Nodal deficiency, spectral flow, and the Dirichlet-to-Neumann map

Abstract

It has been recently shown that the nodal deficiency of an eigenfunction is encoded in the spectrum of the Dirichlet-to-Neumann operators for the eigenfunction’s positive and negative nodal domains. While originally derived using symplectic methods, this result can also be understood through the spectral flow for a family of boundary conditions imposed on the nodal set, or, equivalently, a family of operators with delta function potentials supported on the nodal set. In this paper, we explicitly describe this flow for a Schrödinger operator with separable potential on a rectangular domain and determine a mechanism by which lower-energy eigenfunctions do or do not contribute to the nodal deficiency.

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Acknowledgements

The authors would like to thank Ram Band for interesting discussions and helpful suggestions regarding the manuscript. G.B. acknowledges partial support from the NSF under Grant DMS-1410657. G.C. acknowledges the support of NSERC Grant RGPIN-2017-04259. J.L.M. was supported in part by NSF Applied Math Grant DMS-1312874 and NSF CAREER Grant DMS-1352353.

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Appendix A. Example for a Rectangle

Appendix A. Example for a Rectangle

Let us consider first the one-dimensional eigenvalue problem for the case \(q(x) = 0\) from Sect. 3. Namely, we wish to compute the eigenvalues \(\{ \lambda _n (\sigma ) \}\) for \(\sigma \ge 0\).

A.1. \(\{ Z_k \} = \{ \frac{1}{2} \ell \}\)

The second Dirichlet eigenfunction for the Laplacian the interval \([0,\ell ]\) has a zero at \(\ell /2\). Using this nodal point to define the boundary conditions in \(\sigma \), as in Sect. 3, we look for the eigenvalues \(\lambda _n (\sigma )\). We will use the notation \(\lambda _n (\sigma ; 2)\) to denote the nth eigenvalue that arises from the spectral flow in \(\sigma \) set at the nodal point of the second Dirichlet eigenfunction. Symmetry considerations guarantee that the corresponding lowest eigenfunction, \(u_1 (x)= u_1 (x, \sigma ; 2)\), is symmetric with respect to \(\ell /2\). The eigenvalues \(\lambda _n (\sigma ; 2)\) in this case can be found by taking \(u_1 (x) = \sin (\kappa x)\) on \([0, \ell /2]\) for \(\kappa ^2 = \lambda _n\). Condition (17) gives

$$\begin{aligned} -2 u_1' \left( \frac{\ell }{2} \right) = \sigma u_1 \left( \frac{\ell }{2} \right) , \end{aligned}$$

and hence

$$\begin{aligned} \sigma = - 2 \kappa \cot \left( \kappa \frac{\ell }{2} \right) . \end{aligned}$$
(21)

Thus, \(\lambda _1 (\sigma ; 2) = \kappa ^2\) is given as the implicit solution to (21) for finding the lowest eigenvalue.

A.2. \(\{ Z_k \} = \{ \frac{1}{3} \ell , \frac{2}{3} \ell \}\)

Now, let us consider the next excited state, or the case of the nodal set given by 2 zeros equidistributed throughout the interval. As before, the lowest eigenfunction of \(L_\sigma \), denoted \(u_1 (x)= u_1 (x, \sigma ; 3)\), is symmetric with respect to \(\ell /2\) and we can write

$$\begin{aligned} u_1 (x)= \left\{ \begin{array}{ll} a \sin (\kappa x), &{}\quad x \in [0, \ell /3] \\ b \cos (\kappa (\ell /2-x)) &{}\quad x \in [\ell /3, \ell /2] \end{array} \right. \end{aligned}$$

Hence, conditions (16) and (17) at \(\ell /3\) imply

$$\begin{aligned} a \sin \left( \frac{ \kappa \ell }{3} \right) = b \cos \left( \frac{ \kappa \ell }{6} \right)&= c , \\ - \left( a \kappa \cos \left( \frac{ \kappa \ell }{3} \right) - b \kappa \sin \left( \frac{ \kappa \ell }{6} \right) \right)&= \sigma c, \end{aligned}$$

for \(c = u_1 (\ell /3)\). Solving out for c, we arrive at

$$\begin{aligned} \sigma = \kappa \left( \tan \left( \frac{ \kappa \ell }{6} \right) - \cot \left( \frac{ \kappa \ell }{3} \right) \right) , \end{aligned}$$

which can be solved implicitly for \(\lambda _1 (\sigma ; 3) = \kappa ^2\).

A similar approach applies to find the second eigenfunction \(u_2 (x) = u_2 (x,\sigma ; 3)\), which is anti-symmetric with respect to \(\ell /2\). Following the same logic, we arrive at

$$\begin{aligned} \sigma = -\kappa \left( \cot \left( \frac{ \kappa \ell }{3} \right) + \cot \left( \frac{ \kappa \ell }{6} \right) \right) , \end{aligned}$$

which can be solved implicitly for \(\lambda _2 (\sigma ; 3) = \kappa ^2\).

A.3. An example with nodal deficiency 3 on the rectangle

Let us now consider a rectangle of the form \([0,\pi ] \times [0, \alpha \pi ]\) with \(\alpha < 1\) but such that \(1-\alpha \ll 1\). We observe in this case that for the Laplacian with Dirichlet boundary conditions,

$$\begin{aligned} 1^2 + \left( \frac{1}{\alpha } \right) ^2 = \lambda _{1,1}< \lambda _{2,1}< \lambda _{1,2}< \lambda _{2,2}< \lambda _{3,1} < \lambda _{1,3} = 1^2 + \left( \frac{3}{\alpha } \right) ^2. \end{aligned}$$

Therefore, the sixth eigenvalue \(\lambda _6 = \lambda _{1,3}\) has 3 nodal domains and therefore nodal deficiency 3, see Fig. 6.

Fig. 6
figure6

The nodal deficiency count for the rectangle example in Appendix A

Setting \(\lambda _* = \lambda _6 = \lambda _{1,3}\), we obtain the spectral flow

$$\begin{aligned} \gamma _6 (\sigma )&= \lambda _*, \\ \gamma _5 (\sigma )&= 3^2 + \lambda _1^y (\sigma ; 3), \\ \gamma _4 (\sigma )&= 2^2 + \lambda _2^y (\sigma ; 3), \\ \gamma _3 (\sigma )&= 1^2 + \lambda _2^y (\sigma ; 3), \\ \gamma _2 (\sigma )&= 2^2 + \lambda _1^y (\sigma ; 3), \\ \gamma _1 (\sigma )&= 1^2 + \lambda _1^y (\sigma ; 3), \end{aligned}$$

which was the flow depicted in Fig. 1(right). The above equations can be analyzed to show that \(\gamma _2, \gamma _4, \gamma _5\) all cross \(\gamma _6\) as \(\sigma \rightarrow \infty \), whereas \(\gamma _1\) and \(\gamma _3\) do not.

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Berkolaiko, G., Cox, G. & Marzuola, J.L. Nodal deficiency, spectral flow, and the Dirichlet-to-Neumann map. Lett Math Phys 109, 1611–1623 (2019). https://doi.org/10.1007/s11005-019-01159-x

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Keywords

  • Nodal domains
  • Nodal sets
  • Dirichlet-to-Neumann map
  • Spectral indices

Mathematics Subject Classification

  • 58J50
  • 35B05
  • 35P05
  • 35J10