# Thin triangles

Theorem: the hot spots conjecture is true for acute-angled triangles ABC when $\angle BAC$ is sufficiently small.

Let us write $\varepsilon := \angle BAC$, which we view as being small. We may normalize A = (0,0) and B = (1,0). Since the other two angles $\angle ABC, ACB$ are $\pi/2 - O(\varepsilon)$, we may also normalize $C = (1 + O(\varepsilon^2), \varepsilon + O(\varepsilon^2))$.

## Eigenvalue bound

For the sector of radius 1 and aperture $\varepsilon$, we know (as discussed at the hot spots conjecture) that the second eigenvalue is $j_1^{-2}$, where $j_1=3.8317\ldots$ is the first solution to $J'_0(j_1)=0$ (or equivalently $J_1(j_1)=0$). We claim that for the triangle ABC, the second eigenvalue is $j_1^{-2}+O(\varepsilon)$.

First, the upper bound. We can take the second eigenfunction $J_0(\sqrt{\lambda}r)$ for an inscribed sector to ABC of radius $1-\varepsilon$ and aperture $\varepsilon$, with $\lambda = (1-\varepsilon)^{-2} j_1^{-2}$, and extend it smoothly to a function on ABC obeying the Neumann boundary conditions. It will not quite mean zero, but we can subtract a constant of size $O(\varepsilon)$ to make it of mean zero. The Rayleigh quotient for this object can be shown to be $j_1^{-2} + O(\varepsilon)$, providing the upper bound.

The lower bound can be proven by similar methods once we get good enough C^2 type bounds on eigenfunctions on ABC, but for now let us just establish the weaker bound $\lambda \gg 1$. It suffices to establish the Poincare inequality

$\int_{ABC} |u - c|^2 \ll \int_{ABC} |\nabla u|^2$

for all u on ABC (not necessarily obeying the Neumann boundary condition) and some constant c depending on u. But one can rescale ABC to, say, the unit equilateral triangle in which case the bound is classical, and note that the constants for the rescaling are favorable. (One can also cite here the bounds from [PW1960] or [LS2009] for this bound; the [LS2009] bound in fact gives the lower bound of $j_1^{-2} + O(\varepsilon)$ directly.)

## Regularity of eigenfunctions

Let u be the second eigenfunction on ABC, normalised so that

$\int_{ABC} |u|^2 = |ABC|.$

By repeated integration by parts (as discussed at the hot spots conjecture) we have

$\int_{ABC} |\nabla^j u|^2 = \lambda^j |ABC|$ (*)

for j=0,1,2,3. In particular, $\nabla^j u$ is O(1) on the average on ABC for j=0,1,2,3. In the middle third of the triangle, we can use elliptic regularity (after reflecting the triangle across AB and AC many times so that a disk of radius ~1 sits inside the domain) to in fact conclude that all derivatives are O(1) in this region.

By working in polar coordinates around A, we may expand

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

for some coefficients $c_k$. Because all derivatives of u are O(1) in the middle third of the triangle, we see that $c_k J_{\pi k/\varepsilon}(\sqrt{\lambda} r)$ is rapidly decreasing in k in this middle third, which from the asymptotics of Bessel functions gives excellent regularity on the left third of the triangle; in particular it is not difficult to get uniform bounds on the C^3 norm in this third.

The situation is more delicate on the right third of the triangle. We begin in the region of size $O(\varepsilon)$ around the edge BC. From (*) we know that $\nabla^3 u = O(\varepsilon^{-1/2})$ on average in this region, while from the Poincare inequality (or Sobolev inequality) one can show that $\nabla^2 u = O(1)$ on average in this region. By using Bessel function expansions around B and C one can then show that $\nabla^2 u = O(1)$ pointwise here; in the rest of the triangle one can use elliptic regularity to also get $\nabla^2 u = O(1)$ first on average in regions of diameter $\varepsilon$, and then pointwise by elliptic regularity (and reflection). Thus u is bounded uniformly in C^2.

This is already enough regularity on u to extend u to a circumscribing sector and show that $\lambda = j_0^{-2} + O(\varepsilon)$. Now we can relate u more carefully with a Bessel function. We again work in polar coordinates around the origin A. If we write

$u_0(r) := \frac{1}{\varepsilon} \int_0^\varepsilon u(r,\theta)\ d\theta$

for the averaged radial component of u, defined for $0 \leq r \leq 1-\varepsilon$ (say), then from integration by parts we see that

$u''_0(r) + \frac{1}{r} u'_0(r) + \lambda u_0(r) = 0,$

thus u_0 is a multiple $c_0 J_0(\sqrt{\lambda} r)$ of the Bessel function $J_0(\sqrt{\lambda} r)$ (recall that u_0 has to be continuous at the origin). From the uniform C^2 bounds, we have $u(r,\theta) = u_0(r) + O(\varepsilon)$ for all $0 \leq \theta \leq \varepsilon$ and $0 \leq r \leq 1-\varepsilon$. Given the normalisation of u, this implies that

$c_0^2 \int_0^1 J_0(\sqrt{\lambda} r)^2\ \varepsilon r dr = \frac{1}{2} \varepsilon^2 + O(\varepsilon^3)$

which implies that $c_0 = \pm c_* + O(\varepsilon)$ where $c_*$ is the positive absolute constant

$c_* := (2 \int_0^1 J_0(r/j_1)^2\ r dr)^{-1/2}.$

Without loss of generality we may take the positive sign, thus $c_0 = c_* + O(\varepsilon)$, and so we have the asymptotic

$u(r,\theta) = c_0 J_0(r/j_1) + O(\varepsilon)$ (**)

throughout the triangle. This is already enough to show that the maximum can only occur within $O(\varepsilon^{1/2})$ of A, and the minimum can only occur within $O(\varepsilon^{1/2})$ of B or C.

Now let us show that the maximum can only occur at A. If for contradiction there is a nearby point P to A which also attains the maximum, then $\partial_r u = 0$ at P, but also $\partial_r u=0$ at A. Hence, by Rolle's theorem, one has $\partial_{rr} u = 0$ at some intermediate point Q on the interval AP. But from the Bessel expansion we see that $\partial_{rr} u$ is bounded away from zero at A, and so from the uniform C^3 bounds we obtain a contradiction if $\varepsilon$ is small enough.

It remains to show that that the minimum can only occur on BC. Let n be the unit normal to BC, thus $\partial_n u = 0$ on BC. If for contradiction there is a nearby point Q to BC which also attains the minimum, then $\partial_n u = 0$ at P, so by Rolle's theorem we have $\partial_{nn} u = 0$ for some intermediate point Q on the line segment from P to BC in the direction n. As n differs from the radial direction by $O(\varepsilon)$, we conclude from the uniform C^2 bounds that

$\partial_{rr} u = O(\varepsilon)$ (***)

at Q.

We need to derive a contradiction from (***). To do this, we begin by inserting (***) into the equation

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

which is the eigenfunction equation in polar coordinates. As Q is near to BC, we see from (**) that u is comparable to 1, so $\lambda u$ is comparable to 1 also. From the Neumann boundary condition we have $\partial_r u = O(\varepsilon)$ on BC, and so since Q is within $O(\varepsilon^{1/2})$ of BC we conclude from the uniform C^2 bounds that $\partial u = O(\varepsilon^{1/2})$ at Q. Putting all this together, we see that

$|\partial_{\theta\theta} u| \sim 1$

at Q.

Suppose first that Q is further than $\varepsilon/10$ from BC. On the circular arc in ABC centered at A and passing through Q, we know that $\partial_\theta u=0$ on the ends of this arc, and so by Rolle's theorem we have $\partial_{\theta\theta} u = 0$ somewhere in the interior of this arc. In particular we see that $\partial_{\theta\theta} u = 0$ at some point R within $O(\varepsilon)$ of Q.

By the mean value theorem, we can find a point S on QR where $|\nabla^3 u| \gg \varepsilon^{-1}$. From the geometry of the situation, we see that S is a distance at least $\varepsilon/100$ from B or C. By (*), we thus have $|\nabla^3 u| = O(\varepsilon^{-1/2})$ on average on a ball of radius $\varepsilon/100$ centered at S (extending u by reflection as necessary), which implies by elliptic regularity that $|\nabla^3 u| = O(\varepsilon^{-1/2})$ at S, a contradiction.

Thus we may assume that Q is within $\varepsilon/10$ of BC. To get a contradiction, we will now need a rather delicate analysis of the eigenfunction u in the neighbourhood of BC.

We first look at what u is doing in the $2\varepsilon/3$ neighbourhood of B. We perform a Bessel expansion in polar coordinates around B, obtaining an expansion of the form

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

where $\beta$ is the angle at B, $r=r_B$ is now the distance to B, and $\theta$ is the angle subtended at B relative to the ray BA. Note that $\beta = \pi/2 - O(\varepsilon)$. Evaluating at B we see that $c_0 = O(1)$. Since $\partial_{rr} u = O(1)$ pointwise, and in particular when r is comparable to $\varepsilon$, we conclude using Bessel function asymptotics that

$c_k J_{\pi k/\beta}(\sqrt{\lambda} r) = O( \varepsilon^2 k^{-2} (1.1r/\varepsilon)^{\pi k/\beta} )$

(say) for $r \leq 2\varepsilon/3$ and k positive. Since $\partial_{rrr} u = O(\varepsilon^{-1/2})$ on average in this region by (*), a similar argument gives the improvement

$c_k J_{\pi k/\beta}(\sqrt{\lambda} r) = O( \varepsilon^{5/2} k^{-3} (1.1 r/\varepsilon)^{\pi k/\beta} )$

for k > 1 (in order to keep the exponent $\pi k/\beta$ well away from 2). From these asymptotics, we see in particular that we can bound the Hessian here by

$(u_{xx}, u_{xy}, u_{yy}) = a_0 (1,0,1) + a_1 r_B^{\pi/\beta - 2} (1,0,-1) + O( \varepsilon^{1/2} )$

in the $2\varepsilon/3$-neighbourhood of B for some bounded constants $a_0,a_1$ which are scalar multiples of $c_0,c_1$ respectively. Similarly we will have

$(u_{xx}, u_{xy}, u_{yy}) = a'_0 (1,0,1) + a'_1 r_C^{\pi/\gamma- 2} (1,0,-1) + O( \varepsilon^{1/2} )$

in the $2\varepsilon/3$-neighbourhood of C for some bounded constants $a'_0,a'_1$, where r_C is now the distance to C, and $\gamma$ is the angle at C. Comparing these two asymptotics on the common domain, we see that $a_0 = a'_0 + O(\varepsilon^{1/2})$ and $a_1 = a'_1 + O(\varepsilon^{1/2})$.

Next, recall that on any circular arc in ABC centered at A, the function $\partial_{\theta\theta} u$ has mean zero. Using an arc tangent to BC, and noting from the Neumann boundary conditions and the uniform C^2 bounds that $\partial_r u = O(\varepsilon)$ on this arc, we see that $\partial_{yy} u$ has mean $O(\varepsilon)$ on such an arc. Comparing this with the above two asymptotics, we see that $a_0 - a_1, a'_0 - a'_1 = O(\varepsilon^{1/2})$. On the other hand, evaluating the Hessian of u at B or at C we see that $a_0, a'_0$ are comparable to 1. Putting this together, we see that $\partial_{xx} u$ will be comparable to 1 in the $\varepsilon/10$ neighbourhood of BC, and in particular at Q, giving the desired contradiction.