Annals of Mathematics, 149 (1999), 309–317

A counterexample to the “hot spots” conjecture

By Krzysztof Burdzy and Wendelin Werner*

Abstract

We construct a counterexample to the “hot spots” conjecture; there exists a bounded connected planar domain (with two holes) such that the second eigenvalue of the Laplacian in that domain with Neumann boundary conditions is simple and such that the corresponding eigenfunction attains its strict maximum at an interior point of that domain.

1. Introduction

The “hot spots” conjecture says that the eigenfunction corresponding to the second eigenvalue of the Laplacian with Neumann boundary conditions attains its maximum and minimum on the boundary of the domain. The conjecture was proposed by J. Rauch at a conference in 1974. Our paper presents a counterexample to this conjecture.

Suppose that D is an open connected bounded subset of R^{d}, d ≥ 1. Let {φ_{1}, φ_{2}, … } be a complete set of L^{2}-orthonormal eigenfunctions for the Laplacian in D with Neumann boundary conditions, corresponding to eigenvalues 0 = μ_{1} < μ_{2} ≤ μ_{3} ≤ μ_{4} ≤ . . .. The first eigenfunction φ_{1} is constant.

Theorem 1. There exists a planar domain D with two holes (i.e., conformally equivalent to a disc with two slits) such that the second eigenvalue μ_{2} is simple (i.e., there is only one eigenfunction φ_{2} corresponding to μ_{2}, up to a multiplicative constant) and such that the eigenfunction φ_{2} attains its strict maximum at an interior point of D.

There remains a problem of proving the conjecture under additional assumptions on the geometry of the domain. Since our method does not seem to be able to generate a counterexample with fewer than two holes, it is natural to ask if this failure has causes of fundamental nature.

*Research of the first author partially supported by NSF grant DMS-9700721.

Problem 1. Does the “hot spots” conjecture hold in all planar domains which have at most one hole?

Using a probabilistic coupling argument, Bañuelos and Burdzy (1999) proved the conjecture for some “long and thin” (not necessarily convex) planar domains and for some convex planar domains with a line of symmetry. We know of only one other published result on the conjecture; it is contained in a book by Kawohl (1985).

We are grateful to David Jerison and Nikolai Nadirashvili for telling us about their forthcoming results. They include a proof of the “hot spots” conjecture for convex planar domains and a different counterexample.

See the introduction to Bañuelos and Burdzy (1999) for a detailed review of various aspects of the “hot spots” conjecture, and a complete reference list. Our techniques are very close to those introduced in that paper so we will be rather brief and we ask the reader to consult that paper for more details.

We would like to thank Rodrigo Bañuelos and David Jerison for very useful advice, and the anonymous referee for suggesting a short proof of Lemma 1. The second author had the pleasure of being introduced to the problem by Jeff Rauch, the proposer, at E.N.S. Paris in 1995.

2. Domain construction

Before describing precisely our domain D, let us now give a short intuitive argument that provides some heuristic insight into our counterexample. Consider a planar domain that looks like a bicycle wheel with a hub, at least three very very thin spokes and a tire. Consider the heat equation in that domain with Neumann boundary conditions and an initial temperature such that the hub is “hot” and the tire is “cold.” Due to the fact that the cold arrives in the hub only via the spokes, the “hottest spot” of the wheel will be pushed towards the center of the hub. This implies that the second Neumann eigenfunction in the domain attains its maximum near the center of the hub and therefore not on the boundary of the domain.

For technical reasons that will become apparent in the proof, our domain D does not quite look like a bicycle wheel, but it does have a “hub,” three “spokes” and a “tire.”

We will use \underline{0} as an abbreviation for (0, 0). Let G be the group (containing six elements) generated by the symmetry s with respect to the horizontal axis and the rotation around \underline{0} by the angle 2π/3. We will use the point-to-set mapping T x = {σ(x), σ ∈ G}. Typically, T x contains six points. The meaning of T K for a set K is self-evident.

Suppose ε ∈ (0, 1/200) is a very small constant whose value will be chosen later in the proof. Let us name a few points in the plane,

A_{1} = \underline{0},  A_{2} = (1/7, \sqrt{3}/7),  A_{3} = (5, 1/100),
A_{4} = (11/2, 1/200),  A_{5} = (6, ε),  A_{6} = (13/2, 1/200),
A_{7} = (7, 1/100),  A_{8} = (8, 8\sqrt{3}),  A_{9} = (9, 9\sqrt{3}),
A_{10} = (235, 0).

Let D_{1} be the domain whose boundary is a polygon with consecutive vertices A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}, A_{9}, A_{10} and A_{1}. Let D_{2} be the closure of T D_{1} and let D_{3} be the interior of D_{2}. Finally, we obtain D be removing the line segment between (−18, 0) and (−16, 0) from D_{3}. We will show that D has the properties stated in Theorem 1.

The domain D_{3} has three holes while D has only two, because of the cut between (−18, 0) and (−16, 0).

Let α_{1} and α_{2} be the minimum and maximum of the angles between vectors \overrightarrow{A_{j}A_{j+1}}, j = 1, 2, …, 9, and the horizontal axis. We have chosen the points A_{j}, j = 1, …, 10, in such a way that α_{2} − α_{1} < π/2; this fact will be useful at the end of the proof, when we apply results of Bañuelos and Burdzy (1999).

3. The second eigenvalue is small

In this section, we will prove the following result.

Lemma 1. For every δ > 0, there exists ε_{0} > 0 such that for all ε ∈ (0, ε_{0}), the second Neumann eigenvalue μ_{2} in the domain D = D(ε) defined in Section 2 is not greater than δ.

Proof. Recall the points A_{j} defined in Section 2. Let A_{11} = (6+100ε/(1− 200ε), 0); this point lies at the intersection of the line containing A_{4} and A_{5} and the horizontal axis. Suppose that ε < 1/1600. Let d_{1}(z) denote the Euclidean distance between z and A_{11} and define for all z ∈ D_{1}, the function f_{1} as follows.

• f_{1}(z) = 0 if |z| > 6 or if d_{1}(z) < 400ε,

• f_{1}(z) = log(d_{1}(z)/400ε)/ log(1/800ε) if |z| ≤ 6 and d_{1}(z) ∈ [400ε, 1/2],

• f_{1}(z) = 1 if |z| < 6 and d_{1}(z) > 1/2.

Extend f_{1} into a continous function on D invariant under G. The function f_{1} is equal to 1 in the inside region of D, it is equal to 0 in the exterior region and it slopes from 1 to 0 in the inside parts of the narrow channels connecting the two regions. Note that f_{1} satisfies the Neumann boundary conditions on ∂D.

It is clear that when ε → 0, the two quantities ∫_{D} |f_{1}| and ∫_{D} |f_{1}|^{2} remain bounded and bounded away from 0, and it is elementary to check that ∫_{D} |∇f_{1}|^{2} → 0 when ε → 0.

Similarly, define a function f_{2} that is equal to 1 in the exterior domain, to 0 in the inside part of D and that slopes to 0 in the exterior parts of the narrow necks connecting the two. More precisely, let A_{12} lie at the intersection of the horizontal line and the line containing A_{5} and A_{6}, let d_{2}(z) = dist(z, A_{12}) and for z ∈ D_{1},

• f_{2}(z) = 0 if z < 6 or if d_{2}(z) < 400ε,

• f_{2}(z) = log(d_{2}(z)/400ε)/ log(1/800ε) if |z| > 6 and d_{2}(z) ∈ [400ε, 1/2],

• f_{2}(z) = 1 if |z| > 6 and d_{2}(z) > 1/2.

Extend f_{2} in a continuous fashion to D so that it is invariant under the action of G. Both ∫_{D} |f_{2}| and ∫_{D} |f_{2}|^{2} remain bounded and bounded away from 0 when ε → 0. It is easy to check that f_{2} satisfies the Neumann boundary conditions and that ∫_{D} |∇f_{2}|^{2} → 0 when ε → 0.

Finally, define for all ε > 0, the function f on D by

f(z) = \frac{f_{1}(z)}{∫_{D} f_{1}} − \frac{f_{2}(z)}{∫_{D} f_{2}} .

The function f satisfies the Neumann boundary conditions. As the supports of f_{1} and f_{2} are disjoint, it is clear that ∫_{D} |f|^{2} remains bounded and bounded away from 0 when ε → 0, and that ∫_{D} |∇f|^{2} → 0 when ε → 0. Since f is orthogonal to the constant function 1 (i.e., to the lowest eigenfunction) because ∫_{D} f = 0, we conclude that the second Neumann eigenvalue μ_{2} in D satisfies

0 < μ_{2} ≤ \frac{∫_{D} |∇f|^{2}}{∫_{D} |f|^{2}} .

Hence, μ_{2} = μ_{2}(ε) → 0 when ε → 0.

4. Nodal line of the second eigenfunction

In this section we will show that the nodal lines of any second eigenfunction (i.e., any eigenfunction corresponding to μ_{2}) are confined to a small subset of D when ε is small.

More precisely, consider any second Neumann eigenfunction φ_{2} in D. Let Γ be its nodal line, i.e., Γ = {x ∈ D, φ_{2}(x) = 0} (note that the line Γ is not necessarily connected).

Recall that s denotes the symmetry with respect to the horizontal axis, and define for j = 3, 4, 5, 6, 7 the line segments

K_{j} = \overline{A_{j}s(A_{j})}.

Let M^{o} denote the part of D between K_{3} and K_{7}, and define M = T M^{o} (i.e. “the union of the three bridges”).

The goal of this section is to prove that

Lemma 2. For all small enough ε > 0, Γ ⊂ M.

We divide the proof into several steps.

Step 1. Let Γ_{1} denote a connected component of the nodal line. Suppose that Γ_{1} intersects D\M and that the diameter of Γ_{1} is less than 10^{−10}. We will show that this assumption leads to a contradiction, if ε is sufficiently small.

As the diameter of Γ_{1} is less than 10^{−10} and Γ_{1} ⊄ M, it is easy to see that Γ_{1} has to cut off a domain D_{4} from D of diameter less than 10^{−6} (the boundary of D_{4} would consist of Γ_{1} and a piece of ∂D). It is also easy to prove that the first eigenvalue λ_{1} for the mixed problem in D_{4}, with the Dirichlet conditions on Γ_{1} and the Neumann conditions elsewhere on ∂D_{4}, is larger than some λ_{0} > 0, independent of ε < 1/200 and the shape and location of Γ_{1} (but using the fact that Γ_{1} intersects D \ M). Since λ_{1} = μ_{2}, we can adjust ε to make μ_{2} < λ_{0} using Lemma 1, and we can thus rule out the possibility that the diameter of Γ_{1} is less than 10^{−10}.

Step 2. We now collect some simple facts on reflected Brownian motion in D. We define some further sets: D \ M consists of two connected components, the inner one I (the one containing \underline{0}) and the exterior one E. Also, let M_{i}^{o} (M_{e}^{o}) denote the part of D between the line segments K_{3} and K_{5} (K_{5} and K_{7}). Put M_{e} = T M_{e}^{o} and M_{i} = T M_{i}^{o}.

In the rest of the paper, X_{t} = (X_{t}^{1}, X_{t}^{2}) will denote reflected Brownian motion in D (with normal reflection on ∂D). For all U ⊂ D, τ_{U} will denote the first hitting time of U by X; i.e.,

τ_{U} = inf{t > 0 : X_{t} ∈ U}.

Define Z_{t} = |X_{t}^{1} − 6|. As long as X_{t} stays in M^{o}, the process Z_{t} is a onedimensional Brownian motion reflected at 0, with some local time push always pointing away from 0, due to the normal reflection of X_{t} on the boundary of D. Hence, there is some p_{1} > 0, independent of ε < 1/200, such that Z_{t} may reach 1 within 1/2 unit of time, for any starting point of X_{t} inside M^{o}, with probability greater than 2p_{1}. In other words, if X_{0} ∈ M^{o} then with probability greater than 2p_{1}, the process X_{t} will hit K_{3} ∪ K_{7} before time t = 1/2. By symmetry, the process will be more likely to hit K_{3} first, if it starts to the left of K_{5} (i.e. in M_{i}^{o}), and it will be more likely to hit K_{7} first if it starts in M_{e}^{o}.

The same analysis applies to the other two “bridges” of D. Hence, there exists p_{1} > 0 such that for all ε ∈ (0, 1/200), for all x ∈ M_{i} and all x′ ∈ M_{e},

P (τ_{I} < 1/2 | X_{0} = x) > p_{1}  and  P (τ_{E} < 1/2 | X_{0} = x′) > p_{1}.

Suppose that γ ⊂ D is a connected set of diameter greater than 10^{−10} such that γ ∩ I ≠ ∅. It is easy and elementary to prove the following: There exists p_{2} > 0 such that for all ε ∈ (0, 1/200), for all x ∈ I, for all γ ⊂ D satisfying the above conditions,

P (τ_{γ} < 1/2 | X_{0} = x) ≥ P (τ_{γ} < 1/2, τ_{T K_{4}} > 1 | X_{0} = x) > p_{2}.

Note that p_{2} is independent of ε as the second probability in the last formula depends only on the connected component of D \ T K_{4} containing \underline{0} and this component is independent of ε.

One can also easily state and derive the counterpart of this result for the outer domain E.

Step 3. We are now ready to prove Lemma 2.

Assume first that Γ ∩ E = ∅ and that Γ ∩ I ≠ ∅. By the Courant Nodal Line Theorem (Courant and Hilbert (1953)) the nodal line Γ divides D into two connected components. Under the current assumptions, one of these two components is a subset of M ∪ I; we will call this component D_{c}. The first eigenvalue of the Laplacian in D_{c} with mixed boundary conditions, Dirichlet on Γ and Neumann elsewhere on the boundary of D_{c} is exactly μ_{2}.

Let Γ_{1} denote a connected component of Γ that intersects I. Step 1 implies that the diameter of Γ_{1} is at least 10^{−10}. Hence, using the results of Step 2, we get that for all ε < 1/200, for all x ∈ I ∩ D_{c},

P (τ_{Γ} ≤ 1/2 | X_{0} = x) ≥ p_{2}.

On the other hand, for all x ∈ M_{i}, using the strong Markov property at time τ_{I}, Step 2 and the last inequality, we get that

P (τ_{Γ} ≤ 1 | X_{0} = x) ≥ p_{2}p_{1}.

Finally, for all x ∈ M_{e} ∩ D_{c}, as E ∩ D_{c} = ∅,

P_{x}(τ_{Γ} ≤ 1 | X_{0} = x) ≥ P (τ_{E} ≤ 1 | X_{0} = x) ≥ p_{1}.

Hence, for all x ∈ D_{c},

P (τ_{Γ} ≤ 1 | X_{0} = x) ≥ p_{1}p_{2},

so that the Markov property applied at times n = 1, 2, …, implies that for all n ≥ 1,

P (τ_{Γ} ≥ n | X_{0} = x) ≤ (1 − p_{1}p_{2})^{n}

and consequently that μ_{2} ≥ − log(1 − p_{1}p_{2}). Note that p_{1} and p_{2} are independent of ε < 1/200. Hence, combining this with Lemma 1 shows that for small enough ε, one never has {Γ ∩ I ≠ ∅ and Γ ∩ E = ∅}.

The other two cases, namely {Γ ∩ E ≠ ∅ and Γ ∩ I = ∅} and {Γ ∩ E ≠ ∅ and Γ ∩ I ≠ ∅}, can be dealt with in the same way. Hence, for small ε, Γ ⊂ M.

Remark. In almost exactly the same way, one could prove that the nodal line is in fact confined to an arbitrarily small neighbourhood of T K_{6} when ε is sufficiently small, but Lemma 2 is sufficient for our purposes.

In the rest of the paper ε > 0 is assumed to be small enough so that the nodal line of any second Neumann eigenfunction in D is a subset of M.

5. The second eigenvalue is simple

Our proof of the fact that the second eigenvalue is simple is based on an almost trivial argument. However, this argument seems to be so useful that we state it as a lemma. It originally appeared in the proofs of Propositions 2.4 and 2.5 of Bañuelos and Burdzy (1999).

Lemma 3. Suppose that there exists z_{0} ∈ D such that the nodal line of any second Neumann eigenfunction does not contain z_{0}. Then the second eigenvalue is simple.

Proof. Suppose that φ_{2} and φ̃_{2} are two independent eigenfunctions corresponding to μ_{2}. By assumption, φ_{2}(z_{0}) ≠ 0 and φ̃_{2}(z_{0}) ≠ 0 so the function

x ↦ φ_{2}(x)φ̃_{2}(z_{0}) − φ_{2}(z_{0})φ̃_{2}(x)

is a nonzero eigenfunction corresponding to μ_{2}. Since it vanishes at z_{0}, we obtain a contradiction.

The lemma applies to our domain D because Γ ⊂ M.

6. Gradient direction for the second eigenfunction

This final part of the proof follows the arguments of Bañuelos and Burdzy (1999) so closely that we will only present a sketch and refer the reader to that paper for more details.

Let A denote the disc B(\underline{0}, 1/10), and let u(t, x) be the solution to the Neumann heat problem in D_{1} with the initial temperature u(0, x) = 1_{D_{1}∩A}(x).

We set u(t, y) = u(t, x) for all y ∈ T x and then extend the function u(t, x) to all x ∈ D by continuity. Due to the fact that u satisfies the Neumann boundary conditions in D_{1} and the symmetry, it is clear that u(t, x) solves the Neumann heat equation in D with the initial condition u(0, x) = 1_{A}(x).

Since the nodal line of φ_{2} is confined to M, the sign of 1_{A}(x)φ_{2}(x) is constant. We conclude that

c_{2} = ∫_{D} u(0, x)φ_{2}(x)dx = ∫_{A} φ_{2}(x)dx ≠ 0,

and so the second eigenfunction coefficient c_{2} is nonzero in the eigenfunction expansion for u(t, x),

u(t, x) = c_{1} + c_{2}φ_{2}(x)e^{−μ_{2}t} + ….

With no loss of generality, we can assume that c_{2} > 0, choosing the sign of φ_{2} accordingly. But (see, e.g., Proposition 2.1 of Bañuelos and Burdzy (1999)),

u(t, x) = c_{1} + c_{2}φ_{2}(x)e^{−μ_{2}t} + R(t, x), x ∈ D, t ≥ 0,

where R(t, x) converges to 0 as t → ∞ faster than e^{−μ_{2}t}, uniformly in x ∈ D. Hence, if we can show that for some fixed x, y ∈ D and all t > 0 we have u(t, x) ≥ u(t, y) then we must also have φ_{2}(x) ≥ φ_{2}(y).

Recall that α_{1} and α_{2} denote the minimum and maximum of the angles between vectors \overrightarrow{A_{j}A_{j+1}}, j = 1, 2, …, 9, and the horizontal axis and that α_{2}−α_{1} < π/2. In view of this fact, the arguments of Theorems 3.1 and 3.2 (see also Example 3.2) of Bañuelos and Burdzy (1999) can be easily adjusted to our domain D_{1} and imply that with our choice of the initial condition for u(t, x), we have u(t, x) ≥ u(t, y) whenever the angle between the vector \overrightarrow{xy} and the horizontal axis lies within (α_{2} − π/2, α_{1} + π/2). Hence, we have φ_{2}(x) ≥ φ_{2}(y) for all such x, y ∈ D_{1}. In particular, for every x ∈ D_{1}, φ_{2}(\underline{0}) ≥ φ_{2}(x). In order to prove the strong inequality we observe that for every x ∈ \overline{D_{1}} \ \{\underline{0}\}, we can find a nonempty open set F_{x} ⊂ D_{1} such that for every y ∈ F_{x}, the angles formed by the vectors \overrightarrow{\underline{0}y} and \overrightarrow{yx} with the horizontal axis belong to (α_{2} − π/2, α_{1} + π/2). If φ_{2}(\underline{0}) = φ_{2}(x) then φ_{2}(\underline{0}) = φ_{2}(y) = φ_{2}(x) for all y ∈ F_{x}. The remark following Corollary (6.31) in Folland (1976) may be applied to the operator Δ + μ_{2} to conclude that the eigenfunctions are real analytic and therefore they cannot be constant on an open set unless they are constant on the whole domain D. We see that φ_{2} attains its strict maximum in \overline{D_{1}} at the point \underline{0}. Since the same argument applies to every set σ(D_{1}) for all σ ∈ G, the function φ_{2} attains its strict maximum in D at \underline{0}. This completes the proof of Theorem 1.

University of Washington, Seattle, WA E-mail address: burdzy@math.washington.edu

Université Paris-Sud, Orsay, France

E-mail address: wendelin.werner@math.u-psud.fr

References

[1] R. Ba^{}ñuelos and K. Burdzy, On the “hot spots” conjecture of J. Rauch, to appear in J. of Functional Analysis, 1999.

[2] R. Courant and D. Hilbert, Methods of Mathematical Physics, Interscience Publishers Inc., New York, 1953.

[3] G. B. Folland, Introduction to Partial Differential Equations, Mathematical Notes, Princeton Univ. Press, Princeton, 1976.

[4] B. Kawohl, Rearrangements and Convexity of Level Sets in PDE, Lecture Notes in Mathematics 1150, Springer-Verlag, New York, 1985.

(Received March 3, 1998)