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The Neumann Laplacian has a zero constant mode

Example

Assume the Axiom of Choice and Countable Choice for the Sobolev interfaces. On $(0, π)$ , the real coefficient case of the divergence-form Laplacian (Uniformly elliptic divergence-form operators and their sesquilinear forms) has Neumann form $a_{N} (u, v) = \int_{0}^{π} u^{'} v^{'}$ on $H^{1} (0, π; R)$ (Integer-order Sobolev spaces and their norms). The constant function $u_{0} \equiv 1$ satisfies $a_{N} (u_{0}, v) = 0 = 0 \cdot \int_{0}^{π} u_{0} v$ for every $v \in H^{1} (0, π; R)$ , so $(0, 1)$ is a weak Neumann eigenpair (The Neumann spectrum and the constant zero mode). For every integer $k \geq 1$ , $u_{k} (x) = \cos (k x)$ is a weak Neumann eigenfunction with eigenvalue $k^{2}$ . The Neumann form is nonnegative and the constant mode has Rayleigh quotient $0$ , so the lowest weak Neumann eigenvalue on $H^{1} (0, π; R)$ is $0$ . On the mean-zero subspace ${u \in H^{1} (0, π; R) : \int_{0}^{π} u = 0}$ the first Rayleigh value is positive by Poincare--Wirtinger and at most $1$ , witnessed by $\cos x$ . The constant mode is exactly the zero mode removed by the mean-zero restriction, in contrast with the Dirichlet problem, where constants are not admissible.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; the interval $(0, π)$ ; the Neumann form $a_{N} (u, v) = \int_{0}^{π} u^{'} v^{'}$ on $H^{1} (0, π; R)$ ; the functions $u_{k} (x) = \cos (k x)$ for integers $k \geq 0$ ; and $v \in H^{1} (0, π; R)$ .

[F1]

One-dimensional representatives: every class in $H^{1} (0, π)$ has an absolutely continuous representative $v^{*}$ on $[0, π]$ with $(v^{*})^{'} = v^{'}$ almost everywhere, and the fundamental theorem of calculus holds for it (One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives, Fundamental theorem of calculus for absolutely continuous functions, Integer-order Sobolev spaces and their norms).

[F2]

Derivative calculus: $u_{k}^{'} (x) = - k \sin (k x)$ , $u_{k}^{''} (x) = - k^{2} \cos (k x)$ , and $u_{k}^{'} (0) = u_{k}^{'} (π) = 0$ for every $k \geq 0$ (The derivatives of sine and cosine are cosine and minus sine, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Quarter-turn values and shifts by pi/2 and pi).

[F3]

Neumann form and its zero mode: the natural Neumann weak identity has no boundary condition on the test function, the constant functions are weak Neumann eigenfunctions with eigenvalue $0$ , and the form is nonnegative with kernel the componentwise constants (The Neumann spectrum and the constant zero mode, Uniformly elliptic divergence-form operators and their sesquilinear forms, Zero-boundary Sobolev space as a norm closure).

[F4]

Mean-zero positivity: on the mean-zero subspace of a bounded connected extension domain the first Neumann Rayleigh value is positive, by Poincare--Wirtinger, and is characterised variationally (The first positive Neumann eigenvalue has the mean-zero Rayleigh characterisation, Poincare-Wirtinger on bounded connected extension domains by Rellich compactness, Dirichlet Laplacian eigenpairs on an interval).

Verification

technique · direct

1.1F1F2F3givenalgebra

The weak Neumann identities. By [F2], $u_{k}^{'}$ is continuously differentiable on $[0, π]$ , hence absolutely continuous (its derivative is bounded); with [F1], The product of two absolutely continuous functions is absolutely continuous makes $u_{k}^{'} v^{*}$ absolutely continuous with derivative $u_{k}^{''} v^{*} + u_{k}^{'} v^{'}$ almost everywhere, and the fundamental theorem [F1] gives $\int_{0}^{π} u_{k}^{'} v^{'} = [u_{k}^{'} v^{*}]_{0}^{π} - \int_{0}^{π} u_{k}^{''} v^{*} = 0 + k^{2} \int_{0}^{π} u_{k} v,$ because the boundary term vanishes by [F2]. Thus $a_{N} (u_{k}, v) = k^{2} \int_{0}^{π} u_{k} v$ for every $v \in H^{1} (0, π; R)$ ; for $k = 0$ this reads $0 = 0$ , giving the constant zero mode, and for $k \geq 1$ it says that $\cos (k x)$ is a weak Neumann eigenfunction with eigenvalue $k^{2}$ .

2.1F3step 1.1givenalgebra

Lowest eigenvalue on all of $H^{1}$ . The form is $a_{N} (u, u) = ∥ u^{'} ∥_{L^{2}}^{2} \geq 0$ with $a_{N} (u_{0}, u_{0}) = 0$ for the nonzero constant $u_{0}$ , so the infimum of the Rayleigh quotient over $H^{1} (0, π) ∖ {0}$ is $0$ , attained at the constants; in particular the lowest weak Neumann eigenvalue on $H^{1} (0, π)$ is $0$ .

3.1F3F4step 1.1givenalgebra∎

The mean-zero restriction. On the mean-zero subspace $V = {u : \int_{0}^{π} u = 0}$ Poincare--Wirtinger [F4] gives a positive constant $C$ with $∥ u ∥_{L^{2}} \leq C ∥ u^{'} ∥_{L^{2}}$ , so the Rayleigh quotient on $V ∖ {0}$ is bounded below by $C^{- 2} > 0$ , and by [F4] its infimum is the first positive Neumann Rayleigh value. Taking $v = \cos x$ in step 1.1 gives $a_{N} (\cos x, \cos x) = \int_{0}^{π} \sin^{2} x d x = π / 2$ while $\int_{0}^{π} \cos^{2} x d x = π / 2$ , so this value is at most $1$ ; the constant mode is exactly the element removed by the mean-zero restriction, in contrast with the Dirichlet problem where constants are excluded by the zero-trace domain.

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Sources

Richard S. Laugesen, Spectral Theory of Partial Differential Equations (University of Illinois lecture notes, arXiv:1203.2344, complete 120 pages) (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, 2020, complete 158-page graduate notes) (standard reference, not scraped)
Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer Universitext, 2011, complete 614-page text) (standard reference, not scraped)