Lower bounds for the Steklov eigenvalue problem

Lower bounds for the Steklov eigenvalue problem

Mémoire

Salman Davoudi

Maîtrise en mathématiques - avec mémoire

Maître ès sciences (M. Sc.)

Lower bounds for the Steklov eigenvalue problem

Mémoire

Salman Davoudi

Sous la direction de:

Alexandre Girouard, Research Director Javad Mashreghi, Research Co-Director

Résumé

Le problème de Steklov est un problème spectral qui provient de la mécanique des fluides. C’est un problème de valeur propre dont les paramètres spectraux sont dans la condition au bord. Son spectre coïncide avec celui de l’opérateur de Dirichlet-Neumann. Le spectre du problème de Steklov est discret lorsque l’opérateur de trace est compact, ce qui est le cas lorsque la frontière du domaine est lipschitzienne. Dans ce mémoire, nous prouvons de deux manières différentes l’effondrement vers 0 du spectre de Steklov pour un domaine en forme d’haltère dégénérant vers deux disques. On se concentre par la suite sur les domaines dont la frontière n’est pas uniformément lipschitzienne. Nous donnons deux exemples pour montrer que l’opérateur de trace n’est pas compact pour ces domaines. De plus, nous présentons une borne inférieure pour la première valeur propre σ₁ non nulle du problème de Steklov pour les domaines ayant deux axes de symétrie. Enfin, nous présentons des bornes inférieures pour le problème des valeurs propres Steklov pour les domaines étoilés. Ces résultats sont dus à J. R. Kuttler et V. G. Sigillito. [7,8].

Abstract

The Steklov problem is a spectral problem whose origin lies in the mechanics of fluids. It is an eigenvalue problem with spectral parameters in the boundary conditions, which has various applications. Its spectrum coincides with that of the Dirichlet-to-Neumann operator. The spectrum of the Steklov’s problem is discrete when the trace operator is compact. In this master’s thesis, we prove the collapse of the Steklov spectrum for a dumbbell domain in two manners. We will focus on non-Lipschitz domains. We give two examples to show that the trace operator is not compact for non-Lipschitz domains. Furthermore, we present a lower bound to the first non-zero eigenvalue σ1 of the Steklov problem for domains having two axes

of symmetry. Finally, we present lower bounds for the Steklov eigenvalue problem for star-shaped domains. These results were due to J. R. Kuttler and V. G. Sigillito restrict domains to domains with two axes of symmetry or star-shaped domains [7,8].

Contents

1 Steklov eigenvalue problem 2

1.1 Steklov eigenvalue problem . . . 2

1.2 Eigenfunctions on the disk . . . 3

2 Comparison of Steklov eigenvalues with Dirichlet eigenvalues 7 2.1 Monotonicity property of Dirichlet eigenvalues. . . 7

2.2 Minimax Principle . . . 12

2.3 Counterexample to eigenvalue monotonicity for the Neumann problem . . . 20

2.4 Counterexample to eigenvalue monotonicity for the Steklov problem . . . . 22

3 Collapse of the Steklov spectrum for a dumbbell domain 24 3.1 First proof of the collapse of the Steklov spectrum for a dumbbell domain . 25 3.2 Second proof of the collapse of the Steklov spectrum for a dumbbell domain 29 4 Trace operator for non-Lipschitz domain 33 4.1 Trace operator . . . 33

4.2 Example of a non-Lipschitz domain . . . 34

4.3 A non compact inclusion H1(Ω) −→ L2_{(Ω) . . . . 35}

4.4 A non bounded trace T : H1(Ω) −→ L2(∂Ω) . . . 36

4.5 Rellich theorem and trace theorem . . . 38

5 An inequality for a Steklov eigenvalue by the method of defect 39 5.1 Preliminaries . . . 39

5.2 The nodal-line theorem . . . 40

5.3 The method of defect. . . 44

5.4 Some examples . . . 45

6 Lower bounds for Steklov eigenvalues problem for star-shaped domains 47 6.1 Preliminaries . . . 47

6.2 Derivation of the inequalities . . . 47

Conclusion 50

Acknowledgments

I would like to express my special appreciation and thanks to my advisor professor Alexandre Girouard, you have been a tremendous professor for me. I have been extremely lucky to have a supervisor who cared so much about my work, and who responded to my questions and queries so promptly. I would like to thank my research co-director, Javad Mashreghi. I am so deeply grateful for their help, professionalism, valuable guidance and financial support throughout this project and through my entire program of study. I do not have enough words to express my deep and sincere appreciation. I am indebted to them for their help.

Introduction

The aim of this master’s thesis is to present some aspects of the Steklov problem. This problem was introduced at the beginning of the 20th century, by Vladimir Steklov [14] , in the context of the modeling of the vibrations of a fluid subjected to gravity. Then it attracted the attention of mathematicians both in the search for applications and as a problem at the frontier of analysis and geometry, especially in spectral geometry. This branch of mathematics studies the links between the shape of a space and the eigenvalues of operators naturally defined on it.

The problem of Steklov is as follows. Given a bounded domain Ω ⊂ Rnsufficiently regular, we look for all the real numbers σ for which there exists a non-zero function f ∈ C2(Ω) ∩ C1_(Ω)

which satisfies the following equations: (

∆f = 0 in Ω,

∂f

∂V = σf on ∂Ω,

where ∆ is the Laplacian and _∂V∂ is outward normal derivative. When the trace operator is compact, the spectral theorem for this problem states that the eigenvalues form a sequence 0 = σ0< σ1 ≤ σ2 ≤ ... % ∞. In this master’s thesis, we prove the collapse of the Steklov

spec-trum for a dumbbell domain in two manners (i.e. lim_→0+σ_n(Ω_) = 0 for all n = 1, 2, ... ). We give two examples to show that the trace operator can be non compact for non-Lipschitz do-mains. Furthermore, we give a lower bound to the first non-zero eigenvalue σ2 of the Steklov

problem for plane regions having two axes of symmetry. Finally, we give lower bounds for Steklov eigenvalue problem for star-shaped domains. These results are due to Kuttler and Sigillito [7,8].

Chapter 1

Steklov eigenvalue problem

The Steklov problem is an eigenvalue problem with the spectral parameter in the boundary conditions, which has various applications. Its spectrum coincides with that of the Dirichlet-to-Neumann operator. Over the past years, there has been a growing interest to give lower bounds to the non-zero eigenvalues σ_n of the Steklov problem for domains which are open bounded subsets in RN. These lower bounds dependents on the shape of the domain. A lower bound to the first non-zero eigenvalue σ1 of the Steklov problem for domains having two

axes of symmetry and lower bounds for Steklov eigenvalue problem for star-shaped domains is given by J.R. Kuttler and V. G. Sigillito [7,8].

1.1

Steklov eigenvalue problem

We consider the following definitions from the partial differential equations of L.C. Evans [4]. Definition 1.1.1. A domain Ω is an open bounded connected subset in RN.

Definition 1.1.2. Let Ω be a planar domain with Lipschitz boundary. The Sobolev space W1,2(Ω) consists of all locally summable functions u : Ω −→ R such that, Du exists and belongs to L2_(Ω).

Definition 1.1.3. The support of function g is

supp g ={x ∈ Ω : g(x) 6= 0}

Definition 1.1.4. A function h is weakly harmonic in a domain Ω if Z

Ω

h∆g dx = 0

for all g with compact support in Ω and continuous second derivatives. Definition 1.1.5. We define

The Steklov eigenvalue problem is ( ∆u = 0 in Ω, ∂ ∂Vu = σ u on ∂Ω, (1.1)

where _∂V∂ is the outward normal derivative.

By spectral theorem for Steklov problem [9], we know that the spectrum of the Steklov problem is discrete for Lipschitz domains and the eigenvalues

0 = σ0≤ σ1(Ω) ≤ σ2(Ω) ≤ σ3(Ω) ≤ ... % ∞

satisfy the following variational characterization σn(Ω) = infEsup06=u∈E

R Ω|∇u| 2 _dx R ∂Ωu2ρ ds , n ≥ 1. (1.2)

Here the infimum is taken over all n + 1-dimensional subspaces E of the Sobolev space H1(Ω). Note that, the Steklov spectrum always starts with the eigenvalue σ₀= 0, and the correspond-ing eigenfunctions are constant.

1.2

Eigenfunctions on the disk

In this section we compute the eigenfunctions on the disk.

Example 1. Consider a disk Ω = {x ∈ R2 _{: |x| < 1}. In order to compute the eigenfunctions}

on the disk, we need to separate variables using polar coordinates r and Θ. We first derive a formula for the Laplacian in polar coordinates. Since r =px2_{+ y}2 _{and tan Θ =} y

x, we have ∂u ∂x = ∂u ∂r ∂r ∂x + ∂u ∂Θ ∂Θ ∂x, ∂u ∂y = ∂u ∂r ∂r ∂y+ ∂u ∂Θ ∂Θ ∂y. Thus, we have ∂u ∂x = ∂u ∂r 2x 2px2_{+ y}2 + ∂u ∂Θ −y x2 1 + (y_x)2, ∂u ∂y = ∂u ∂r 2y 2px2_{+ y}2 + ∂u ∂Θ 1 x 1 + (y_x)2. Thus, ∂u ∂x = ∂u ∂r x r − ∂u ∂Θ y r2, ∂u ∂y = ∂u ∂r y r + ∂u ∂Θ x r2.

Furthermore, We have ∂2u ∂x2 = ∂ ∂x( ∂u ∂x) = ∂ ∂x( ∂u ∂r x r − ∂u ∂Θ y r2) (1.3) = ( ∂ ∂x( ∂u ∂r)).( x r) + ( ∂ ∂x( x r)). ∂u ∂r − ( ∂ ∂x( ∂u ∂Θ)). y r2 − ∂u ∂Θ( ∂ ∂x. y r2) (1.4) = x 2 r2 ∂2u ∂r2 − 2 xy r3 ∂2u ∂r∂Θ+ y2 r4 ∂2u ∂Θ2 + y2 r3 ∂u ∂r + 2 xy r4 ∂u ∂Θ. (1.5) Therefore, ∂2u ∂x2 = x2 r2 ∂2u ∂r2 − 2 xy r3 ∂2u ∂r∂Θ+ y2 r4 ∂2u ∂Θ2 + y2 r3 ∂u ∂r + 2 xy r4 ∂u ∂Θ and similarly ∂2u ∂y2 = y2 r2 ∂2u ∂r2 + 2 xy r3 ∂2u ∂r∂Θ+ x2 r4 ∂2u ∂Θ2 + x2 r3 ∂u ∂r − 2 xy r4 ∂u ∂Θ. It follows that the Laplacian has the form

∆u = ∂ 2_u ∂x2 + ∂2u ∂y2 = ∂2u ∂r2 + 1 r2 ∂2u ∂Θ2 + 1 r ∂u ∂r. Therefore, the Laplacian in polar coordinates takes the form

∆ = ∂ 2 ∂r2 + 1 r ∂ ∂r + 1 r2 ∂2 ∂Θ2.

We now look for the solutions of the form u(r, θ) = R(r)Φ(θ). From ∆u = 0, we get R(r)00Φ(θ) + 1

rR(r)

0_{Φ(θ) +} 1

r2R(r)Φ

00_{(θ) = 0.}

Thus, by multiplying both sides of equation by _R(r)Φ(θ)r2 , we have r2 R(r)Φ(θ)  R(r)00Φ(θ) + 1 rR(r) 0 Φ(θ) + 1 r2R(r)Φ 00 (θ)  = 0, and therefore r2R 00_(r) R(r) + r R0 R(r)+ Φ00(Θ) Φ(Θ) = 0. This means that there exists k such that

r2R 00_(r) R(r) + r R0 R(r) = k 2 _{= −}Φ00(Θ) Φ(Θ) . Thus, we have −Φ00(Θ) = k2Φ(Θ) (1.6)

and r2R 00_(r) R(r) + r R0 R(r) = k 2_. So, r2R00(r) + rR0(r) − k2R(r) = 0 (1.7) Furthermore, R(r)00Φ(Θ) +1 rR(r) 0_{Φ(Θ) =} k2 r2R(r). R(r)00Φ(Θ) +1 rR(r) 0 Φ(Θ) −k 2 r2R(r) = 0.

The equation (1.6) has the general solution

Φ(θ) = A cos(kΘ) + B sin(kΘ), (1.8)

where A and B are two constants. The equation (1.7) is Euler equation. We consider R(r) = rm. Then we have

r2(m(m − 1)rm−2) + r(mrm−1) − k2.rm= 0 =⇒ (m(m − 1) + m − k2)rm= 0 =⇒ (m2− k2)rm = 0

=⇒ m = ±k.

Thus, we have R(r) = ark+ br−k, where a and b are constants. If k = 0, then r2R00+ rR0 = 0. We consider S(r) = R0(r). So, r2S0(r) + rS(r) = 0 =⇒ 1 S ds dr = −r =⇒ ds s = − dr r =⇒ Z _ds s = − Z _dr r + c =⇒ log |S| = − log |r| + c. Thus, we have log |S| = − log |r| + c.

Setting c = log |c|, we have S = _rc. Finally, we integrate S to recover R R(r) = Z R0dr + d = Z Sdr + d = Z _c rdr + d = c log |r| + d. In conclusion the general solution of (1.7) is

R(r) = (

crk+ dr−k if k 6= 0.

Since the function log is bounded, c = 0 in (1.9). Thus, we have R(r) = crk_{+ dr}−k _{if k 6= 0.}

Putting (1.8) and (1.9) together, we obtain

uk(r, Θ) =     Akcos(kΘ) + Bksin(kΘ)  rk+  Ckcos(kΘ) + Dksin(kΘ)  rk if k 6= 0, A0+ C0 log |r| + d if k = 0. (1.10)

Since the function log is bounded, c = 0 in (1.9). Thus, we have uk(r, Θ) =  Akcos(kΘ) + Bksin(kΘ)  rk+  Ckcos(kΘ) + Dksin(kΘ)  rk

We should have u(r, Θ) = u(r, Θ + 2π). All points (r cos Θ, r sin Θ) are the same as the points with coordinates (r cos(Θ + 2π), r sin(Θ + 2π)). But

k = n ∈ Z ⇐⇒ (

cos(kΘ) = cos(k(Θ + 2π)),

sin(kΘ) = sin(k(Θ + 2π)). (1.11)

Thus, k must be an integer. We may as well as take to be non negative since the solution where k = n and k = −n are actually coincide. Therefore, the eigenvalues of the unit disk are

0, 1, 1, 2, 2, ..., k, k, ...,

and the corresponding eigenfunctions in polar coordinates (r, Θ) are given by 1, r sin Θ, r cos Θ, ..., r sin kΘ, r cos kΘ, ... .

We know that we have all eigenfunctions because (cos kΘ, sin kΘ) is an orthonormal basis of L1(∂Ω).

Chapter 2

Comparison of Steklov eigenvalues

with Dirichlet eigenvalues

2.1

Monotonicity property of Dirichlet eigenvalues

In this chapter we consider the eigenvalue problem with Dirichlet boundary conditions before the Steklov eigenvalue problem. If we understand property of Dirichlet eigenvalues, it helps to understand better Steklov eigenvalue problem. These results are due to J. Levandosky [12]. 2.1.1 Eigenvalues of the Laplacian

Let Ω ⊂ RN be a domain defined in (1.1.1). We consider the following eigenvalue problem with Dirichlet boundary conditions,

(

−∆v = λv, x ∈ Ω,

v = 0, x ∈ ∂Ω. (2.1)

In general, it is difficult to explicitly compute eigenvalues for a given domain Ω ⊂ RN. We prove that eigenvalues are minimizers of a certain functional. This fact will allow us to approximate eigenvalues for given domains Ω ⊂ RN.

Let 0 < λ1 ≤ λ2 ≤ ... be the eigenvalues of (2.1). For a given function w ∈ H1(Ω), we define

the Rayleigh quotient of w on Ω as ||∇w||2 L2_(Ω) ||w||2 L2_(Ω) = Z Ω |∇w|2 dx Z Ω |w|2 dx .

We consider the following theorem from [3].

Proof. Let v be an eigenfunction with corresponding eigenvalue λ. Then λ Z Ω v2 dx = − Z (∆v)v dx = Z Ω |∇v|2 dx − Z ∂Ω v∂v ∂V dS(x) = Z Ω |∇v|2 dx.

Where V is outward normal vector on the boundary Ω and the third equality holds because in (2.1) we have v = 0 in ∂Ω. Therefore, λ Z Ω v2 dx = Z Ω |∇v|2 dx ≥ 0.

Further, we claim that

Z Ω |∇v|2 _{dx > 0.} Suppose Z Ω

|∇v|2 dx = 0, then |∇v| = 0 which implies v is constant since Ω is a connected domain. But, by assumption v = 0 on ∂Ω. Therefore, if v is constant on Ω and v = 0 on ∂Ω, then v ≡ 0. However, the zero function is not an eigenfunction. Therefore,

λ Z Ω v2dx = Z Ω |∇v|2 dx > 0. which implies λ > 0. Definition 2.1.1. Let Y ≡ {w ∈ C2(Ω) : w 6≡ 0, w(x) = 0 f or x ∈ ∂Ω}.

We call Y the set of trial functions for (2.1). We consider the following theorem from [3]. Theorem 3. (Minimum principle for the first eigenvalue) Suppose there exists a function u ∈ Y such that u minimizes the Rayleigh quotient over all trial functions w ∈ Y . That is,

m := ||∇u||

2

||u||2 = minw∈Y

(

||∇w||2

||w||2

) .

Then m is the first eigenvalue of (2.1). That is, m = λ1and u is a corresponding eigenfunction.

Proof. Suppose u is the minimizer of the Rayleigh quotient and m is the Rayleigh quotient of u. That is, m = Z Ω |∇u|2 dx Z Ω |u|2 dx . (2.2)

Pick a function v ∈ Y . Let f () ≡ Z Ω |∇(u + v)|2 dx Z Ω |(u + v)|2 dx .

If u minimizes the Rayleigh quotient, then f must satisfy f0(0) = 0. Taking the derivative of f , we see that f0() = Z Ω (u + v)2 dx  2 Z Ω |∇u.∇v + |∇v|2 dx  − Z Ω 2(u + v)v dx Z Ω |∇(u + v)2|  Z Ω (u + v)2 dx 2 . Therefore, f0(0) = Z Ω u2 dx  2 Z Ω |∇u.∇v dx  −  2 Z Ω uv dx Z Ω |∇u|2 _dx  Z Ω u2 dx 2 . Now, f0(0) = 0 implies Z Ω u2 dx Z Ω ∇u.∇v dx  = Z Ω uv dx Z Ω |∇u|2 dx  which implies Z Ω ∇u.∇v dx = Z Ω |∇u|2 dx Z Ω u2 dx Z Ω uv dx = m Z Ω uv dx.

Using the divergence theorem, we have

− Z Ω (4u)v dx + Z ∂Ω ∂u ∂vv dS(x) = m Z Ω uv dx.

By assumption, v = 0 on ∂Ω. Therefore, the boundary term vanishes. Therefore, − Z Ω (4u)v dx = m Z Ω uv dx

for all v ∈ Y . Now, as this is true for all trial functions v ∈ C2(Ω), v 6≡ 0, and v(x) = 0 for x ∈ ∂Ω, we conclude that

which means that u is an eigenfunction of (2.1) with corresponding eigenvalue m.

It only remains to show that m is the smallest eigenvalue. Suppose v is another eigenfunction of (2.1) with corresponding eigenvalue λi. We just need to show that λi ≥ m. Using the

Divergence theorem and the fact that v vanishes on the boundary, we have

m = ||∇u|| 2 ||u||2 ≤ ||∇v||2 ||v||2 = Z Ω |∇v|2 dx Z Ω |v|2 dx = − Z Ω (4v)v dx Z Ω |v|2 dx = λi Z Ω |v|2 dx Z Ω |v|2 dx = λi.

Therefore, the theorem is proved.

Theorem 4. (Minimum principle for the nth eigenvalue) Fix an integer n ≥ 1. Let vi,

i = 1, ..., n − 1 be the first n − 1 eigenfunctions of (2.1). Without loss of generality, these eigenfunctions may be chosen to be orthogonal in L2_{(Ω). Let}

Yn≡ {w : w ∈ C2(Ω), w 6≡ 0, w = 0 f or x ∈ ∂Ω, hw, vii = 0 f or i = 1, ..., n − 1},

where hw, vii =

R

Ωwvidx is inner product in L

2_{(Ω). Suppose there exists a function u} n ∈ Yn

which minimizes the Rayleigh quotient over all functions w ∈ Yn. That is, suppose

mn:= ||∇un||2 ||un||2 = min w∈Yn ||∇w||2 ||w||2 .

Then mnis the nth eigenvalue of (2.1). That is, λn= mn and un is an eigenfunction of (2.1)

with eigenvalue mn.

Proof. Suppose un∈ Yn is the minimizer of the Rayleigh quotient over all functions w ∈ Yn.

That is, mn= ||∇un||2 ||un||2 = min w∈Yn ||∇w||2 ||w||2 .

Fixing v ∈ Yn, defining f () as before and using the fact that f0(0) = 0, we see that

Z

Ω

(∆un+ mnun)v dx = 0.

This is true for any v ∈ Yn. Therefore, we conclude that

Z

Ω

(∆un+ mnun)v dx = 0. (2.3)

for all trial functions v which satisfy hv, v_ii = 0 for i = 1, ..., n − 1. To conclude that

we need to show that (2.3) is true for all trial functions (not just those trial functions which are orthogonal to the first n − 1 eigenvalues).

Now let h be an arbitrary trial function. Let

v(x) := h(x) − n−1 X k=1 ckvk(x) where, ck= hh, vki hv_k, vki ,

and the vi are the first n − 1 eigenfunctions. We claim that

Z Ω (∆un+ mnun)h dx = 0. Indeed, we have Z Ω (∆un+ mnun)h dx = Z Ω (∆un+ mnun)[v + n−1 X k=1 ckvk] dx = Z Ω (∆un+ mnun)v dx + n−1 X k=1 ck Z Ω (∆un+ mnun)vk dx.

Now, first, we know that v is orthogonal to vi for i = 1, ..., n − 1, and, therefore, the first

term on the right-hand side above vanishes. We verify this as follows. Let vi be an arbitrary

eigenfunction for i = 1, ..., n − 1. Then

hv, v_ii = Z Ω vvi dx = Z Ω (h − n−1 X k=1 ckvk)vi dx = Z Ω hvi dx − n−1 X k=1 ck Z Ω vkvi dx = Z Ω hvi dx − ci Z Ω vividx = Z Ω hvi dx − Z Ω hvidx = 0

using the definition of c_i and the fact that eigenfunctions are orthogonal. Therefore, Z

Ω

(∆un+ mnun)v dx = 0.

Next, we claim that for all eigenfunctions vi, i = 1, ..., n − 1,

Z

Ω

(∆un+ mnun)vi dx = 0.

We prove this claim as follows. Fix an eigenfunction v_i. Let λ_ibe its corresponding eigenvalue. Then Z Ω (∆un+ mnun)vi dx = − Z Ω (∇un.∇vi) dx + Z ∂Ω ∂un ∂V vi dS(x) + Z Ω mnunvidx = Z Ω (un.∇vi) dx − Z ∂Ω un ∂vi ∂V dS(x) + Z ∂Ω ∂un ∂V vi dS(x) + Z Ω mnunvidx

= (−λi+ mn)

Z

Ω

unvidx.

By assumption, un∈ Ynwhich implies unis orthogonal to the first n−1 eigenvalues. Therefore,

Z Ω unvidx = 0 for i = 1, ..., n − 1. Hence, Z Ω (∆un+ mnun)vi dx = 0.

Consequently, we conclude that Z

Ω

(∆un+ mnun)h dx = 0,

where h is an arbitrary trial function. Finally, we conclude that ∆un+ mnun= 0.

and, therefore, un is an eigenfunction with eigenvalue mn.

Now, clearly, mn ≥ λn−1 ≥ λn−2 ≥ ... because Yn⊂ Yn−1⊂ .... We can prove that the other

eigenvalues λ_n+1, λn+2, ... are larger than mn using the same technique as in the previous

theorem, and the fact that the eigenfunctions v_k for k ≥ n + 1 satisfy hv_k, vii = 0 for i =

1, ..., n − 1.

2.2

Minimax Principle

In this section, we present another theorem regarding the eigenvalues of (2.1). This theorem is known as the minimax principle. It will allow us to prove a relationship between eigenvalues of a domain contained within a larger domain. In particular, we will show that if Ω₁ ⊂ Ω₂, then λ_k(Ω1) ≥ λk(Ω2), where λk is the kth eigenvalue of (2.1). This fact will give us another

means of approximating eigenvalues of arbitrary domains Ω. Before we get to these results, we present some motivation for the minimax principle.

Rayleigh-Ritz approximation

The Rayleigh-Ritz method is a method of finding approximations to eigenvalue equations that cannot be solved explicitly. We use the Rayleigh–Ritz method to compute the approximate solutions to the Dirichlet eigenvalue problem 2.1. These results are due to Kelly [6]. Let w1, ..., wn be n arbitrary trial functions. (w is a trial function if it is C2(Ω) and vanishes on

∂Ω, but is not identically zero.) Let

w =

n

X

k=1

ckwk,

be a linear combination of these trial functions. Suppose we made a really good choice of trial functions, and, in particular, choose w in such a way that w was an eigenfunction of (2.1) with

eigenvalue λ. Of course, this is not likely by randomly guessing w, but we will use this idea to find a way of approximating eigenvalues.

Now, if w was an eigenfunction of (2.1), then we know that λhwj, wi = λ Z Ω wjwdx = − Z Ω wj∆wdx = Z Ω ∇w_j.∇wdx = h∇wj, ∇wi,

using the fact that w, wj are trial functions, and therefore, vanish on the boundary of Ω.

Furthermore, using the definition of w, we have λhwj, n X k=1 ckwki = h∇wj, ∇( n X k=1 ckwk)i, which implies λ n X k=1 ckhwj, wki = n X k=1 ckh∇wj, ∇wki. Define ajk = h∇wj, ∇wki bjk = hwj, wki. (2.4) Therefore, if w was actually an eigenfunction, we would have

λ n X k=1 ckbjk = n X k=1 ckajk for i = 1, ..., n.

In other words, defining the n × n symmetric matrices A = (aij), B = (bjk), and letting c =         c1 . . . cn         , we would have λBc = Ac =⇒ [A − λB]c = 0.

In particular, this would mean A − λB is a singular matrix, and therefore, det[A − λB] = 0. We use this technique to approximate our first n eigenvalues. We now state our approximation technique.

Let w1, ..., wn be any n trial functions. For this choice of trial functions, define ajk, bjk as

in (2.4) and let A, B be the corresponding n × n symmetric matrices with entries ajk, bjk,

respectively. Then the n roots of the characteristic equation det(A − λB) = 0

are approximations to the first i eigenvalues λ₁, ..., λi. See [15].

We now turn to the minimax principle for eigenvalues of (2.1). We use the Rayleigh-Ritz approximation method to motivate the minimax principle. First, we prove the following lemma.

Lemma 5. Let A, B be n × n symmetric matrices. In addition, let B be positive definite. For i = 1, ..., n, let λ∗_i be the n roots of the characteristic equation det(A − λB) = 0.

Then the roots λ∗_i are all real, and the largest root λ∗_n satisfies λ∗_n= max  Ac·c Bc·c | c ∈ RN, c 6= 0  . (2.5)

Remark 6. For A = (ajk) and B = (bjk), where ajk and bjk are defined as in (2.4) for some

choice of trial functions w_i, A and B will satisfy the hypotheses of this lemma. We will use this lemma to motivate the minimax principle.

In order to prove Lemma (5), we first prove the following claim.

Claim 7. Let A, B be n × n symmetric matrices that satisfies in the hypotheses of the lemma

5. Then, there exists a set of vectors {v_i} which form a basis for RN _{and such that each v} i

satisfies the equation

Avi= λ∗iBvi

for some λ∗_i. Furthermore,

Bvi.vj = 0 f or i 6= j.

Remark 8. In the proof below, we will use the following fact regarding positive definite, real symmetric matrices. If B is a positive definite, real symmetric matrix, then there exists a lower triangular matrix L whose diagonal entries are positive and such that B = LLT . This is called the Cholesky decomposition [11].

Proof. (Proof of claim 7) First, we will show that the roots of the characteristic equation det(A − λB) = 0.

are all real. Assume λ∗_i is a root of this equation. Then, using the fact that B = LLT for some lower triangular matrix L whose diagonal entries are positive, we have

det(A − λ∗_iLLT) = 0 ⇐⇒ det(L−1A(LT)−1− λ∗_iI) = 0.

Therefore, λ∗_i is a root of the characteristic equation det(A − λB) = 0 if and only if λ∗_i is an eigenvalue of the matrix

But, we see that M is real symmetric, and, therefore, all its eigenvalues are real. In addition, M has an orthonormal eigenbasis {ui}.

We will now use this orthonormal eigenbasis {ui} to construct a basis for RN consisting of

solutions of Av = λBv for some λ ∈ R. Let

vi = (LT)−1ui.

As the {u_i} form a basis for RN _{and (L}T₎−1 _{has rank n, we see that the set {u}

i} form a basis

for RN. We now need to show that each v_i satisfies the equation Avi= λ∗iBvi

for some λ∗_i. By assumption, u_i is an eigenvector of M with corresponding eigenvalue λ∗_i. Therefore,

M ui = λ∗iui =⇒ L−1A(LT)−1ui = λ∗iui =⇒ (LT)−1L−1A(LT)−1ui = λ∗i(LT) −1

ui

=⇒ (LLT)−1Avi = λ∗ivi=⇒ Avi = λ∗iBvi.

Therefore, we have found a set of vectors {vi} which form a basis for RN and such that each

vi satisfies the desired equation for some λ∗i ∈ RN.

It remains only to show that

Bvi.vj = 0 f or i 6= j.

Using the definition of v_j and the fact that B = LLT for some lower triangular matrix L whose diagonal entries are positive, we have

Bvi.vj = LLT(LT)−1ui.(LT)−1uj = Lui.(LT)−1uj

= uiTLT(LT)−1uj = uTi .uj = ui.uj = 0,

using the fact that the set {ui} is orthonormal.

We now have the required ingredients to prove Lemma 5. Proof. (proof of Lemma 5)

By hypothesis, we have that B is positive definite. Therefore, applying Claim (7), we see that all roots of the characteristic equation

det(A − λB) = 0

are real. Let λ∗_i be the largest of these roots. We now need to prove (2.5). We will begin by showing that max  Ac·c Bc·c ≤ λ ∗ n | c ∈ RN, c 6= 0  (2.6)

Let c ∈ RN_{. By Claim (7), we can write c as a linear combination of the v}

i where each vi is

a solution of

Avi= λ∗iBvi

for one of the λ∗_i. Therefore, writing

c = a1v1+ ... + anvn,

and using the fact that B is positive definite, we have

Ac.c = A(a1v1+ ... + anvn).c = (a1λ∗1Bv1+ ... + anλ∗nBvn).(a1v1+ ... + anvn)

= a2₁λ∗₁Bv1.v1+ ... + a2nλ∗nBvn.vn ≤ a2₁λ∗_nBv1.v1+ ... + a2nλ ∗ nBvn.vn = λ∗_n(B(a1v1+ ... + anvn)).(a1v1+ ... + anvn) = λ∗_n(Bc.c) Consequently, we have shown that

Ac.c Bc.c ≤ λ

∗ n.

Taking the maximum of both sides over all c ∈ RN, we have proven (2.6). It remains only to show that

λ∗_n≤ max  Ac·c Bc·c | c ∈ R N_{, c 6= 0}  . (2.7)

We do so by finding a specific c ∈ RN such that λ∗_n≤ Ac.c

Bc.c.

We know that λ∗_n is a root of the characteristic equation det(A − λB) = 0. Therefore, there exists a vector v 6= 0 ∈ R such that (A − λ∗_nB)vn= 0. Let c = vn. Therefore,

Avn.vn Bvn.vn = λ ∗ nvn.vn Bvn.vn = λ∗_n and we have proved (2.5).

We now return to motivating the minimax principle. Recall from our Rayleigh-Ritz approxi-mation that for a given domain Ω ⊂ RN, we can approximate the first n eigenvalues of (2.1) by looking for the roots of the characteristic equation

det(A − λB) = 0,

where A = (ajk) and B = (bjk) for ajk, bjk defined in (2.4) for some choice of trial functions

{w_i} for Ω. From Lemma 5, we have shown that the largest root of det(A − λB) is given by max Ac.c

Bc.c = λ

∗

Therefore, for a fixed set of trial functions w₁, ..., wn, for a given domain Ω ⊂ RN, we have

the following formula for the approximation of the nth eigenvalue of (2.1). Let A = (a_jk) = (h∇wj, ∇wki), B = (bjk) = (hwj, wki). We see that

Ac.c = h n X k=1 cj∇wj, n X k=1 ck∇wki, Bc.c = h n X k=1 cjwj, n X k=1 ckwki.

Therefore, for A, B defined in terms of the trial functions w₁, ..., wn, we see that the largest

root of the characteristic equation det(A − λB) is given by λ∗_n(w1, ..., wn) = maxc∈RN_\{0} {

||∇w||2

||w||2 : w = Pn

i=1ciwi}. (2.8)

This value λ∗_n(w1, ..., wn) gives us an approximation to the nth eigenvalue of (2.1). We will show

below that if we take the minimum of λ∗_n(w1, ..., wn) over all possible sets of trial functions,

then we will get the exact value of the nth eigenvalue of (2.1). Theorem 9. (Minimax Principle)

Let Y denote the set of trial functions associated with (2.1). The nth eigenvalue of (2.1) is given by λn= min  λ∗_n(w1, ..., wn) | w1, ..., wn∈ Y  . (2.9)

That is, the minimum is taken over all possible sets of n linearly independent trial functions.

Proof. First, we will show that λn ≤ min λ∗n. Fix n linearly independent trial functions

w1, ..., wn. Let w = n X j=1 cjwj

be a linear combination of the n trial functions such that w is orthogonal to the first n − 1 eigenfunctions v₁, ..., vn−1 of (20). That is, choose cj such that

hw, v_ki =

n

X

j=1

hwj, vki = 0 f or k = 1, ..., n − 1.

We know we can solve this system, because we have only n − 1 equations for our n unknowns c1, ..., cn. Now, from the minimum principle for the nth eigenvalue, we know that

λn≤ ||∇w||2 ||w||2 , because λn= min{ ||∇v||2 ||v||2 , v ∈ Yn},

where Y_n is as defined in Theorem (4), and w ∈ Y_n. Therefore, λn≤ ||∇w||2 ||w||2 ≤ maxc∈RN{ ||∇Pn j=1cjwj||2 ||Pn j=1cjwj||2 } = λ∗_n(w1, ..., wn).

Now taking the minimum of both sides over all possible sets of n linearly independent trial functions, we see that

λn≤ min(w1,...,wn)∈Yλ

∗

n(w1, ..., wn).

Now, we need to show that λn≥ min λ∗n. In particular, we will show there exists a particular

choice of trial functions w1, ..., wnsuch that λ∗n(w1, ..., wn) ≤ λn. Let w1, ..., wn be the first n

eigenfunctions of (2.1) with corresponding eigenvalues λ₁, ..., λn. Without loss of generality,

we may assume they are orthogonal and normalized. By definition, λ∗_n(w1, ..., wn) = maxc∈RN{ ||∇Pn j=1cjwj||2 ||Pn j=1cjwj||2 }.

Now, using the fact that the wj are eigenfunctions, orthogonal, and normalized, we have

||∇ n X j=1 cjwj||2 = h n X j=1 cj∇wj, n X j=1 cj∇wji = −h n X j=1 cjwj, n X j=1 cj∆wji = h n X j=1 cjwj, n X j=1 λjcjwji = n X j=1 hcjwj, λjcjwji = n X j=1 λjc2jhwj, wji = n X j=1 λjc2j.

Again, using the fact that the wj are orthogonal and normalized, we have

|| n X j=1 cjwj||2= h n X j=1 cjwj, n X j=1 cjwji = n X j=1 hcjwj, cjwji = n X j=1 c2_jhwj, wji = n X j=1 c2_j. Therefore, we have λ∗_n(w1, ..., wn) = maxc∈RN Pn j=1λjc2j Pn j=1c2j ≤ max_c∈RN Pn j=1λnc2j Pn j=1c2j = λn.

Therefore, for this choice of trial functions w1, ..., wn, we have

λ∗_n(w1, ..., wn) ≤ λn

and, consequently,

min λ∗_n(w1, ..., wn) ≤ λn(w1, ..., wn) ∈ Y.

Therefore, our theorem is proved.

Theorem 10. If Ω₁ ⊂ Ω₂, then λ_n(Ω1) ≥ λn(Ω2). Where λn(Ωi) is the nth eigenvalue of the

Dirichlet problem (2.1) on Ωi.

Proof. Let Y (Ωi) be the set of trial functions for Ωi, i = 1, 2. Recall

Y (Ωi) = {w : w ∈ C2(Ωi), w 6≡ 0, w = 0 for x ∈ ∂Ωi}.

For emphasis, we let

λ∗_n(w1, ..., wn)|Ωi = λ

∗

n(w1, ..., wn),

where the L2 norms are taken over Ωi.

By the minimax principle, we know that λn(Ωi) = min  λ∗_n(w1, ..., wn)|Ωi w1, ..., wn∈ Y (Ωi)  = min w1,...,wn∈Y (Ωi) max c∈RN ||∇w||2 L2_(Ω i) ||w||2 L2_(Ω i) : w = n X j=1 ciwi  .

For a fixed set of trial functions w1, ..., wn for Ω1, let c∗ = c∗(w1, ..., wn) be the vector in RN

which maximizes the Rayleigh quotient. That is,

λ∗_n(w1, ..., wn)|Ωi ≡ ||∇w∗||2 L2_(Ω i) ||w∗_||2 L2_(Ω i) where w∗ ≡ n X j=1 c∗_iwi.

Now, we can extend each of the trial functions wi to be a trial function for Ω2 by extending

wi to be zero outside Ω1. Let ˜wi denote wi extended to Ω2 in this way. Therefore, it is clear

that λ∗_n( ˜w1, ..., ˜wn)|Ω2 ≡ ||∇ ˜w∗_||2 L2_(Ω i) || ˜w∗||2 L2_(Ω i) , where w˜∗_≡ n X j=1 c∗_iw˜i.

As the functions ˜wi are zero outside Ω1, we see that

λ∗_n(w1, ..., wn)|Ω1 = λ

∗

n( ˜w1, ..., ˜wn)|Ω2. (2.10) Now suppose v₁, ..., vnare the n trial functions for Ω1 which minimize λn∗(w1, ..., wn)|Ω1. Then using (2.10), we see that

λ∗_n( ˜v1, ..., ˜vn)|Ω2 = λ ∗ n(v1, ..., vn)|Ω1, and therefore, λn(Ω2) = minv1,...,vn∈Y (Ω2) λ ∗ n|Ω2 ≤ λ ∗ n(v1, ..., vn)|Ω1 = λn(Ω1) as claimed.

Corollary 11. For domain Ω ⊂ RN_{, the eigenvalues of the Dirichlet problem,}

(

−∆v = λv, x ∈ Ω,

u = 0, x ∈ ∂Ω, (2.11)

form an infinite sequence {λn} such that λn−→ ∞ as n −→ ∞.

Proof. For domain Ω ⊂ RN_{, let}

R = {(x1, ..., xn) ∈ RN : |xi| < M, i = 1, ..., n}

for M sufficiently large such that Ω ⊂ R. Now we can explicitly calculate the eigenvalues of R. In particular, the eigenvalues are given by

λm1,...,mn(R) = n X j=1 (miπ 2M) 2_,

where the mi are positive integers. We see that these eigenvalues form an infinite sequence

which goes to infinity. By the above theorem, we know that the corresponding eigenvalues for Ω satisfy λm(Ω) ≥ λm(R). Therefore, we conclude that the eigenvalues of Ω form an infinite

sequence which goes to infinity.

2.3

Counterexample to eigenvalue monotonicity for the

Neumann problem

In this section, we give a counterexample to eigenvalue monotonicity for the Neumann prob-lem. See [2]. Let Ω and ˜Ω be two domains in RN, and denote the eigenvalues of the Neumann problem on these domains by λj(Ω) and ˜λj( ˜Ω) respectively. In this section, we show that

If Ω ⊃ ˜Ω 6=⇒ λ_j(Ω) ≤ ˜λj( ˜Ω), ∀j ≥ 1.

We consider the following eigenvalue problem with Neumann boundry conditions, (

−∆ϕ = λϕ, x ∈ Ω, ∂Vϕ|∂Ω= 0, x ∈ ∂Ω,

(2.12)

where ∂V is outward normal derivative. We consider a rectangle Ω = [0, l] × [0, m]. We use

the method of separation of variables for cartesian coordinates x and y. That is, look for solutions of the form ϕ(x, y) = f (x)g(x). With Neumann boundary conditions ∂Vϕ|∂Ω = 0,

the eigenfunctions are

Figure 1: Eigenfunctions on a square. Refer to reference [2]. and have corresponding eigenvalues

λjk = (jπ_l )2+ (kπ_m)2 f or j, k ≥ 0. (2.14)

Indeed, the operator ∆ act on C∞(Ω) that simply differentiates a function ϕ ∈ C∞(Ω) twice with respect to each position variable

∆ϕjk = ∂2_ϕ jk ∂x2 + ∂2_ϕ jk ∂y2 . (2.15)

The function ϕ_jk(x, y) has two variables. The first and second derivatives of ϕjk(x, y)

accord-ing to variable x are

∂ϕjk(x, y) ∂x = −( jπ l ) sin( jπ l x) cos( kπ my) ∂2ϕjk(x, y) ∂x2 = −( jπ l ) 2_cos(jπ l x) cos( kπ my),

respectively, and the first and second derivatives of ϕ_jk(x, y) according to variable y are ∂ϕjk(x, y) ∂y = −( kπ m) cos( jπ l x) sin( kπ my) ∂2ϕjk(x, y) ∂y2 = −( kπ m) 2_cos(jπ l x) cos( kπ my), respectively. Therefore,

∆ϕjk(x, y) = −(jπ_l )2cos(jπ_l x) cos(kπ_my) − (kπ_m)2cos(jπ_l x) cos(kπ_my)

= −((jπ_l )2+ (kπ_m)2)(cos(jπ_l x) cos(kπ_my))

= −λjk.ϕjk(x, y)

Figure 2: Rectangle contained in a square

Thus,

−∆ϕ_jk(x, y) = λjkϕjk(x, y).

If we suppose that l = m in the rectangle Ω = [0, l] × [0, m], then we have a square. In the figure 1, the eigenfunctions on a square are shown as if they were seen from above. (see figure 1) [2].

We consider a rectangle contained in a square. We consider square Ω where l = 1, m = 1, j = 1 and k = 0. According to (2.15), we have λ₂ = π2 . Then, we consider rectangle ˜Ω with side length l = √2(0.9), j = 1 and k = 0. Therefore, ˜λ2 = π2

(√2(0.9))2 =

π2

1.62, which is smaller

than λ2 = π2. (See figure 2).

Although a smaller rectangle ˜Ω is inscribed inside a larger square Ω (i.e,Ω ⊃ ˜Ω), the second eigenvalue ˜λ2( ˜Ω) = π

2

(√2(0.9))2 =

π2

1.62 is smaller than the second eigenvalue λ2(Ω) = π 2_.

Therefore,

If Ω ⊃ ˜Ω 6=⇒ λ_j(Ω) ≤ ˜λj( ˜Ω) ∀j ≥ 1.

2.4

Counterexample to eigenvalue monotonicity for the

Steklov problem

Let Ω and ˜Ω be two domains in RN, and denote the eigenvalues of the Steklov problem on these domains by σj(Ω) and ˜σj( ˜Ω) respectively. In this section we show that

Figure 3. Rectangle contained in the unit disc

Ω ⊃ ˜Ω 6=⇒ σj(Ω) ≤ ˜σj( ˜Ω) ∀j ≥ 1.

We consider the Steklov eigenvalue problem for rectangle ˜Ω and unit disc Ω ,where rectangle Ω contained in a unit disc ˜Ω. (see figure 3). When the width of rectangle goes to zero, eigenvalues of the rectangle tends to zero (see proof of the collapse of the Steklov spectrum for a dumbbell domain in sections3.1and3.2). By example1, we know the eigenvalues of the unit disc are 0, 1, 1, 2, 2, ... Therefore,

Chapter 3

Collapse of the Steklov spectrum for a

dumbbell domain

In this chapter we prove in two manners the collapse of the Steklov spectrum for a dumbbell domain with a thin passage (see figure 4). These results are due to A. Girouard and I. Polterovich [5].

Lemma 12. For any real numbers a₁, a2 and positive numbers b1, b2 we have

a1+ a2 b1+ b2 ≤ max{a1 b1 ,a2 b2 }. Proof. We have a1+ a2 b1+ b2 = b1 a1 b1 + b2 a2 b2 b1+ b2 , which is convex combination of a1

b1 and a2 b2. Thus, a1+ a2 b1+ b2 = b1 a1 b1 + b2 a2 b2 b1+ b2 ≤ max{a1 b1 ,a2 b2 }.

Corollary 13. For any real numbers a1, a2, ..., an and positive numbers b1, b2, ..., bn we have

a1+ a2+ ... + an b1+ b2+ ... + bn ≤ max{a1 b1 ,a2 b2 , ...,an bn }. Proof. We have n P i=1 ai n P i=1 bi = n P i=1 bia_bi_i n P i=1 bi = n X i=1 ci ai bi ,

Figure 4. The domain Ω with a thin passage where c_i= bi n P i=1 bj and n P i=1

ci= 1. This is convex combination of a_b11, ...,a_bnn. Thus,

a1+ a2+ ... + an b1+ b2+ ... + bn ≤ max{a1 b1 ,a2 b2 , ...,an bn }.

3.1

First proof of the collapse of the Steklov spectrum for a

dumbbell domain

In this section we consider a dumbbell domain: two disks joined by a thin channel. When the width of the channel tends to 0, we prove that each eigenvalues σn of the Steklov spectrum

tends to 0.

Let Ω_ = D1∪P∪D1, where D1 and D2 are two copies of the unit disk joined by a rectangular

passage P of length  and width 3 where 0 <   1 (See figure 4):

P= {(x, y)| x ∈ ( − 2 , + 2 ) , y ∈ ( −3 2 , +3 2 )}.

What is essential in this construction is that the width of the passage tends to zero much faster than its length. We prove that,

lim

→0+σn(Ω) = 0 for all n = 1, 2, ... . (3.1) Proof. Let un: Ω−→ R be defined by

un(x, y) =    sin 2πnx_
in P_ 0 elsewhere.

These functions are orthogonal in L2_(∂Ω

). Indeed, for n 6= m we have

Z ∂Ω sin 2πnx   sin 2πmx   dx = 2 Z  2 − 2 sin 2πnx   sin 2πmx   dx = 4 Z  2 0 sin 2πnx   sin 2πmx   dx = 2 Z  2 0 cos 2 (n − m) πx   − cos 2 (n + m) πx   dx = 2   2 (n − m) πsin  2 (n − m) πx   −  2 (n + m) πsin  2 (n + m) πx  ₂ 0 =  (n − m) πsin ((n − m) π) −  (n + m) πsin ((n + m) π) = 0.

Furthermore, for n = m we have Z ∂Ω sin 2πnx   sin 2πnx   dx = 2 Z  2 − 2 sin2 2πnx   dx = 4 Z  2 0 sin2 2πnx   dx = 2 Z  2 0 1 − cos 4πnx 
dx = 2  x −  4πn  sin 4πnx 
₂ 0 = .

We use of min-max characterization (1.2). We consider the subspace E ⊂ H1_(Ω

) defined by

E = span (u1, u2, ..., un+1) .

It follows from (1.2) that

σn(Ω) ≤ supu∈E R Ω|∇u| 2 R ∂Ωu 2 .

We prove that this is bounded above by a quantity which tends to zero as  −→ 0. Indeed, we consider the function u ∈ E. It is a linear combination

u = c1u1+ c2u2+ ... + cn+1un+1. It follows that σn(Ω) ≤ sup c1,...,cn+1∈R R Ω| Pn+1 i=1 ci∇ui|2 R ∂Ω  Pn+1 i=1 ciui 2.

Because the functions u_i are perpendicular in L2_{(∂Ω) the denominator is} Z ∂Ω u2 ds = Z ∂Ω n+1 X i=1 ciui !2 ds = Z ₂ − 2 n+1 X i=1 c2_i sin2 2πix 
dx = n+1 X i=1 c2_i Z  2 − 2 sin2 2πix 
dx = n+1 X i=1 c2_i  2. Moreover the numerator is

Z Ω |∇u|2= Z P n+1 X i=1 (ci∂xui) !2 dxdy. We have ∂xui(x, y) = ∂ui(x, y) ∂x = ( 2πi  ) cos( 2πix  ).

These functions cos(2πix_ ) are orthogonal in (−₂ ,₂). Indeed, for n 6= m we have Z  2 − 2 cos 2πnx   cos 2πmx   dx = 2 Z  2 0 cos 2πnx   cos 2πmx   dx = Z  2 0 cos 2 (n − m) πx   + cos 2 (n + m) πx   dx =   2 (n − m) πsin  2 (n − m) πx   +  2 (n + m) πsin  2 (n + m) πx  ₂ 0 =  2 (n − m) πsin ((n − m) π) +  2 (n + m) πsin ((n + m) π) = 0.

13 in the second following inequality. Thus, we have σn(Ω) ≤ supu∈E R Ω|∇u| 2 R ∂Ωu 2 = sup c1,...,cn+1∈R R Ω| Pn+1 i=1 ci∇ui|2 R ∂Ω  Pn+1 i=1 ciui 2 = sup c1,...,cn+1∈R R3₂ −3 2 R₂ − 2     Pn+1

i=1 ci 2πi_
cos 2πix_

    2 dxdy 2R₂ − 2  Pn+1

i=1 cisin 2πix_

2 dx = R3₂ −3 2 R₂ − 2 Pn+1 i=1 2πi 
2

cos2 2πix_
dxdy 2R₂

− 2

Pn+1

i=1 sin2 2πix
dx

= Pn+1 i=1 R3₂ −3 2 R₂ − 2 2πi 
2

cos2 2πix_
dxdy 2Pn+1 i=1 R₂ − 2 sin2 2πix_
dx ≤ max i=1,2,...,n+1 n R3₂ −3 2 R₂ − 2 2πi 
2 cos2 2πix 
dxdy 2R₂ − 2 sin2 2πi_
dx o = max i=1,2,...,n+1 n 3R₂ − 2 2πi 
2 cos2 2πix_
dx 2R  2 − 2 sin2 2πi_
dx o = max i=1,2,...,n+1 n 3 2πi 
2R₂ − 2 1+cos(4πix  ) 2 dx 2R₂ − 2 1−cos(4πix  ) 2 dx o = max i=1,2,...,n+1 n 3 2πi 
2_x 2 +  8πisin 4πix 
₂ − 2 2x₂ −_8πi sin 4πix_


 2 − 2 o = max i=1,2,...,n+1 n3 2πi 
2 _ 2
2 ₂
o = max i=1,2,...,n+1 n  (πi)2o −→ 0

when  −→ 0+. Thus, we have lim

→0+σn(Ω) = 0 for all n = 1, 2, ... . (3.2)

Figure 5: Graph of function h

3.2

Second proof of the collapse of the Steklov spectrum for a

dumbbell domain

In this section, we provide the second proof of the collapse of the Steklov spectrum for a dumbbell domain with use of bump functions that are supported in the channel P.

3.2.1 Construction of bump functions Lemma 14. (Smoothly joining zero)

Let h : R −→ R be defined by

h(t) := (

e−1/t f or t > 0, 0 f or t ≤ 0.

Then h belongs to C∞(R), 0 ≤ h ≤ 1, and h(t) > 0 if and only if t > 0. (See figure 5)

We now begin our constructions in the smooth case.

(i) Basic bump f unction : An explicit example of a function ψ ∈ C_c∞(RN) with 0 ≤ ψ ≤ 1, supp(ψ) = B1(0), and ψ(x) > 0 when |x| < 1, is given by

ψ(x) := (

e−1/(1−|x|2) for |x| < 1

0 for otherwise

(see figure 4).

Smoothness of ψ follows from the chain rule upon noticing that ψ(x) = h(1 − |x|2) with h as in Lemma14.

Figure 6: Graph of function ψ(x)

(ii) N ormalization : Since ψ ∈ C_c∞(RN), from (i), satisfiesR_RNψ(y)dy > 0, we may set ρ(x) := R ψ(x)

RNψ(y)dy

and obtain a function ρ ∈ C_c∞(RN), supp(ρ) = B1(0), 0 ≤ ρ ≤ 1, ρ(x) > 0 when |x| < 1 , and

such that in addition R_RNρ(x)dx = 1. (iii) Scaling : For  ∈]0, 1] we set

ρ(x) := 1 nρ( x ), (x ∈ R N_).

Then we have for every  ∈]0, 1], ρ_ ∈ C∞

c (RN), ρ≥ 0, supp(ρ) = B(0), and

R

RNρ= 1.

(iv) T ranslation : Let x0∈ RN be arbitrary. Defining the functions ϕ,x0(0 <  ≤ 1) on R

N by ϕ,x0(x) := ρ(x − x0) = 1 nρ( x − x0  ), we obtain ∀ ∈]0, 1]: ϕ ∈ Cc∞(RN), ϕ ≥ 0, supp(ϕ) = B(x0), and

R

RNϕ= 1.

Thus, we constructed smooth normalized non-negative bump functions in D(RN_{), where}

D(RN) = C_c∞(RN) = {φ∈ C∞(RN) | supp(φ) is compact},

Figure 7: Graph of bump functions u_i(x) in the channel P

3.2.2 Each eigenvalues σn of the Steklov spectrum tends to 0 in the

channel P = (−1, 1) × (−, )

We consider x1, ...xm+1 ∈ (−1, 1) for m ∈ N, where xi = −1 + _m+12 i. Let un : Ω −→ R be

defined by

ui(x, y) := ρδ(x − xi)

for small δ = _4m1 . Thus,

supp(ui) = (xi− δ, xi+ δ) × (−, ).

We consider the subspace E ⊂ H1(Ω) defined by n + 1 bump functions with disjoint supports.

E = span (u1, u2, ..., un+1) ,

It follows from (1.2) that

σn(Ω) ≤ supu∈E R Ω|∇u| 2 R ∂Ωu 2 .

We prove that this is bounded above by a quantity which tends to zero as  −→ 0. Indeed, we consider the function u ∈ E. It is a linear combination

We take the supremum among all possible choices of c₁, c2, ..., cn+1and we use of the Corollary 13 in the second following inequality. Since supp u_i∩ supp u_j = ∅ if i 6= j, we have

σn(Ω) ≤ supu∈E R Ω|∇u| 2 R ∂Ωu 2 = sup u∈E R Ω     Pn+1 i=1 ci∇ui     2 R ∂Ω  Pn+1 i=1 ciui 2 = sup c1,...,cn+1∈Rn R1 −1 R −     Pn+1 i=1 ci  ∂ρδ(x−xi) ∂x     2 dydx 2R1 −1  Pn+1 i=1 ciρδ(x − xi) 2 dx = sup c1,...,cn+1∈Rn R1 −1 R − Pn+1 i=1 c2i  ∂ρδ(x−xi) ∂x 2 dydx 2R1 −1 Pn+1 i=1 c2i (ρδ(x − xi))2dx = Pn+1 i=1 R1 −1 R −  ∂ρδ(x−xi) ∂x 2 dydx 2Pn+1 i=1 R1 −1(ρδ(x − xi))2dx = 2Pn+1 i=1 Rδ −δ  ∂ρδ(x−xi) ∂x 2 dx 2Pn+1 i=1 Rδ −δ(ρδ(x − xi)) 2 dx ≤ max 1≤i≤m+1   Rδ −δ  ∂ρδ(x−xi) ∂x 2 dx Rδ −δ(ρδ(x − xi))2dx  −→ 0 when  −→ 0+. Thus, we have lim

→0+σn(Ω) = 0 f or all n = 1, 2, ... . (3.3)

Chapter 4

Trace operator for non-Lipschitz

domain

In this chapter, we introduce the trace operator. It is compact for bounded Lipschitz domains. We give two examples to show that the trace operator is not compact for a non-Lipschitz open set.

4.1

Trace operator

Theorem 15. (Trace theorem)

Assume Ω is bounded, ∂Ω is C1 _{and W}1,2_{(Ω)is defined in (1.1.2). Then there exists a bounded}

linear operator T : W1,2(Ω) −→ L2(∂Ω) such that (i) T u = u|∂Ω if u ∈ W1,2(Ω) ∩ C(Ω) and (ii) ||T u||_L2_(∂Ω)≤ C||u||_W1,2_(Ω)

for each u ∈ W1,2_{(Ω), and the constant C depending only on L}2 _{and Ω.}

Proof. See reference [9].

Definition 4.1.1. We call T u the trace of u on ∂Ω.

Definition 4.1.2. Suppose T ∈ B(X, Y ), where X and Y are Banach spaces. We say that T is a compact operator if, for all bounded sequence (xn)n≥1⊂ X, the sequence (T (xn))n≥1⊂ Y

has a convergent subsequence in Y .

Theorem 16. Let Ω ⊂ RN be a bounded domain with Lipschitz boundry. Then the trace operator defined in theorem 15 is compact.

Proof. Refer to reference [9].

4.2

Example of a non-Lipschitz domain

The definition of Lipschitz domain is in terms of Lipschitz function describing the boundary. In this section, we give an example of non-Lipschitz domain.

Definition 4.2.1. A bounded domain Ω ⊂⊂ RN (Ω compactly embedded in RN) is called Lipschitz if locally the boundary ∂Ω is the graph of a Lipschitz function. Let ψ : RN −1−→ R be a function which satisfies the Lipschitz condition

|ψ(x) − ψ(y)| ≤ M|x − y|, ∀ x, y ∈ RN −1_{. (4.1)}

A bounded domain Ω is called Lipschitz if near every boundary point p ∈ ∂Ω there exists a neighborhood U of p such that after a rotation

Ω ∩ U := {(x0, xn) ∈ U |xn> ψ(x0)}

for some Lipschitz function ψ.

Thus, a Lipschitz domain is an open and connected domain with a compact boundary ∂Ω by definition which is locally parameterize with Lipschitz continuous functions. We consider a domain Ω ⊂ R2 with a sharp point in zero. (See figure 8).

We consider function f (x) =√x on [0, 1]. Define f (x) =√x which is uniformly continuous on [0, 1]. To see this observes

(i) (√x −√y)(√x +√y) = x − y

(ii) since −√y ≤√y we have (√x −√y) ≤ (√x +√y) Now let  > 0 and x > y. Choose δ := 2 then

(√x −√y)2 ≤ (√x −√y)(√x +√y) = x − y < δ =⇒ |√x −√y| <√δ = 

Thus, f (x) =√x is uniformly continuous but its not Lipschitz continuous on [0, 1]. Given any M > 0, choose c = 0 and 0 < x < 1/M2.

So that √1 x > M. Then we have |f (x) − f (c)| |x − c| = |√x| |x| = 1 √ x > M

Figure 8. Domain Ω with a sharp point in zero

Since M is arbitrary, this show that f (x) =√x is not Lipschitz continuous function on [0, 1], because √1

x is not bounded at zero and hence f (x) =

√

x is not a Lipschitz function on Ω. Therefore, ∂Ω is not a Lipschitz boundary .

4.3

A non compact inclusion H

(Ω) −→ L

(Ω)

In this section, I give an example where the inclusion of H1_{(Ω) −→ L}2_{(Ω) is not compact,}

where bounded open set Ω is not a Lipschitz. I will construct a sequence of functions {f_n} bounded in H1(Ω) = W1,2(Ω) that does not admit any convergent subsequence in L2(Ω), providing an example to show that H1(Ω) ⊂ L2(Ω) is not compact when the bounded open set Ω is not Lipschitz.

We take ak:= _C1k for a constant C > 10, and we consider the set Ω = S∞

k=1Bk, where for any

k ∈ N, Bk denotes the ball of radius a2k centered in ak. Notice that

ak−→ 0 as k −→ ∞ and ak− a2k> ak+1+ a2k+1.

Thus, Ω is the union of disjoint balls, it is bounded and it is not Lipschitz because it’s not connected. (See figure 9). For any n ∈ N, we define the function f_n: Ω −→ R as follows

fn(x) =

(

π−1/2a−2_n for x ∈ Bn,

0 for x ∈ Ω − Bn.

We observe that {f_n} is a bounded sequence in W1,2_{(Ω) but we cannot extract any subsequence}

convergent in L2(Ω) from the sequence of functions {fn}, because for x ∈ Ω

||f_n||2 W1,2_(Ω) = ||∇fn||2L2_(Ω)+ ||fn||2L2_(Ω) = 0 + Z Ω |fn(x)|2 dx

Figure 9: Open set Ω is union of disjoint balls Bk, where k=1,2,3,.... = Z Bn π−1a−4_n dx = π−1a−4_n Z Bn 1 dx = π−1a−4_n π(a2_n)2 = π−1a−4_n πa4_n= 1 < ∞.

We prove by contradiction that fnis not converging in L2(Ω). If fn is converging, then there

exist f such that f_n−→ f in L2_{(Ω). Let G}

k= f χB1+...+f χBk. It follows that G ≡ 0. Indeed, ||Gk||L2_(Ω)= ||(f − f_n)χ_B1+ ... + (f − f_n)χ_Bk||_L2_(Ω)

≤ ||(f − fn)||L2_(Ω) , if n > k −→ 0.

Moreover, Gk−→ f pointwise. Thus,

f = lim

n−→∞Gk = 0.

This is contradiction.

4.4

A non bounded trace T : H

(Ω) −→ L

(∂Ω)

In this section, we give an example where the trace T : H1(Ω) −→ L2(∂Ω) is not bounded, where Ω is not a Lipschitz domain. We will construct a sequence of functions {fn} bounded in

H1(Ω) = W1,2(Ω) that does not admit any convergent subsequence in L2(∂Ω), providing an example to show that H1(Ω) ⊂ L2(∂Ω) is not compact when the open set Ω is not Lipschitz.

We make sequence a_k as mentioned in section 4.3. For any n ∈ N, we define the function fn: Ω −→ R as follows fn(x) = ( √₂ 2 π −1/2_a−1 n for x ∈ Bn, 0 for x ∈ Ω − Bn.

We observe that {f_n} is a bounded sequence in W1,2_{(Ω) and ∀n ∈ N, f}

n ∈ L2(∂Ω) but we

cannot extract any subsequence convergent in L2(∂Ω) from the sequence of functions {fn}.

We have ||f_n||2 W1,2_(Ω) = ||∇fn||2L2_(Ω)+ ||fn||2L2_(Ω) = 0 + Z Ω |fn(x)|2 dx = Z Bn 1 2π −1_a−2 n dx = 1 2π −1_a−2 n Z Bn 1 dx = 1 2π −1_a−2 n π(a2n)2 = 1 2a 2 n< ∞. Thus, ||fn||2_L2_(Ω) −→ 0 as n −→ ∞.

Therefore, {fn} ⊂ W1,2(Ω) and is not converging sequence in W1,2(Ω). We have

||f_n||2_L2_(∂Ω) = Z ∂Ω |f_n(x)|2 dx = Z ∂Bn 1 2π −1_a−2 n dx = 1 2π −1_a−2 n Z ∂Bn 1 dx = 1 2π −1_a−2 n (2πa2n) = 1 < ∞. Thus, ∀n ∈ N, f_n∈ L2_{(∂Ω) and {f}

n} is a sequence in L2(∂Ω). The trace is not continuous.

4.5

Rellich theorem and trace theorem

In this section, we present Rellich theorem and trace theorem. The proof of these two theorems are presented in master’s thesis of Marc-Antoine Labrie [9].

Theorem 17. (Rellich Theorem)

Suppose that Ω ⊂ RN _{is a bounded domain with continuous boundary ∂Ω. Then the embedding}

identity

S : H1(Ω) −→ L2(Ω) is compact.

Proof. See reference [9].

Theorem 18. (Trace Theorem)

Let Ω be a bounded domain with Lipschitz boundary. Then there exist a unique linear and continuous mapping T : H1_{(Ω) −→ L}2_{(∂Ω) such that for all f ∈ C}∞_{( ¯Ω), we have}

T (f ) = f |∂Ω.

Chapter 5

An inequality for a Steklov eigenvalue

by the method of defect

In this chapter, we give a lower bound to the first non-zero eigenvalue σ1 of the Steklov

problem for plane domains having two axes of symmetry. We first prove a nodal-line theorem of some interest in itself. This nodal-line theorem does not seem to appear anywhere in the literature, although its proof is a straightforward application of familiar methods. The results of the nodal-line theorem then permit us to use the method of defect to obtain our bound by integrating an easily obtained one-dimensional version of the desired inequality [7].

5.1

Preliminaries

Let Ω be a domain of the x1, x2-plane with piecewise smooth boundary ∂Ω. We consider the

Steklov eigenvalue problem

∆u = 0 in Ω, ∂u

∂n = σu on ∂Ω, (5.1)

where ∆ = Σn_i=1∂x∂22 i

is the Laplacian and n the unit outer normal on ∂Ω. The problem has a discrete spectrum of eigenvalues

0 = σ0< σ1≤ σ2≤ ...,

with corresponding eigenfunctions u1= constant, u2, u3, .... The eigenvalues can be

character-ized by σn= min R Ω|grad v| 2_dx H ∂Ωv2ds , (5.2)

where the minimum is taken over all continuous and piecewise continuously differentiable functions v satisfying

I

∂Ω

The minimum of (5.2) is attained when and only when v is an eigenfunction of (5.1) associated with σ_n.

5.2

The nodal-line theorem

Definition 5.2.1. (nodal-line)

A curve in Ω along which an eigenfunction u_n vanishes is called a nodal line of u_n. That is N = {x ∈ C|un(x) = 0}.

Definition 5.2.2. (harmonic function)

A function u(x1, x2) which has continuous second partial derivatives and solve Laplace‘s

equa-tion ∆u = ∂_∂x2u2 1

+∂_∂x2u2 2

= 0 is called a harmonic function. That is

u(x1, x2) is harmonic function ⇐⇒ ∆u = 0.

Example 19. (i) All constant u(x1, x2) = c are harmonic function of degree 0.

(ii) All linear polynomials u(x1, x2) = ax1+ bx2 are harmonic functions of degree 1.

(iii) The quadratics polynomials u(x1, x2) = x21− x22 and u(x1, x2) = x1x2 are harmonic

func-tions of degree 2.

Theorem 20. (Green’s first identity) If u is harmonic in a domain containing R and f(x, y) is continuously differentiable in R, and closed curve C bounding the domain R, then

I C f∂u ∂nds = Z Z R 5f. 5 u dA.

Theorem 21. Let u be harmonic in a domain containing R. Then u = 0 on curve C implies u = 0 on R.

Proof. We use Green‘s first identity theorem, taking f = u. This gives I C u∂u ∂nds = Z Z R | 5 u|2dA. If u = 0 on C, then we have I C u∂u ∂n = 0 =⇒ Z Z R | 5 u|2dA = 0 =⇒ | 5 u|2= 0 everywhere in R.

Since | 5 u| is continuous and bigger than or equal to zero everywhere, being a square; =⇒ 5u = 0 in R, since its magnitude is 0

=⇒ ux= 0 and uy = 0 in R.

Thus u = c. Since u = 0 on the boundary C, the constant c must be 0.

Theorem 22. (Unique continuation theorem for harmonic functions)

If u vanishes of finite order in the normal direction at x0 and u(x) ≥ 0 for x ∈ R, then the

following hold.

(i) u vanishes identically in a neighborhood of x0 in R.

(ii) u vanishes identically in the normal direction at x0 , i.e, u(tx0) = 0 for all t ≤ 1 such

that the segment [tx0, x0]is contained in ¯R.

Theorem 23. (Nodal line theorem)

The nodal lines of un divide Ω into no more than n subdomains, and no nodal line is a closed

curve.

Proof. We prove the second part of the theorem first. If a nodal line were closed, we would have u_n= 0 in the interior of the curve since ∆un= 0 in Ω (see theorem21). But then un= 0

would vanish identically in B by the Unique Continuation Theorem for harmonic functions (see Theorem22). Thus no nodal line can be a closed curve. Now suppose that the nodal lines of u_n divide B into more than n subdomains. Let D₁, D2, ..., Dn be n of these subdomains.

Note that ∂Ω ∩ ∂Di is not empty. Indeed, let A1 = int(Di) ∩ Ω and A2 = int(Dci) ∩ Ω. Thus

A1, A2 are open sets. We have A1∩ A2 = ∅ and A1∪ A26= Ω because Ω is a connected domain.

Thus

∂Di∪ ∂Ω = ∂(Ω − ¯Dic) 6= ∅.

Define wi to agree with un on Di and vanish on Ω − Di i = 1, 2, ..., n. Notice that wi 6= 0 on

∂Ω ∩ ∂Di, otherwise wi = 0 in Di (since ∆wi = 0). Indeed, if wi= 0 on ∂B ∪ ∂Di =⇒ wi = 0

on ∂D_i

We use Green‘s first identity theorem, takinhg f = u = w_i. This gives I ∂Di wi ∂wi ∂n ds = Z Z R | 5 wi|2dA wi = 0 on ∂Di, then we have I ∂Di wi ∂wi ∂n = 0 =⇒ Z Z Ω | 5 w_i|2dA = 0 =⇒ | 5 wi|2 = 0 everywhere in Ω.

Since it is continuous and greater than or equal to zero everywhere, being a square 5wi= 0 inDi , since its magnitude is 0

=⇒ wix = 0 and wiy = 0 in Di. Since u = 0 on the boundary ∂Di, the constant c must be 0.

Thus, wi = 0 in Di. this is contradiction since wi = un 6= 0 in ∂Di (by definition of wi).

Therefore, w_i 6= 0 in ∂D_i.

Thus, we can find a linear combination, say Σn_i=1aiwi, such that

I

∂Ω

v2 ds = 1 and v satisfies (5.3). Indeed,

hv, vi = I ∂Ω v2ds = Z ∂Ω a2₁w1.w1+ a22w2.w2+ ... + a2nwn.wnds = (a2₁+ ... + a2_n) I ∂Ω 1ds = (a2₁+ ... + a2_n)L

we can choose a₁, ..., an such that (a21+ ... + a2n) = L1, thus

I ∂Ω v2ds = (a2₁+ ... + a2_n)L = L.1 L = 1. We have I vukds = 0, k = 1, 2, ..., n − 1. Indeed, v = Σn_i=1aiwi = a1.w1+ ... + an.wn. Thus, I ∂Ω vukds = I ∂Ω (a1w1+...+anwn).uk= I ∂Ω a1(w1.uk)+...+an(wn.uk) = 0 for k = 1, ..., n−1.

Moreover, the wi hence v, are continuous and piecewise continuously differentiable. Since ∂v

∂n = σnv on ∂Ω, it follows from Green’s first identity that

R Ω|grad v| 2_dx H ∂Ωv2ds = σn.

Indeed, ∂v ∂n = ∂(a1w1+ ... + anwn) ∂n = (a1+ ... + an)∂un ∂n = (a1+ ... + an)∂σnun = σnv. Thus, _∂n∂v = σnv on the ∂Ω.

Indeed, by first green identity when u = v and f = v then I ∂Ω v∂v ∂n ds = Z Ω 5v. 5 vdx =⇒ I ∂B v∂v ∂n ds = Z Ω 5v. 5 vdx =⇒ I ∂Ω vσnv = Z Ω | 5 v|2_dx =⇒ σn I ∂Ω v2ds = Z Ω | 5 v|2 =⇒ σn= R Ω| 5 v| 2 H ∂Ωv2 ds .

Thus, v minimizes (5.2) and is therefore an eigenfunction of (5.1). Indeed, we have

σn= min R Ω| 5 v|2 H ∂Ωv2 ds , with (5.2) and σn= R Ω| 5 v| 2 H ∂Ωv2 ds .

When v = Σn_i=1ai.wi. Thus we minimize (5.2) and therefore an eigenfunction of (5.1).

Hence, v is harmonic and vanishes on the nonempty subdomain Ω − D₁ ∪ ....D_n. Using the unique continuation theorem we arrive at a contradiction. This completes the theorem. Indeed, by the unique continuation theorem,

If v vanishes of finite order at x0 and u(x) ≥ 0 for x ∈ R ⊂ Ω where R = Ω − D1∪ .... ∪ Dn,

then v vanishes identically in Ω. Thus, v = 0. This is contradiction

The application we wish to make of this theorem is the following. Since an eigenfunction u₁ associated with σ1 satisfies

H

∂Ωu1ds = 0, we see that u1 must have a nodal line. By the

5.3

The method of defect

Theorem 24. (Schwarz reflection principle)

Let B ⊂ C be an open symmetric domain with respect to the real axis, and let B+_{= B∩{Imz >}

0} be the part of it in the upper half plane. Moreover, assume that u ∈ Har(B), and that u(z) −→ 0 as z ∈ B+ tends to any point on the real axis {Im z = 0}. Then u extends to be harmonic on B, and the extension satisfies

u(¯z) = −u(z) z ∈ B. 5.3.1 Method of defect

Suppose our domain B has two distinct axes of symmetry. They may be assumed perpendic-ular, and, with no loss of generality, we take them to be the x₁ and x₂ axes. We will call a function defined on B even-even, odd-odd, even-odd, or odd-even as u is respectively even in both x1 and x2, odd in both x1 and x2, even in x1 and odd in x2, or odd in x1 and even in x2.

Every eigenfunction of (5.1) can be assumed to belong to one of these symmetry classes. From the previous section, u2 must be in the even-odd or odd-even symmetry classes. Otherwise,

u2 has an even number of nodal lines. Hence, the nodal line is an axis of symmetry.

Indeed,

(i) As u is even in both x1 and x2, we have

u(x1, −x2) = u(x1, x2) = −u(x1, x2) = −u(x1, −x2).

Thus, u(x1, −x2) = 0 and axis x2 is nodal-line for u(x1, x2).

Similarly,

u(−x1, x2) = −u(x1, x2) = −u(−x1, −x2) = −u(−x1, x2).

Hence, u(−x₁, x2) = 0 and axis x1 is nodal-line for u(x1, x2).

Thus, when u is even in both x1, x2, u has even number of nodal lines.

(ii) Similarly, When u is odd in x1 and x2, we don’t have nodal line.

(iii) When u is even in x1 and odd in x2, there is exactly one nodal line.

(iv) When u is odd in x1 and even in x2, there is exactly one nodal line. The one-dimensional

inequality [u(0)]2+ [u(l)]2 ≤ l 2 Z l 0 [u0(y)]2dy. (5.4)

Indeed, u(x) =R₀xu0(y)dy. Hence,

|u(x)| =  Z x 0 u0(y)dy =⇒ |u(x)|2 = | Z x 0 u0(y)dy|2

=⇒ |u(x)|2 ≤ Z x 0 |u0(y)|2dy =⇒ |u(x)|2 ≤ C Z x 0 |u0(y)|2dy. Where C > 1. If we suppose that C = L₄, then

|u(0)|2 _≤ L 4 Z x 0 |u0(y)|2dy |u(x)|2≤ L 4 Z x 0 |u0(y)|2dy. Therefore, [u(0)]2+ [u(l)]2≤ l 2 Z l 0 [u0(y)]2dy

for continuous and piecewise continuously differentiable functions on the interval (0, l) which satisfy u(0) = ¯u(l), is easily shown by solving the Euler equation.

Let us first consider the case when the eigenfunction u2 is odd across the x1− axis. Suppose

the boundary ∂B can be expressed by x2 = ±f1(x1), −a1 ≤ x1≤ a1. Then, employing (5.4),

H ∂Bu 2 2 ds = 2 Z −a1 a1 [u2(x1, f1(x1))]2(1 + [f10(x1)]2) 1 2dx₁ ≤ Z (1 + [f₁0(x1)]2) 1 2[f₁(x₁) Z f1(x1) −f1(x1) [∂u2(x1, x2) ∂x2 ]2dx2]dx1

≤ [max−a1≤x1≤a1f1(x1)(1 + f

0 1(x1))]2) 1 2] Z a1 −a1 Z f1(x1) −f1(x1) (∂u2 ∂x2 )2dx2dx1

≤ [max−a1≤x1≤a1f1(x1)(1 + f

0 1(x1))]2) 1 2] Z B |grad u₂|2_dx.

By similarly treating the case when u₂ is odd across the x₂− axis, we have our inequality: σ−1₁ ≤ max i=1,2[max fi(xi)(1 + [f 0 i(xi)]2) 1 2], (5.5)

where x₁= ±f2(x2) , −a2 ≤ x2≤ a2, is another representation of ∂B.

5.4

Some examples

We give a few examples of the application of (5.5) to particular domains. For an upper bound, we use the isoperimetric inequality of Weinstock, which says

σ1≤

2π

where L is the length of ∂Ω. First, consider the rhombus

|x1 a1 | + |x2 a2 | ≤ 1. We have L = 4(a2₁+ a2 2)1/2 and fi(xi) = aj(1 − | xi ai |) j 6= i. Thus, from (5.5) and (5.5), we have

min(a2 a1 ,a1 a2 ) ≤ σ1(a21+ a22)1/2≤ π 2. (5.7) When a₁ = a2 = S/ √

2, the rhombus is a square of side S, and (6.1) become 1 ≤ σ₁S ≤ π₂, whereas the exact value to five places is σ1S = 1.3765.... Indeed,

|x1 a1 | + |x2 a2 | ≤ 1 We consider |x1 a1| + | x2 a2| = 1. Thus, we have x1 = a1(1 − | x2 a2|). Hence, f2(x2) = a1(1 − | x2 a2|). similarly, f₁(x1) = a2(1 − |x_a1₁|). We have, f10(x1) = a_a2₁ and similarly, f20(x2) = a_a1₂. Thus,

f1(x1)(1 + [f10(x1)]2) 1 2 = a2 a1 (a1+ a2)1/2 . Similarly, f2(x2)(1 + [f20(x2)]2) 1 2 = a1 a2 (a1+ a2)1/2 . Therefore, we have that

min(a2 a1 ,a1 a2 ) ≤ σ1(a21+ a22)1/2≤ π 2 Next, consider the ellipse (x1

a1) 2_{+ (}x2 a2) 2_{≤ 1 for which} fi(xi) = aj(1 − ( xi ai )2)1/2 j 6= i.

Indeed, we have f₁(x1)(1 + f10(x1)2)1/2= a2. Similarly, f2(x2)(1 + f20(x2)2)1/2= a1.

For simplicity, we worsen (5.6) by combining it with the classical isoperimetric inequality

L2 ≥ 4πA, (5.8)

where A, the area of the ellipse, is πa1a2. Thus, we have

[max(a1, a2)]−1 ≤ σ1 ≤ (a1.a2)−1/2. (5.9)