On nodal lines of Neumann eigenfunctions

We present a new method for locating the nodal line of the second eigenfunction for the Neumann problem in a planar domain. The technique is based on the `mirror coupling' of reflected Brownian motions.

of the second Neumann eigenfunction lies on the line of symmetry.Then one can replace the original Neumann problem on the whole domain with the mixed Dirichlet-Neumann problem on a nodal subdomain.As it often happens in analysis, the Dirichlet problem (in this case, the mixed Dirichlet-Neumann problem) turns out to be considerably easier to deal with than the original Neumann problem.Hence, one would like to be able to find the location of the nodal line for the second Neumann eigenfunction in domains which are not necessarily symmetric.If this goal is achieved, some further progress on the "hot spots" conjecture may be expected.
Very few methods for finding the nodal line are known; the following short list is probably complete.
(a) If a domain has a line of symmetry, the nodal line lies on that line, subject to some natural extra assumptions.
(b) There are a handful of classes of domains, such as rectangles and ellipses, for which explicit formulae for eigenfunctions are known.
(c) One can find an approximate location of the nodal line in domains which are "long and thin" in the asymptotic sense; see Jerison (2000).
The purpose of this note is to add the following item to this list.
(d) In some domains, one can use the probabilistic method of "mirror couplings" to delineate a region which the nodal line must intersect.
It should be emphasized that one cannot expect to find the location of the nodal line of the second Neumann eigenfunction in the sense of an explicit formula, except in some trivial cases.What one can realistically hope to achieve is to obtain sufficiently accurate information about the nodal line so that this information can in turn be used to prove some other results of interest; the combination of Lemmas 2 and 3 of Burdzy and Werner (1999) is an example of such a result on nodal lines.
Finally, we would like to mention the results of Melas (1992)  Recall that the first eigenvalue for the Laplacian in D with Neumann boundary conditions is equal to 0 and the corresponding eigenfunction is constant.The second eigenvalue need not be simple but its multiplicity can be only 1 or 2 (Nadirashvili (1986(Nadirashvili ( , 1988))).Proof.Suppose that B is one of the nodal domains for a second eigenfunction and a.s.Since the stationary measure for X t is the uniform distribution in D, X t will hit C 2 with probability one, and so it will leave B * with probability 1. Recall that there are exactly two nodal domains-B is one belong to the connected set B * , their line of symmetry K τ Y (B 1 ) must also intersect B * .
We have and so µ * 2 ≥ µ 2 , but this contradicts the fact that µ 2 > µ * 2 .Our initial assumption that x ∈ B ⊂ D \ A for some nodal domain B must be false.
We will illustrate our main result with two examples.Before we do that, we recall a few crucial facts about mirror couplings from Burdzy and Kendall (2000) and Bañuelos and Burdzy (1999).
Suppose that D is a polygonal domain and I is a line segment contained in its boundary.Let J denote the straight line containing I and recall that K t denotes the mirror line.Let H t denote the "hinge," i.e., the intersection of K t and J.Note that H t need not belong to ∂D.Suppose that for all t in [t 1 , t 2 ], the reflected Brownian motions X t and Y t do not reflect from any part of ∂D except I. Let α t denote the smaller of the two angles formed by K t and J. Then all possible movements of K t have to satisfy the following condition.
(M) The hinge does not move within the time interval [t A domain D with piecewise smooth boundary can be approximated by polynomial domains D n .Mirror couplings in the approximating domains D n converge weakly to a mirror coupling in D. One can deduce which motions of the mirror K t in D are possible by analyzing all allowed movements of mirrors in D n 's and then passing to the limit.We will not present the details of this limit theorem here.The readers who are concerned about full rigor should add an extra assumption to Example 2 that D is a polygonal domain.
Example 1. Suppose that D is an obtuse triangle with vertices C 1 , C 2 and C 3 .Let C 3 be the vertex with an angle greater than π/2.Let C j C k denote the line segment with endpoints C j and C k , and let d(C j , C k ) denote the distance between these points.Let  (ii) The nodal line lies within A 1 .
Let α t denote the angle formed by the mirror K t with C 1 C 2 .Let β 1 and β 2 denote the angles such that if α t = β 1 then K t is perpendicular to C 2 C 3 and when α t = β 2 then K t is perpendicular to C 1 C 3 .An argument totally analogous to the proof of Theorem 3.1 of Bañuelos and Burdzy (1999) shows that if with X 0 = x and Y 0 = y.We will show that K t ∩ D ⊂ A for all t ≥ 0.
We have K 0 ∩D ⊂ A, by the definition of y.Assume that it is not true that K t ∩D ⊂ A for all t.We will show that this assumption leads to a contradiction.If follows from this assumption that for some ε > 0 there exists Hence, the processes X t and Y t can reflect only on We will argue heuristically that it is impossible to obtain much sharper estimates for the nodal line location without imposing some extra assumptions on the obtuse triangle.
To see this, consider an obtuse triangle with two angles very close to π/2-one slightly less than the right angle and another one slightly larger than that (see Figure 2).Such a triangle has a shape very close to a thin circular sector.Let us assume that the diameter of the triangle is equal to 1.The nodal line in a circular sector is an arc with center at its vertex C. The nodal line distance from C is equal to a 0 /a 1 ≈ 0.63, where a 0 is the first positive zero of the Bessel function of order 0 and a 1 is the first positive zero of its derivative.This follows from known results on eigenfunctions in discs and estimates for Bessel function zeroes, see Bandle (1980)  Little is known about eigenfunctions in triangles different from equilateral; see Pinsky (1980Pinsky ( , 1985) ) for that special case.

Theorem 1 .
The nodal set is the set of points in D where the second eigenfunction vanishes.If D is not simply connected, the nodal set need not be a connected curve.The nodal set of any second Neumann eigenfunction divides the domain D into exactly two nodal domains (connected and open sets).Suppose that A ⊂ D is closed, x ∈ D\A, and D 1 is the connected component of D \A which contains x.Assume that for some y ∈ D \D 1 , the mirror K t for the coupling starting from (X 0 , Y 0 ) = (x, y) satisfies K t ∩ D ⊂ A for all t ∈ [0, ζ] a.s.Then for any second Neumann eigenfunction in D, none if its nodal domains can have a closure which is a subset of D \ A containing x.
and the following pairs of line segments are perpendicular: C 4 C 8 and C 2 C 3 , C 6 C 9 and C 1 C 3 , C 5 C 10 and C 1 C 3 , C 7 C 11 and C 2 C 3 .See Figure 1.
for all t ≥ 0 a.s.Consider any point x ∈ D \ A to the left of the arc C 4 C 5 .Let y ∈ D be such that C 1 , x and y lie on the same line and d(C 1 , y) = d(C 1 , C 3 ).Consider a mirror coupling (X t , Y t )

C 1 C 2 and T 1 ,
2 and C 1 C 3 on the time interval [T 2 , T 1 ].According to (M), the distance from C 1 to K t cannot decrease between times T which contradicts the definition of these times.Claim (i) now follows from Theorem 1.The second claim is a consequence of the first one and the bounds on the direction of the gradient of the second eigenfunction proved in Theorem 3.1 of Bañuelos and Burdzy (1999).

C 3 is
perpendicular to C 1 C 2 , and let D 1 be the larger of the two connected components of D \ C 12 C 3 .Then the nodal line for the second Neumann eigenfunction is contained in D 1 .

Figure 3
Figure 3 shows an example of D and the corresponding rectangle A.
on the nodal lines in the Dirichlet case.2.Mirror couplings and nodal lines.Suppose that D is a planar domain (open and connected set) with a piecewise smooth boundary.Informally speaking, a mirror coupling is a pair (X t , Y t ) of reflecting Brownian motions in D, such that the line of symmetry K t between X t and Y t does not change on any interval (s, u) such that X t / which is a contradiction.We conclude that τ X (B * ) < τ Y (B 1 ) a.s.Let µ 2 > 0 denote the second Neumann eigenvalue.Then µ 2 is the first eigenvalue for the mixed problem in the nodal domain B, with the Neumann boundary conditions on ∂D and the Dirichlet boundary conditions on the nodal line.The fact that B * is strictly larger than B easily implies that µ 2 > µ * 2 , where µ * 2 is the analogous mixed eigenvalue for B * .Using the well known identification of Brownian motion density with the heat equation solution, we obtain from Proposition 2.1 of Bañuelos and Burdzy (1999) that lim t→∞ 1 , t 2 ], i.e., H t = H t 1 for all t ∈ [t 1 , t 2 ].The angle α t is a non-decreasing function of t on [t 1 , t 2 ].
Suppose that D is a triangle with vertices C 1 , C 2 and C 3 and that the angle at C 3 is greater than π/2.Let C 12 ∈ C 1 C 2 be the unique point such that C 12 , p. 92.Our methods yield 0.5 as the lower bound for the distance of the nodal line from C; we see that this estimate cannot be improved beyond 0.63.