2012 @ 3:01 am

• yes, too fine a mesh will likely give poor results. Not rounding, I suspect the quality of the mesh degenerates, unless one puts in some checks on the quality of the mesh. One must also be a bit cautious: the eigenfunctions are not $C^\infty$, so there’s not reason to assume taking very high degree polynomials will work (in fact, already quartic approximants show bad results).

Comment by Nilima Nigam — June 23, 2012 @ 4:25 am

• It seems that in both cases you are getting the third eigenvalue.

Comment by Bartlomiej Siudeja — June 23, 2012 @ 3:13 am

• I could not reply to your latest post. With old data even the third eigenfunction was good for hot-spots conjecture. I am not sure what vertices you are using for 40-60-80 triangle, but the old ones where giving me an obtuse triangle. You have right isosceles vertex for 40-60-80 in the new file (aa, bb values).

Comment by Bartlomiej Siudeja — June 23, 2012 @ 6:14 am

• fixed, thanks. I’m going to take a break from coding now, since I’m making silly errors.

Comment by Nilima Nigam — June 23, 2012 @ 6:36 am

• You deserve a break. You are doing a great job with the numerical side of the project.

Comment by Bartlomiej Siudeja — June 23, 2012 @ 6:38 am

• thanks are due to you- you’ve been incredibly patient and careful, and I appreciate your looking over the results. I fixed the print statement for the vertices. Who knows, maybe introduced some other error. I’ll stop for now.

Comment by Nilima Nigam — June 23, 2012 @ 6:51 am

• Your 40-60-80 triangle is a bit obtuse, when plotted. It seems aa and bb should be different. In fact we should probably switch from 40-60-80 to something with rational vertices.

Comment by Bartlomiej Siudeja — June 26, 2012 @ 12:50 am

• hmm, it plots fine for me.

Comment by Nilima Nigam — June 26, 2012 @ 1:57 am

• Assuming you have 0,0 and 1,0 as other 2 vertices 0.83, 0.3 gives obtuse triangle using Pythagorean theorem.

Comment by Bartlomiej Siudeja — June 26, 2012 @ 2:29 am

14. Just some other thoughts on the analytic approach (I should also say that I have been following the comments the rigorous-numerics approach but I haven’t had much to contribute):

—————————-
Thought 1:

I was thinking about how to argue rigorously that the Nodal line for an arbitrary triangle shouldn’t be too “squiggly”. Intuitively it seems to me that this is a consequence of the Raleigh quotient interpretation: A squiggly nodal line would seem to make $u$ have a large $H^1$-norm, whereas “straightening out the squiggle” would lower its $H^1$-norm while leaving the $L^2$-norm unchanged.

More concretely, I’ve been looking at the problem of minimizing the Raleigh quotient for mean-zero $u$ subject to the constraint $u(x)\in\{-1,1\}$. Since $u$ can only take one of two values, we divide the region into a “hot region” and “cold region”. The $L^2$-norm of $u$ is $1$ and the $H^1$-norm of $u$ should be proportional to the length of the curve dividing the two sub-regions. Therefore, minimizing the Raleigh quotient is the same as minimizing the length of a curve which divides the triangle into two sub-regions of equal area. I believe that such a line should not be squiggly…

Then, we might consider the case that $u$ takes on one of $n$ values and minimizing the Raleigh quotient would again correspond to minimizing the lengths of the curves which divide the different sub-regions (where $u$ takes on different values).

Then, somehow we could maybe show that as $u$ is allowed to take more values, it approximates the true $u$.

One thing that concerns me is that I don’t see how this approach would account for the known counterexamples to the hot-spots conjecture in non-simply connected domains. In fact, I don’t have any intuition behind those counterexamples so if someone does have intuition I would love to hear it :-)
——————————————-
Thought 2:

From the comment thread started by David (meditationnate): For the interval $[0,1]$, if $u$ solves the Neumann problem then $u'$ solves the Dirichlet problem. So one could first find information about the solution to the Dirichlet problem and then integrate it to get information about the solution to the Neumann problem.

I wonder if any information could be gained from studying $\nabla u$. I don’t have any particular idea… but such an argument might be nice because recovering $u$ from $\nabla u$ requires the domain to be simply connected… which is precisely the largest class of domains where the conjecture is conjectured to hold!
———————————————-
Thought 3:

Since reflected Brownian motion is dual to the heat flow, and since the Laplacian is self adjoint, the heat flow started from a point $x$ looks the same as the probability density function (p.d.f.) of a reflected Brownian motion started at the point $x$.

It therefore ought to be the case that we have an ergodicity result that says: If you consider *one* path of reflected Brownian motion started from any point $x$, then if you look at some long term average of where it spends its time it ought to correspond somehow to the pdf/second eigenfunction.

Then it would suffice to show that a reflected Brownian motion doesn’t spend much time in the sharpest corners (which is intuitively true since the boundary reflection is perpendicular to the boundary, giving it a shove away from the corner, and sharper corners make the Brownian motion hit the boundary more frequently).

(Edit: On second thought, I think my intuition about reflected Brownian motion having a hard time going into a sharp corner is wrong. For one thing a perpendicular push at the boundary doesn’t push you “away” but rather doesn’t affect your radial distance from the corner one way or another. More to the point, if you consider reflected Brownian motion in an infinite wedge, it should be the same as if you reflect the infinite wedge many times… and so the chance of the reflected Brownian motion getting close to the corner should be the same as that of a free Brownian motion getting close to the origin)

Comment by Chris Evans — June 22, 2012 @ 3:11 am

15. Hi Chris,

Your ideas are nice. Regarding the second thought, here’s one way to think about the eigenvalue problem.
Right now, we look for $u\in H^1$ so that $-\Delta u - \lambda u=0$ on the domain, and $\frac{\partial u}{\partial n}=0$ on the boundary. We also enforce that the mean of $u=0$.
Now let us rewrite the problem as $\sigma - \nabla v =0$, and $\nabla \cdot \sigma + \lambda v=0$ on the domain. On the boundary, enforce $\sigma \cdot n=0$. Also enforce $u$ to have mean zero. This system will have solutions $\sigma \in H(div)$ and $v\in L^2$.

Such a reformulation is called a mixed method in the numerical analysis literature. Maybe the same in the PDE literature? Mixed methods are very well studied. Here’s a nice paper by Boffi and Gastaldi introducing these methods in a general setting for parabolic problems http://www.ing.unibs.it/~gastaldi/paper/evo.pdf

Comment by Nilima Nigam — June 22, 2012 @ 4:52 am

16. I was thinking about a geometric interpretation for the Laplacian, when we start the heat equation with the second eigenfunction $u_{2}$, looking at the evolution for this particular $u_{2}$ as time goes to infinity. If $v\left( t \right)$ is the evolved function/heat distribution at time $t$, then as $t$ goes to infinity the sup norm of $v \left( t \right)$ goes to zero. With $t$ fixed, we can look at the mean curvature of the graph of $v(t)$ as a surface embedded in $R^3$. If $h$ is the mean curvature at a point on that surface, then by definition $h = 1/2 \left( \kappa_{1} + \kappa_2 \right)$, where $\kappa_{1}$ and $\kappa_{2}$ are the principal curvatures of the surface, at some point in the graph in $R^3$ of $v \left( t \right)$. The principal curvatures were studied by Gauss, at the beginnings of differential geometry. As $t$ goes to infinity, the normal to the surface is almost perpendicular to the $x$ , $y$ plane. If the normal is parallel to the $z$-axis, the mean curvature $h$ is twice the Hessian matrix of $v \left( t \right)$ as a function of $x$ and $y$. The trace of the Hessian is both twice the mean curvature $h$, as well as the Laplacian of $v\left( t \right)$ at the reference point, say $\left(x_{0}, y_{0} \right)$. So as $t$ goes to infinity, the Laplacian at the reference point should be very close to to the twice the mean curvature, so Laplacian ~= $2h$ as $t$ goes to infinity. Positive curvature belongs to surfaces such as $z = x^2 + y^2$ (we need to say which normal to a surface serves as reference to determine mean curvature, with sign included). So, Laplacian ~= $2h$ as $t$ goes to infinity. I’ve thought about that when considering the evolution of $v\left( t \right)$ in a very short time interval . I’ve also thought about the heat kernel and Green’s function methods. The heat-reflecting property of the sides of the triangles make things very complicated, it seems.

Comment by meditationatae — June 23, 2012 @ 5:55 am

17. I’ve just rolled the thread over again, as this one is hitting the 100-comment mark: https://polymathprojects.org/2012/06/24/polymath7-research-threads-3-the-hot-spots-conjecture/

Comment by Terence Tao — June 24, 2012 @ 7:23 pm