Nodal Sets of Steklov Eigenfuntions
by
Katarı́na Bellová
A dissertation submitted in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
Department of Mathematics
New York University
May, 2012
Professor Fanghua Lin
Abstract
We study the nodal set of the Steklov eigenfunctions on the boundary of a smooth
bounded domain in Rn – the eigenfunctions of the Dirichlet-to-Neumann map Λ. For a
bounded Lipschitz domain Ω ⊂ Rn , this map associates to each function u defined on
the boundary ∂Ω, the normal derivative of the harmonic function on Ω with boundary
data u.
Under the assumption that the domain Ω is C 2 , we prove a doubling property for the
eigenfunction u. The main goal of this Thesis is to estimate the Hausdorff Hn−2 -measure
of the nodal set of u|∂Ω in terms of the eigenvalue λ as λ grows to infinity, provided Ω is
fixed. In case that the domain Ω is analytic, we prove a polynomial bound (Cλ6 ).
My methods, which build on the work of Lin, Garofalo and Han [Garofalo and Lin,
CPAM 40 (1987), no. 3; Lin, CPAM 42 (1989), no. 6; Han and Lin, JPDE 7 (1994),
no. 2], can be used also for solutions to more general elliptic equations and/or boundary
conditions.
iii
Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iii
Introduction
0.1 Problem setting and main results . . . . . . . . . . . . . . . . . . . . . . .
0.2 Other related problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
2
3
4
1 Steklov Eigenfunctions and the Doubling Condition
6
1.1 Global Integral quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2 The Frequency Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Doubling Condition for Steklov Eigenfunctions . . . . . . . . . . . . . . . 18
2 Size of the Nodal Sets of Steklov Eigenfunctions
39
2.1 Nodal Sets of Steklov Eigenfunctions on Analytic Domains . . . . . . . . 39
2.2 Nodal Sets of Steklov Eigenfunctions on Smooth Domains . . . . . . . . . 44
Bibliography
45
iv
Introduction
Many classical results in linear elliptic partial differential equations are motivated by
complex analysis: the maximum principle (for modulus of holomorphic functions), the
unique continuation properties, the Cauchy integral representation formula, and the
interior gradient estimates of holomorphic functions were all generalized to the solutions
of linear second order elliptic PDEs with suitably smooth coefficients. A lot is known
about the nodal (zero) sets of holomorphic functions in complex plane, and it is an
important general research topic to find analogues for the nodal and critical point sets
of solutions to PDEs. In some cases, properties of nodal sets of solutions are themselves
the primary concern: in a study of moving defects in nematic liquid crystals by Lin [28],
the singular set of optical axes (i.e. defects) of liquid crystals in motion can be described
precisely by the nodal set of solutions to certain parabolic equations. In other cases,
nodal sets provide important information in the study of other properties of solutions.
In this Thesis, we are studying the nodal set of the Steklov eigenfunctions on the
boundary of a smooth bounded domain in Rn – the eigenfunctions of the Dirichlet-toNeumann map Λ. For a bounded Lipschitz domain Ω ⊂ Rn , this map associates to each
function u defined on the boundary ∂Ω, the normal derivative of the harmonic function
on Ω with boundary data u. More generally, one can consider an n-dimensional smooth
Riemannian manifold (M, g) instead of Ω, and replace the Laplacian by the LaplaceBeltrami operator ∆g . Our methods, which build on the work of Lin, Garofalo, and Han
([18, 29, 23]), can be used also for solutions to more general elliptic equations and/or
boundary conditions.
The Steklov eigenfunctions were introduced by Steklov [34] in 1902 for bounded domains in the plane. He was motivated by physics – the functions represent the steady
state temperature on Ω such that the flux on the boundary is proportional to the temperature. The problem can also be interpreted as vibration of a free membrane with
the mass uniformly distributed on the boundary. Note that the eigenfunction’s nodal set
represents the stationary points on the boundary. The Steklov problem is also important
in conformal geometry, Sobolev trace inequalities, and inverse problems.
The Thesis is organized as follows. The remainder of the introduction summarizes
the main results, describes related problems and fixes some basic notation. In Chapter 1
we prove a doubling condition for Steklov eigenfunctions on a C 2 -domain Ω, which will
serve as a cornerstone for the nodal set estimate in both analytic and smooth setting.
In Chapter 2 we prove an explicit estimate for the nodal set of Steklov eigenfunctions in
the case that Ω has analytic boundary. We also discuss the case when Ω is smooth, but
1
not analytic, and outline a possible direction of research in that setting.
0.1
Problem setting and main results
Let Ω ⊂ Rn be a Lipschitz domain. The Dirichlet-to-Neumann operator Λ : H 1/2 (∂Ω) →
H −1/2 (∂Ω) is defined as follows. For f ∈ H 1/2 (∂Ω), we solve the Laplace equation
∆u = 0
in Ω,
u=f
on ∂Ω.
This gives a solution u ∈ H 1 (Ω), and we set (Λf ) to be the trace of ∂u
∂ν on ∂Ω, where ν
is the exterior unit normal. We obtain a bounded self-adjoint operator from H 1/2 (∂Ω)
to H −1/2 (∂Ω). It has a discrete spectrum {λj }∞
j=0 , 0 = λ0 < λ1 ≤ λ2 ≤ λ3 ≤ ...,
limj→∞ λj = ∞. The eigenfunctions of Λ (called Steklov eigenfunctions) corresponding
to eigenvalue λ can be identified with the trace on ∂Ω of their harmonic extensions to
Ω, which satisfy
∆u = 0
in Ω,
(1)
∂u
= λu on ∂Ω.
∂ν
The main goal of this Thesis is to estimate the size (the Hausdorff Hn−2 -measure) of
the nodal set of u|∂Ω in terms of λ as λ grows to infinity, provided Ω is fixed. In Chapter
2.1 we prove the following bound in case that Ω has analytic boundary:
Theorem 1. Let Ω ⊂ Rn be an analytic domain. Then there exists a constant C
depending only on Ω and n such that for any λ > 0 and u which is a (classical) solution
to (1) there holds
Hn−2 ({x ∈ ∂Ω : u(x) = 0}) ≤ Cλ6 .
(2)
The scaling λ6 in (2) is not optimal. Actually, our proof gives approximately λ5.6 , but we
believe that the optimal scaling is λ. However, even this polynomial bound is valuable
in a problem like this – the main difficulty in the estimate is to avoid an exponential
bound eCλ .
As in [29] and [23], we use a doubling condition, i.e. a control of the L2 -norm of u on
a ball B2r (x) by the L2 -norm of u on a smaller ball Br (x), as the crucial tool to estimate
the nodal set. We prove a doubling condition in the following form in Chapter 1:
Theorem 2. Let Ω ⊂ Rn be a C 2 domain. Then there exist constants r0 , C > 0
depending only on Ω and n such that for any r ≤ r0 /λ, x ∈ ∂Ω, and u which is a
(classical) solution to (1), there holds
Z
Z
2
Cλ5
u2 .
u ≤2
B(x,r)∩∂Ω
B(x,2r)∩∂Ω
To put our work into context, note that our problem is similar in nature to the
classical question of estimating the size of nodal sets of eigenfunctions of the Laplace
operator in a compact manifold. The following conjecture was proposed by Yau in [39]:
2
Conjecture 0.1.1. Suppose (M n , g) is a smooth n-dimensional connected and compact
Riemannian manifold without boundary. Consider an eigenfunction u corresponding to
the eigenvalue λ, i.e. ,
∆g u + λu = 0 on M.
Then there holds
√
√
c1 λ ≤ Hn−1 ({x ∈ M ; u(x) = 0}) ≤ c2 λ,
where c1 and c2 are positive constants depending only on (M, g).
This conjecture was proved in case that (M n , g) is analytic by Donnelly and Fefferman
in [10]. It is still open whether Conjecture 0.1.1 holds if (M n , g) is only smooth. The
known results for the smooth case are far from optimal, the upper bound remaining
exponential (see [25]).
Another similar problem has been studied for the Neumann eigenfunctions on a
piecewise analytic plane domain Ω ⊂ R2 in [36]. This paper is concerned about the
asymptotics of the number of nodal points of the eigenfunctions on the boundary ∂Ω, as
the eigenvalue λ increases to infinity. It proves that this number is bounded above by
CΩ λ.
0.2
Other related problems
In this section we briefly present some other problems related to the Dirichlet-to-Neumann
map and its eigenvalues/eigenfunctions. They are classical problems which illustrate the
importance of this map.
The classical Yamabe Problem consists in showing that every Riemannian compact
manifold, without boundary, admits a conformally related metric with constant scalar
curvature. It was formulated by Yamabe [38] in 1960 and proved by Aubin ([4], 1976)
and Schoen ([33], 1984). A related problem for a manifold (M, g) with boundary asks
to find a conformally related metric with zero scalar curvature on M and constant mean
curvature on the boundary. It was solved in most cases by Escobar ([12]) and by Marques
([31]) in the remaining ones, and can be thought of as a generalization of the Riemann
mapping theorem to higher dimensions. This problem can be reformulated as finding a
solution to
∆g u = 0
in M,
n
du n − 2
on ∂M,
+
hg u = λu n−2
dν
2
u>0
in M,
where hg is the mean curvature on the boundary. That corresponds to solving
Λg (u) + f (u) = 0
n
(3)
n−2 . Equations of this type often come
on ∂M with a nonlinearity f (u) = n−2
2 hg u − λu
up when studying the geometry of a manifold. Much like in [7] for a different problem,
the understanding of Steklov eigenfunctions is helpful in the study of such equations.
3
Another well-known example is looking for the optimal constant in the Sobolev trace
inequality. The first non-zero eigenvalue for the Steklov problem, λ1 , has a variational
characterization:
R
|∇f |2
λ1 = R min RM
.
2
∂M f =0
∂M f
From this characterization follows this Sobolev trace inequality: for all functions f ∈
H 1 (M ), we have
Z
Z
1
2
|f − f | ≤
|∇f |2 ,
(4)
λ
1
∂M
M
where f is the mean value of f when restricted to the boundary. The last inequality
is fundamental in the study of existence and regularity of solutions of some boundary
value problems. Hence it is important to know the dependence of λ1 on the geometry of
M . It has been studied e.g. in [37, 32, 13, 16]. Estimates for the higher eigenvalues and
their distribution have also been studied, see e.g. [5, 26, 9].
Inequality (4) implies the embedding H 1 (M ) ֒→ L2 (∂M ). This is not the optimal
embedding. For simplicity, consider the half-space Rn+ . For any function f on Rn+ ,
sufficiently smooth and decaying fast enough at infinity, there holds
||f ||L2(n−1)/(n−2) (∂Rn ) ≤ C(n)||∇f ||L2 (Rn+ ) ,
+
(5)
where C(n) is a known constant. The functions f for which equality holds in (5) satisfy
the Euler-Lagrange equation
∆f = 0
∂f
− Q(Rn+ )f n/(n−2) = 0
∂ν
on Rn+ ,
on ∂Rn+ ,
where Q(Rn+ ) = 1/C(n)2 . This is again an equation of type (3). See [11] for details, and
[27] for a generalization for manifolds.
There are many more problems related to the Steklov eigenfunctions. The understanding of the problem could have applications in inverse conductivity problems ([8, 35]),
cloaking ([2]), and modeling of sloshing of a perfect fluid in a tank ([15, 6]).
0.3
Notation
Throughout this Thesis, Br = B(0, r) will denote the open ball in Rn with center 0 and
radius r, and B(x, r) the open ball in Rn with center x and radius r. Similarly we use
k
B n−1 (x, r) for ball in Rn−1 and B C (z, r) for ball in Ck (k = n, n − 1).
We denote the coordinates of a vector b ∈ Rn by bi , i = 1, 2, . . . , n. We abbreviate
2w
∂w
the partial derivatives as ∂x
= wxi , ∂x∂i ∂x
= wxi ,xj .
i
j
We use the letter C as a constant (usually depending only on n and Ω), which can
change from line to line.
For a given Ω, there are only finitely many eigenvalues λ of the Dirichlet-to-Neumann
map which are less than 1. Hence, when we are proving upper bounds in the form of
4
Cλk with specific k and C depending on Ω, without loss of generality we can assume
that the eigenvalue λ is larger or equal to 1, and we often do so without pointing it out.
5
Chapter 1
Steklov Eigenfunctions and the
Doubling Condition
1.1
Global Integral quantities
In this section, we will derive some properties of the gradient of Steklov eigenfunctions.
Although we will not use them in the proofs of the main results, they illustrate the
special structure of the Steklov eigenfunctions.
Let u be a solution of (1) on a Lipschitz domain Ω. By normalization, assume
Z
u2 = 1.
(1.1)
∂Ω
Proposition 1.1.1.
Z
Ω
|∇u|2 = λ.
(1.2)
Proof. By integration by parts and using (1) and (1.1), we have
Z
Z
Z
Z
Z
∂u
∂u
2
u
|∇u| =
u
u∆u =
u2 = λ.
−
=λ
∂Ω ∂ν
Ω
∂Ω ∂ν
Ω
∂Ω
Next, we are interested in the size of |∇u|2 integrated over ∂Ω. For the gradient in
the normal direction we have
2
Z
∂u = λ2
u 2 = λ2 .
(1.3)
∂Ω ∂ν
∂Ω
We will show that the gradient in the tangential directions is of comparable size.
Z
Theorem 3.
2
∂u ≈ λ2 ,
(1.4)
∂Ω ∂τ
where the sign ≈ means that there exist constants C1 , C2 depending on Ω and n such
R 2
≤ C 2 λ2 .
that C1 λ2 ≤ ∂Ω ∂u
∂τ
Z
6
Proof. To simplify the notation, we will use the summation convention – summing over
the indices i and j (from 1 to n) wherever they come up.
By multiplying the equation ∆u = 0 by x · ∇u and integrating by parts, we get
Z
Z
Z
Z
∂u
2
0=
x i u xi u xj xj =
x · ∇u
−
|∇u| −
x i u xi xj u xj .
(1.5)
∂ν
Ω
∂Ω
Ω
Ω
Further integration by parts yields
Z
Z
Z
Z
2
2
x i u xi xj u xj ,
x · ν|∇u| − n |∇u| −
x i u xi xj u xj =
Ω
ΩZ
∂Ω
ZΩ
Z
1
n
x i u xi xj u xj =
x · ν|∇u|2 −
|∇u|2 .
2 ∂Ω
2 Ω
Ω
Hence from (1.5) we get
Z
Z
2
(n − 2) |∇u| =
Ω
2
∂Ω
x · ν|∇u| − 2
Z
∂Ω
2 Z
2
∂u ∂u x · ν −
=
x · ν − 2
∂τ
∂ν
∂Ω
∂Ω
Z
∂u
∂ν
n−1
XZ
x · ∇u
k=1
∂Ω
x · τk
∂u ∂u
,
∂τk ∂ν
(1.6)
2
where τ1 , . . . , τn−1 are the mutually perpendicular tangential vectors on ∂Ω, and | ∂u
∂τ | =
Pn−1 ∂u
2
′
k=1 | ∂τk | . Note that (1.6) is valid also if we replace Ω by any subdomain Ω ⊂ Ω.
Let us first assume that Ω is star-shaped. Then there exists x0 ∈ Ω such that for each
x ∈ ∂Ω the segment x0 x lies in Ω, and (x − x0 ) · ν(x) ≥ δ for some δ > 0 independent of
x. Without loss of generality, we can assume x0 = 0 (otherwise we can derive (1.6) with
x − x0 in place of x). Since Ω is bounded, we also have |x| ≤ K for x ∈ ∂Ω and some
constant K. Hence from (1.6), using (1.2) and (1.3) and the Cauchy-Schwartz inequality,
we get
2
Z
Z 2
∂u ∂u 1
≤
x · ν ∂τ δ ∂Ω
∂τ
∂Ω
!
2
Z
Z
n−1
XZ
∂u
∂u
1
∂u
x · τk
=
(n − 2) |∇u|2 +
x · ν + 2
δ
∂ν
∂τk ∂ν
Ω
∂Ω
k=1 ∂Ω
Z 2
2
∂u K
1
n−2
+ 2(n − 1)K λ2 .
λ + λ2 +
≤
δ
δ
2 ∂Ω ∂τ δ2
Therefore,
Z
∂Ω
2
∂u ≤ C 2 λ2 ,
∂τ where C2 is a constant depending only on Ω and n.
7
On the other hand, analogously
2
Z 2
Z
∂u ∂u ≥ 1
x
·
ν
∂τ K ∂Ω
∂Ω ∂τ
!
2
Z
Z
n−1
XZ
∂u 1
∂u
∂u
2
x · ν + 2
=
(n − 2) |∇u| +
x · τk
K
∂ν
∂τk ∂ν
∂Ω
Ω
∂Ω
k=1
Z
∂u 2
δ
2(n − 1)K
n−2
− δ λ2 ,
λ + λ2 −
≥
K
K
δ
2K
∂Ω ∂τ
whence
Z
2
∂u ≥ C 1 λ2 ,
∂Ω ∂τ
where C1 > 0 is a constant depending only on Ω and n. We have thus proved (1.4) for
star-shaped Ω.
In the general case, we cover the boundary ∂Ω by p pieces ∂Ω = ∪pα=1 Dα such that
each Dα = ∂Ω ∩ A′α , where A′α is a rectangular box and Dα is connected. Moreover, we
require that Aα = A′α ∩ Ω is star-shaped (this can be assured because Ω is Lipschitz),
and
2n−1
[
Sα,i ∪ Dα ,
∂Aα =
i=1
where each Sα,i lies in a hyperplane Hα,i and if we shift every Hα,i parallelly by ǫ > 0
in the direction away from Aα , the body formed by these shifted hyperplanes and ∂Ω
containing Dα stays in Ω. Such covering of ∂Ω exists, with ǫ > 0 depending on Ω.
β
Now take one Dα and denote by Hα,i
the hyperplane parallel to Hα,i in the distance
of β away from A′α , 0 ≤ β ≤ ǫ. By Fubini’s theorem and (1.2),
Z
Z ǫZ
2
|∇u|2 = λ,
|∇u| dS dβ ≤
0
β
Hα,i
∩Ω
Ω
so there exists β ∈ [0, ǫ] such that
Z
β
Hα,i
∩Ω
|∇u|2 dS ≤ Cλ,
where C depends on Ω (actually, C = 1/ǫ). Hence, by possibly shifting the walls Sα,i of
Dα , we can assume
Z
|∇u|2 ≤ Cλ.
(1.7)
Sα,i
Now apply (1.6) to Aα instead of Ω (see the remark under (1.6)):
(n − 2)
Z
Aα
|∇u|2 =
Z
2 Z
2
n−1
XZ
∂u ∂u ∂u ∂u
x · ν −
x · τk
. (1.8)
x · ν − 2
∂τ
∂ν
∂τ
k ∂ν
∂Aα
∂Aα
∂Aα
k=1
8
For the left-hand side we have
Z
Z
2
(n − 2)
|∇u| ≤ (n − 2) |∇u|2 = (n − 2)λ.
Aα
Ω
If
we split the integrals on S
the right-hand side of (1.8) into integrals over Dα and
S2n−1
2n−1
i=1 Sα,i , all integrals over
i=1 Sα,i are at most of order λ by (1.7). Since
Z 2 Z 2
∂u ∂u ≤
= λ2
∂ν Dα
∂Ω ∂ν
and for some α
Z
2
Z 2
∂u ∂u 1
= 1 λ2 ,
≥
∂ν p
p
∂Ω ∂ν
Dα
the same argument as in the star-shaped case gives us the existence of constants C1 , C2 >
0 depending on Ω such that
Z 2
∂u ≤ C 2 λ2
for all α,
Dα ∂τ
Z 2
∂u ≥ C 1 λ2
for some α.
∂τ Dα
The theorem follows.
1.2
The Frequency Function
In this section we will review the theory developed in [17] and [18] about the frequency
functions for both harmonic functions and solutions to general elliptic equations. We
will heavily rely on these results in the following section (Doubling Condition for Steklov
Eigenfunctions).
1.2.1
Frequency function and doubling condition for harmonic functions
We use the frequency function as the main tool to derive the doubling condition, as
e.g. in [17], [18], [29], [20]. Since we will use the theory also for harmonic functions and
the exposition is easier in this special case, let us present it first in this setting.
Definition. For a harmonic function u on ball B1 and r < 1, the frequency N (r) is
defined as
rD(r)
,
(1.9)
N (r) =
H(r)
where
Z
D(r) =
|∇u|2 dx,
Z Br
H(r) =
u2 dσ,
∂Br
9
where σ is the surface measure. For a harmonic function on a ball B(a, r), N (a, r),
D(a, r) and H(a, r) are defined analogously.
The frequency is a way how to measure the growth of a harmonic function. If u is
a homogeneous harmonic polynomial, its frequency is exactly its degree. See [20] or [22]
for more examples. Let us list some important properties. We refer to the survey paper
[20] for proofs, although they are known much longer and sketches of the proofs can be
found e.g. in [17] or [29]. The following monotonicity property of the frequency function
is attributed to F. J. Almgren, Jr., [1].
Proposition 1.2.1. Let u be a harmonic function in B1 . Then N (r) is a nondecreasing
function of r ∈ (0, 1).
Proof. See [20], Theorem 1.2.
Corollary 1.2.2. The limit limr→0 N (r) exists and equals to the vanishing order of u
at 0.
Proof. See [20], Corollary 1.3.
Proposition 1.2.3. Let u be a harmonic function in B1 . For any r ∈ (0, 1), there holds
d
H(r)
N (r)
log n−1 = 2
.
(1.10)
dr
r
r
Integrating (1.10), we obtain that for any 0 < r1 < r2 < 1, there holds
Z r2
H(r2 )
H(r1 )
N (r)
.
= n−1 exp 2
r
r2n−1
r1
r1
(1.11)
Using the monotonicity of N (Proposition 1.2.1), it follows that
H(r2 )
≤
r2n−1
r2
r1
2N (r2 )
H(r1 )
.
r1n−1
(1.12)
Proof. See [20], Corollary 1.4.
Corollary 1.2.4. Let u be a harmonic function in B1 . Then the function
Z
r 7→ − u2
(1.13)
∂Br
is increasing with respect to r, r ∈ (0, 1), and
Z
Z
2
− u ≤−
Br
u2 .
(1.14)
∂Br
Proof. It follows directly from (1.10) that the function (1.13) is increasing. By integration, we get (1.14).
10
The following result, if we take η = 1/2, is called the doubling condition. It is a
counterpart of Corollary 1.2.4.
Corollary 1.2.5. Let u be a harmonic function in B1 . For any R, η ∈ (0, 1), there holds
Z
Z
2
−2N (R)
− u ≤η
−
u2 ,
(1.15)
∂BR
∂B
Z
Z ηR
− u2 ≤ η −2N (R) − u2 .
BR
(1.16)
BηR
Proof. Taking r1 = ηR, r2 = R in (1.12), we obtain
H(R)
H(ηR)
≤ η −2N (R)
,
n−1
R
(ηR)n−1
which is (1.15). Integrating (1.15) from 0 to R, and using the monotonicity of N (Proposition 1.2.1), we obtain (1.16):
Z
Z
Z RZ
Z R
−2N (r) 1
2
2
u2 dσ∂Bηr dr
η
u =
u dσ∂Br dr ≤
n−1
η
∂Bηr
BR
0
∂Br
0
Z ηR Z
Z
1
1
u2 dσ∂Bρ dρ = η −2N (R) n
u2 .
≤ η −2N (R) n
η 0
η
∂Bρ
BηR
Next, we show that not only having a bound on the frequency implies a doubling
condition (Corollary 1.2.5), but also knowing a doubling condition to be true implies a
bound on the frequency.
Lemma 1.2.6. Let u be a harmonic function in Br , r > 0. Let 0 < α < γ < 1, and
assume
Z
Z
2
− u ≥ κ− u2
(1.17)
Bαr
for some κ > 0. Then
N (αr) ≤
Br
− log (κ(1 − γ n ))
.
2 log (γ/α)
In particular, for any β < α, there holds
Z
Z
log(α/β)
2
n log(γ/α)
− u ≥ (κ(1 − γ ))
−
Bβr
(1.18)
u2 .
Bαr
Proof. Using Corollary 1.2.4 and the assumption (1.17), we obtain
Z rZ
Z
Z
Z
1
u2 dσdρ
−
u2 ≥ − u2 ≥ κ− u2 ≥ κ
n
ω
r
n
γr ∂Bρ
∂Bαr
Bαr
Br
Z r n−1 Z
Z
ρ
1
2
n
dρ
u = κ(1 − γ )−
u2 .
≥κ
ωn rn γr γr
∂Bγr
∂Bγr
11
(1.19)
Using (1.11) and the monotonicity of N , we further obtain
Z γr
H(γr)/(γr)n−1
N (ρ)
n
− log(κ(1 − γ )) ≥ log
≥ 2N (αr) log(γ/α),
=2
n−1
H(αr)/(αr)
ρ
αr
which is (1.18). The inequality (1.19) follows from this bound on N (αr) and the doubling
condition (1.16):
2N (αr) Z
− log(κ(1−γ n )) Z
log(γ/α)
β
β
− u2
u2 ≥
− u2 ≥
α
α
Bαr
Bβr
Bαr
Z
log(α/β)
= (κ(1 − γ n )) log(γ/α) − u2 .
Z
−
Bαr
Next, we recall a result estimating the frequency in a given point by the frequency in
a different point. This is an important property whose adaptation we will use to obtain
global estimates.
Proposition 1.2.7. Let u be a harmonic function in B1 . For any R ∈ (0, 1), there
exists a constant N0 = N0 (R) ≪ 1 such that the following holds. If N (0, 1) ≤ N0 , then
u does not vanish in BR . If N (0, 1) ≥ N0 , then there holds
1
N p, (1 − R) ≤ CN (0, 1), for any p ∈ BR ,
2
where C is a positive constant depending only on n and R. In particular, the vanishing
order of u at any point in BR never exceeds c(n, R)N (0, 1).
Proof. See [20], Theorem 1.6.
1.2.2
Frequency and doubling condition for solutions of general elliptic
equations
We will need a generalization of the frequency (1.9) for more general elliptic equations.
Let us recall the theory developed in [17] and [18].
In the unit ball B1 ⊂ Rn , consider the equation
Lw = div(A(x)∇w) + b(x) · ∇w + c(x)w = 0,
(1.20)
where A(x) = (aij (x))ni,j=1 is a real symmetric matrix-valued function on B1 , b(x) is a
vector valued and c(x) is a scalar function on B1 . We assume
(i) there exists α ∈ (0, 1) such that, for every x ∈ B1 and ξ ∈ Rn ,
α|ξ|2 ≤
n
X
i,j=1
12
aij (x)ξi ξj ;
(1.21)
(ii) there exists Γ > 0 such that, for every x, y ∈ B1 ,
|aij (x) − aij (y)| ≤ Γ|x − y|,
i, j = 1, . . . , n;
(iii) there exists K > 0 such that
X
X
||aij ||L∞ (B1 ) +
||bj ||L∞ (B1 ) + ||c||L∞ (B1 ) ≤ K.
i,j
(1.22)
(1.23)
j
As in [17] and [18], we introduce a Riemannian metric associated with the principal
part of the operator L in (1.20). Set
ḡij (x) = aij (x)(det A(x))1/(n−2) ,
where (aij (x)) = A−1 (x). Let (ḡ ij )(x) = (ḡij (x))−1 ,
r(x)2 = ḡij (0)xi xj ,
η(x) = ḡ kl (x)
∂r
∂r
(x)
(x).
∂xk
∂xl
Finally introduce a new metric tensor gij (x)dxi ⊗ dxj by defining
gij = η(x)ḡij (x).
Then gij are Lipschitz functions with a Lipschitz constant depending on Γ in (1.22). Let
M be the Riemannian manifold (B1 , gij ). As proved in Section 3 of [3], in the intrinsic
geodesic polar coordinates with pole at zero, the metric tensor gij dxi ⊗ dxj becomes
dr ⊗ dr + r2 bij (r, θ)dθi ⊗ dθj , where
∂
bij (0, 0) = δij , bij (r, θ) ≤ Λ,
i, j = 1, 2, . . . , n − 1,
(1.24)
∂r
and Λ depends on n, α, Γ. Hence the geodesic ball of radius r and center at x = 0
coincides with the Euclidean ball Br with center at zero. Notice that if A(x) = I
(identity matrix), then g(x) = I and b(r, θ) = I in the point (r, θ) corresponding to x.
In general, from (1.24) we obtain that there exists r0 > 0 depending on n, α, Γ such that
1
I ≤ (bij (r, θ)) ≤ 2I
2
i.e.
for 0 ≤ r ≤ r0 .
(1.25)
Denote by ∇M and divM , respectively, the intrinsic gradient and divergence on M ,
∇M w =
n
X
g ij
i,j=1
∂w ∂
,
∂xi ∂xj
n
1 X ∂ √
divM X = √
( gXi ),
g
∂xi
i,j=1
13
where (g ij (x)) is the inverse matrix of G = (gij (x)) and g(x) = | det(gij (x))|. Then
(1.20) can be rewritten as
divM (µ(x)∇M w) + bM (x) · ∇M w + cM (x)w = 0.
(1.26)
Here µ(x) = η 1−n/2 (x) is a Lipschitz function. Written in polar coordinates, µ(x) =
µ(r, θ) satisfies
∂
µ(0, 0) = 1, µ(r, θ) ≤ Λ,
∂r
C1 ≤ µ(x) ≤ C2 ,
(1.27)
where C1 , C2 are positive constants depending only on n, α, Γ. Finally
√
bM = G(b/ g),
√
cM = c/ g.
√
Since ( g)−1 is a bounded Lipschitz factor whose bounds depend only on n, α, and Γ,
the coefficients bM and cM satisfy the same bounds (1.23) as b, c, up to a multiplicative
constant depending on n, α, Γ.
1,2
Notation 1.2.8. For a solution w ∈ Wloc
(B1 ) of (1.20) in B1 and 0 < r < 1, let
Z
H(r) =
µw2 dV∂Br
Z ∂Br
D(r) =
µ|∇M w|2 dVM ,
Br
Z
I(r) =
(µ|∇M w|2 + ubM · ∇M w + cM w2 ) dVM ,
Br
N (r) =
rI(r)
,
H(r)
the last quantity being defined only if H(r) > 0. Analogously we define these quantities
not only for solutions w on the unit ball B1 = B(0, 1), but also for solutions on any ball
B(x0 , r0 ) with center x0 and radius r0 . Then H(r), D(r), I(r) and N (r) also depend
on w and x0 , and if the function and the center are not clear from the context, we will
denote them Hw (x0 , r), Dw (x0 , r), Iw (x0 , r), Nw (x0 , r) (we will skip either x0 or w if just
one is not clear from the context).
Note that unlike in the harmonic case, it is not clear anymore where is N (r) defined
and whether N (r) > 0: we have D(r) > 0, H(r) ≥ 0, but the sign of I(r) is not
obvious. However, it can be shown that at least on some interval, N (r) is defined and
N (r)/r ≥ −C for some constant C.
1,2
Lemma 1.2.9. Let w ∈ Wloc
(B1 ) be a nonzero solution to (1.20) in B1 . Then there
exist constants r0 , C > 0, depending on n, α, Γ, K, such that for any r ∈ (0, r0 ),
D(r) ≤ 2I(r) + CH(r),
1
D(r) ≥ I(r) − CH(r),
2
H(r) > 0.
14
(1.28)
Proof. The lemma can be found in [22] as Lemma 3.2.3 and Corollary 3.2.5. In [18], the
last inequality is Lemma 2.2. and the previous two easily follow from the estimates done
in the proof of this lemma.
1,2
Corollary 1.2.10. Let w ∈ Wloc
(B1 ) be a nonzero solution to (1.20) in B1 . Then there
exist constants r0 , C > 0, depending on n, α, Γ, K, such that for any r ∈ (0, r0 ), N (r) is
defined and
N (r)
≥ −C.
r
Proof. Take r0 from Lemma 1.2.9. Then for r ∈ (0, r0 ), H(r) is positive, so N (r) is
defined and by (1.28), we obtain
N (r)
I(r)
D(r)
=
≥
− C ≥ −C.
r
H(r)
2H(r)
Corresponding to the monotonicity of N in the harmonic case, there holds the following bound in the general elliptic setting.
1,2
Theorem 4. Let w ∈ Wloc
(B1 ) be a nonzero solution to (1.20) in B1 . Then there exist
constants r0 , c1 , c2 > 0, depending on n, α, Γ, K, such that
N (R1 ) ≤ c1 + c2 N (R2 )
for any 0 < R1 < R2 ≤ r0 .
(1.29)
Proof. This theorem follows from Theorem 2.1 in [18], using our bounds (1.23) instead
of their more general assumptions on b and c, and checking the L∞ bound on N in the
proof. It can also be found in a more similar form in [22] as Theorem 3.2.1. The only
difference in our formulation is that we state the estimate for any R2 ≤ r0 , not just
R2 = r0 . However, going through the proofs of the theorems in [18] or [22], one can
check that the estimate is true with our formulation without any changes to the proofs.
Another way how to verify inequality (1.29) for R2 < r0 if we know that it holds for
R2 = r0 is to consider a solution wr0 /R2 = w( Rr20x ) of a scaled equation (which satisfies the
assumptions (1.21), (1.22) and (1.23) with the same constants as the original equation),
f2 = r0 , and then use the scaling of N explained in the next
use (1.29) for wr0 /R2 and R
section to deduce (1.29) for w and the original R2 < r0 .
Next, we will recall a few results for solutions of general elliptic equations corresponding to Proposition 1.2.3, Corollary 1.2.4 and Corollary 1.2.5. Instead of Proposition 1.2.3,
we have the following.
1,2
Proposition 1.2.11. Let w ∈ Wloc
(B1 ) be a nonzero solution to (1.20) in B1 . Then
H(r)
N (r)
d
log n−1 = O(1) + 2
,
(1.30)
dr
r
r
15
where O(1) denotes a function bounded by a constant depending on n, α, Γ and the L∞ bound on the leading order coefficients A. Integrating (1.30), we obtain that for any
0 < R1 < R2 < 1, there holds
Z R2
H(R2 )
H(R1 )
N (r)
= n−1 exp O(1)(R2 − R1 ) + 2
.
(1.31)
r
R2n−1
R1
R1
Using Theorem 4 (the bound for N ), it follows that if R2 ≤ r0 (where r0 comes from
Theorem 4 and depends on n, α, Γ, K), then
c1 +c2 N (R2 )
H(R1 )
H(R2 )
(R2 −R1 ) R2
,
(1.32)
n−1 ≤ C
R1
R2
R1n−1
where C, c1 , c2 depend on n, α, Γ, K.
Proof. Equation (1.30) can be found in [18] as equation (2.16). The fact that O(1) does
not depend on the coefficients b and c can be verified by going back to the proof of this
equation. The O(1) is the same as in [17] for equations with no lower order terms.
The following corollary corresponds to Corollary 1.2.4.
1,2
Corollary 1.2.12. Let w ∈ Wloc
(B1 ) be a nonzero solution to (1.20) in B1 . Then there
exist constants r0 , C > 0 depending only on n, α, Γ, K such that
Z
Z
2
− w ≤ C− w2 for any 0 < s < r < r0
(1.33)
∂Bs
and
∂Br
Z
Z
2
− w ≤ C−
Br
w2
∂Br
for r ∈ (0, r0 ).
(1.34)
Proof. Using Corollary 1.2.10, i.e. N (r)/r ≥ −C, and (1.30), we obtain
H(r)
N (r)
d
≥ −C
log n−1 = O(1) + 2
dr
r
r
for some constants r0 , C and r < r0 . Hence the Rfunction eCr H(r)/rn−1 is increasing.
Next, the function µ in the definition of H(r) = ∂Br µw2 dV∂Br is bounded both from
below and above by positive constants ((1.27)), and by (1.25) we can adjust r0 so that
the measure dV∂Br , r < r0 is comparable with Hn−1 . Then we easily obtain the estimate
(1.33) for the average integrals. By integration, we get (1.34).
Next, we state the doubling condition.
1,2
Theorem 5. Let w ∈ Wloc
(B1 ) be a nonzero solution to (1.20) in B1 . Then there exist
constants r0 , C, c1 , c2 > 0, depending on n, α, Γ, K, such that for any 0 < R1 < R2 < r0 ,
c1 +c2 N (R2 ) Z
Z
R2
2
w2 ,
(1.35)
−
w ≤C
−
R
1
∂BR1
∂BR2
c1 +c2 N (R2 ) Z
Z
R2
− w2 ≤ C
− w2 .
(1.36)
R
1
BR
BR
2
1
16
Proof. This can be found in a slightly different form in [18] as Theorem 1.2, or inR [22] as
Theorem 3.2.7. It also follows from (1.32) using that H(r) is comparable with ∂Br w2
for r small enough.
Finally, let us show the equivalent of Lemma 1.2.6: knowing a doubling condition to
be true implies a bound on the frequency.
1,2
Lemma 1.2.13. Let w ∈ Wloc
(B1 ) be a nonzero solution to (1.20) in B1 . Assume
Z
Z
2
− w ≥ κ− w2
(1.37)
Bζr
Br
for some κ, ζ ∈ (0, 1) and r ∈ (0, r0 ], where r0 depends on n, α, Γ, K and is chosen so
that the previous properties in this subsection hold. Then there exists a constant Cζ > 0
depending on n, α, Γ, K and ζ such that
N (ζr) ≤ Cζ (1 − log κ).
(1.38)
In particular, for any β < ζ, there exist constants C1 , C2 depending on n, α, Γ, K and
ζ, β such that
Z
Z
1 C2
κ − w2 .
(1.39)
− w2 ≥
C1
Bζr
Bβr
Proof. We copy the proof of Lemma 1.2.6. Choose γ such that ζ < γ < 1, e.g. γ = 1+ζ
2 .
Using Corollary 1.2.12 and the assumption (1.37), we obtain
Z
Z
Z
Z rZ
1
1
1
1
2
2
2
−
w ≥ − w ≥ κ− w ≥ κ
w2 dσdρ
n
C
C
C
ω
r
n
γr ∂Bρ
Bζr
Br
∂Bζr
Z
Z r n−1 Z
1
ρ
1
1
n
2
≥ κ
dρ
w = κ(1 − γ )−
w2 .
C ωn rn γr γr
C
∂Bγr
∂Bγr
R
2
Using (1.31), Theorem 4 and the comparability of H(r) and ∂Br w , we further obtain
H(γr)/(γr)n−1
n
C − log(κ(1 − γ )) ≥ log
H(ζr)/(ζr)n−1
Z γr
N (ρ)
= O(1)(γ − ζ)r + 2
ρ
ζr
1
≥ −C + N (ζr) log(γ/ζ),
C
and (1.38) follows. The inequality (1.39) follows from this bound on N (ζr) and the
doubling condition (1.36):
f f
Z
Z
Z
1 β C1 −C2 log κ
1 β c1 +c2 N (ζr)
2
2
− w2
− w ≥
− w ≥
C ζ
C ζ
B
Bζr
Bβr
Z ζr
1 C2
κ − w2 .
=
C1
Bζr
17
1.3
Doubling Condition for Steklov Eigenfunctions
In this section, Ω ⊂ Rn will denote a C 2 domain, and u will be a Steklov eigenfunction
corresponding to eigenvalue λ, harmonically extended to Ω, i.e satisfying (1):
∆u = 0
∂u
= λu
∂ν
in Ω,
on ∂Ω.
The doubling condition, i.e. controlling the L2 -norm of u on a ball B2r (x) by the
L2 -norm of u on a smaller ball Br (x), is an important property when trying to estimate
the nodal set. The main result of this section is the following theorem.
Theorem 2. There exist constants r0 , C > 0 depending only on Ω and n such that for
any r ≤ r0 /λ and x ∈ ∂Ω,
Z
Z
2
Cλ5
u2 .
(1.40)
u ≤2
B(x,r)∩∂Ω
B(x,2r)∩∂Ω
Remark 1.3.1. The exponent λ5 in (1.40) is not optimal. The proof below essentially
gives approximately λ4.6 , but we believe that the optimal scaling is λ. However, even
this polynomial bound is good in a problem like ours – by a more direct application of
the results in [18], one gets only exponential bound (i.e. eCλ instead of our λ5 ).
1.3.1
Reduction to Neumann boundary condition and reflection across
boundary
Notation 1.3.2. Since ∂Ω is compact and C 2 , it has bounded curvature and there exists
δ > 0 such that the map
(y, t) 7→ y + tν(y)
is one-to-one from ∂Ω × (−δ, δ) onto δ-neighborhood of ∂Ω. For any ρ ≤ δ, denote
{y + tν(y) | y ∈ ∂Ω, t ∈ (−ρ, 0)} = {x ∈ Rn | dist(x, ∂Ω) < ρ} ∩ Ω = Ωρ ,
{y + tν(y) | y ∈ ∂Ω, t ∈ (0, ρ)} = {x ∈ Rn | dist(x, ∂Ω) < ρ} \ Ω̄ = Ω′ρ .
This means that for each x ∈ Ωδ , there exists a unique closest point on the boundary ∂Ω,
and for each y ∈ ∂Ω there exists x ∈ Ωδ such that dist(x, y) = δ and B(x, δ) ⊂ Ω. Furthermore, since Ω\Ωδ is connected, for each two points y1 , y2 ∈ ∂Ω and the corresponding
points x1 , x2 ∈ Ωδ such that dist(x1 , y1 ) = dist(x2 , y2 ) = δ and B(x1 , δ), B(x2 , δ) ⊂ Ω,
there exists a curve Γx1 ,x2 in Ω \ Ωδ with endpoints x1 , x2 . If we look at Γ as a set of
points in Ω, then {x ∈ Rn : dist(x, Γ) < δ} ⊂ Ω. We will need these properties later.
Fix this δ (depending on Ω) for the rest of this section.
We want to extend the function u defined on Ωδ ∪ ∂Ω to
Ωδ ∪ ∂Ω ∪ Ω′δ = D,
18
so that the boundary ∂Ω becomes a hypersurface in D. In order to do so, define
v(x) := u(x)eλd(x)
for x ∈ Ωδ ∪ ∂Ω,
where d(x) = dist(x, ∂Ω) is the distance function. Fix this notation throughout the rest
of this section.
Note that v(x) = 0 if and only if u(x) = 0 and v(x) = u(x) on ∂Ω. Since ∂Ω is C 2 ,
so is d(x) on Ωδ . For y ∈ ∂Ω and x = y + tν(y) ∈ Ωδ , we have
∇d(x) = ∇d(y) = −ν(y),
∆d(x) = −
n−1
X
i=1
(1.41)
κi (y)
,
1 − κi (y)d(x)
where {κi } are the principal curvatures of ∂Ω, see the Appendix, pp. 381-383 in [19].
From (1) we get that v satisfies the equation
div(A(x)∇v) + b(x) · ∇v + c(x)v = 0
∂v
=0
∂ν
where
in Ωδ ,
on ∂Ω,

A=I

b = −2λ∇d
in Ωδ .

c = λ2 − λ∆d
(1.42)
(1.43)
Write each x ∈ Ωδ as x = y + tν(y), y ∈ ∂Ω and t ∈ (0, −δ), and consider the
reflection map
Ψ : Ωδ → Ω′δ
Ψ(x) = y − tν(y).
Since ∂Ω is C 2 , Ψ is C 2 . We will also use the notation Ψ(x) = x′ . For x′ ∈ Ω′δ define
v(x′ ) := v(Ψ−1 (x′ )) = v(x).
∂v
Then v ∈ C 2 (Ω′δ ), and since ∂ν
= 0 on ∂Ω, we also have that ∇v is Lipschitz in D, and
hence v is twice weakly differentiable in D. By (1.42), v satisfies
div(A∇v) + b · ∇v + cv = 0
in Ω′δ ,
(1.44)
with A = (aij )ni,j=1 ,
aij (x′ ) =
n
X
∂Ψi
k=1
bi (x′ ) = −
∂xk
(x)
∂Ψj
(x)
∂xk
∂ ij ′
a (x ) + ∆Ψi (x) + ∇Ψi (x) · b(x)
∂x′j
c(x′ ) = c(x).
19
(1.45)
Hence v satisfies
div(A∇v) + b · ∇v + cv = 0
(1.46)
a.e. in D, where by (1.43) and (1.45) we have the bounds
||A||L∞ (D) ≤ C,
||b||L∞ (D) ≤ Cλ,
(1.47)
2
||c||L∞ (D) ≤ Cλ ,
with the constant C depending only on the domain Ω. Since v is also twice weakly
differentiable, it satisfies equation (1.46) in the strong sense (as in [19, Chapter 9]). We
will see soon that v is also a weak solution of (1.46) (this is not yet clear since we do not
know how A behaves across the hypersurface ∂Ω).
Note that
A(x′ ) = ∇Ψ(x)(∇Ψ)T (x) for x′ ∈ Ω′δ .
(1.48)
Proposition 1.3.3. A is uniformly Lipschitz in D, with the Lipschitz constant depending
on the boundary ∂Ω.
Proof. This is clear in Ωδ and Ω′δ ; what we are concerned about is the hypersurface ∂Ω.
First let us show that
∇Ψ(x)(∇Ψ)T (x) = I
for x ∈ ∂Ω,
(1.49)
where we take Ψ(x) = x for x ∈ ∂Ω (and consider the gradient of Ψ in points on ∂Ω
only from the side of within Ω).
Take any x0 ∈ ∂Ω. If the tangent plane to ∂Ω at x0 is parallel to the plane {xn = 0},
then it is easy to see that


1 0 ··· 0 0
0 1 · · · 0 0 



.. 
..
∇Ψ(x) =  ...
.
. 


0 0 · · · 1 0 
0 0 · · · 0 −1
and indeed ∇Ψ(x0 )(∇Ψ)T (x0 ) = I. If the tangent plane to ∂Ω at x0 is general, there
exists a rotation R which maps this tangent plane to a tangent plane parallel to {xn = 0}.
Then by the change of coordinates y = Rx, the mapping Ψ̃(y) = RΨ(RT y) will be exactly
like Ψ in the special case discussed, and we will have
I = ∇Ψ̃(y0 )(∇Ψ̃)T (y0 ) = R∇Ψ(x0 )RT R(∇Ψ)T (x0 )RT =
= R∇Ψ(x0 )(∇Ψ)T (x0 )RT ,
I = ∇Ψ(x0 )(∇Ψ)T (x0 )
(we are using the rotation property R−1 = RT a few times).
20
Hence (1.49) is true, and we can define A(x) = I = ∇Ψ(x)(∇Ψ)T (x) for x ∈ ∂Ω.
Together with (1.43) we get that A is Lipschitz in Ωδ . Since ∂Ω is smooth, so is ∇Ψ and
together with (1.45) we get that A is Lipschitz in Ω′δ . Hence, A is Lipschitz in the whole
D: for x ∈ Ωδ and y ∈ Ω′δ , the segment from x to y intersects ∂Ω in some z, and then
|A(x) − A(y)| ≤ |A(x) − A(z)| + |A(z) − A(y)| ≤ K(|x − z| + |z − y|) = K|x − y|.
Remark 1.3.4. Note that if we start with a more general elliptic matrix A than the
Laplacian in our domain Ωδ , then after reflexion we do not necessarily end up with a
Lipschitz extension of A. In general, the reflected solution will satisfy (1.44) with
A(x′ ) = ∇Ψ(x)A(x)(∇Ψ)T (x).
For instance, if we consider the problem in R2 and take
1 0
0 1
A(x0 ) =
,
∇Ψ(x0 ) =
0 2
1 0
for some x0 ∈ ∂Ω (having the tangent line in x0 parallel to the line x1 − x2 = 0 gives us
this ∇Ψ), we get
2 0
T
∇Ψ(x0 )A(x0 )(∇Ψ) (x0 ) =
6= A(x0 ),
0 1
so A will be discontinuous at x0 .
Proposition 1.3.5. Equation (1.46) is uniformly elliptic in D.
This is clear for all x ∈ Ωδ , and for x ∈ Ω′δ we have by (1.48)
ξ T A(x′ )ξ = ξ T ∇Ψ(x)(∇Ψ)T (x)ξ = |(∇Ψ)T (x)ξ|2 ≥
1 2
|ξ| ,
η2
where
η = sup |(∇Ψ)(x)−1 | = sup |∇Ψ−1 (x′ )|
x∈Ωδ
x′ ∈Ω′δ
depends only on Ω.
Proposition 1.3.6. The function v satisfies (1.46) in the weak sense, i.e. for all ϕ ∈
C0∞ (D) we have
Z
−A∇v · ∇ϕ + b∇vϕ + cvϕ = 0.
D
21
Proof. Since v ∈ C 2 (Ωδ ) ∩ C 2 (Ω′δ ) ∩ C 1 (D) and it satisfies (1.46) in Ωδ and Ω′δ pointwise,
we have
Z
(−A∇v · ∇ϕ + b∇vϕ + cvϕ)
DZ
Z
(−A∇v · ∇ϕ + b∇vϕ + cvϕ)
=
(−A∇v · ∇ϕ + b∇vϕ + cvϕ) +
=
Z
Ωδ
(div(A∇v)ϕ + b∇vϕ + cvϕ) +
Ωδ
+
=
Z
Z
∂Ω
Z
Ω′δ
∂Ωδ
(div(A∇v)ϕ + b∇vϕ + cvϕ) +
Ω′δ
A∇vϕ · ν −
Z
∂Ω
Z
A∇vϕ · νΩδ
∂Ωδ
A∇vϕ · νΩ′δ
A∇vϕ · ν = 0.
We used that ϕ = 0 on ∂D ⊃ (∂Ωδ \ ∂Ω) ∪ (∂Ω′δ \ ∂Ω), the exterior normals for Ωδ and
Ω′δ on ∂Ω satisfy νΩδ = ν = −νΩ′δ , and A is continuous across ∂Ω.
Doubling condition for v:
To prove Theorem 2, we will first derive an analogous doubling condition on the solid
D for v, and then use the approach as in [29] (used for parabolic equations) to deduce
Theorem 2:
Theorem 7. There exist constants r0 , C > 0 depending only on Ω and n such that for
any r ≤ r0 /2λ and x ∈ ∂Ω,
Z
Z
5
v2.
(1.50)
v 2 ≤ 2Cλ
B(x,r)
B(x,2r)
The main difficulty we encounter is that for a general elliptic equation, an adaptation
of the results in [18] gives us a bound for the frequency only on small balls of radius of
order 1/λ. That globally translates into the suboptimal exponential bound mentioned
in Remark 1.3.1. Therefore we move on from the more general equation (1.46), which is
satisfied by v in the neighborhood of ∂Ω, back to the original Laplace equation inside of
Ω, and then back to the neighborhood of the boundary with the equation (1.46) again.
1.3.2
Frequency for v
Near the boundary ∂Ω, we will consider the equation (1.46). Hence we use the generalization of the frequency (1.9) for more general elliptic equations from Subsection 1.2.2.
Let us apply the results from that subsection to our function v.
Recall Theorem 4: for a solution w to a general elliptic equation on B1 , it gives us
the bound for the frequency of the form N (R1 ) ≤ c1 + c2 N (R2 ), if 0 < R1 < R2 ≤ r0 .
However, it does not give an explicit bound for r0 , c1 , c2 in terms of the L∞ bounds on
the coefficients of the equation. If we use the bounds (1.47) instead of (1.23) and go
22
through the proof of the theorem, we obtain that (1.29) holds for r0 ∼ 1/λ, c1 , c2 ∼ 1.
We can obtain these bounds easier from Theorem 4 by scaling.
Scaling the equation:
Recall that the function v satisfies the equation (1.46):
div(A∇v) + b · ∇v + cv = 0 a.e. in D,
with A elliptic and Lipschitz, the ellipticity and Lipschitz constants depending only on
Ω. By (1.47) we have
||A||L∞ (D) ≤ C,
||b||L∞ (D) ≤ Cλ,
||c||L∞ (D) ≤ Cλ2 ,
with the constant C depending only on the domain Ω. Consider this equation in a ball
B(x0 , r1 /λ) ⊂ D and define
vx0 ,λ (x) := v(x0 + x/λ)
for x ∈ B(0, r1 ).
Then vx0 ,λ satisfies
div(Ax0 ,λ ∇vx0 ,λ ) + bx0 ,λ · ∇vx0 ,λ + cx0 ,λ vx0 ,λ = 0 a.e. in B(0, r1 ),
with
(1.51)
Ax0 ,λ (y) = A(x0 + y/λ),
bx0 ,λ (y) = λ−1 b(x0 + y/λ),
cx0 ,λ (y) = λ−2 c(x0 + y/λ),
so Ax0 ,λ (y), bx0 ,λ (y) and cx0 ,λ (y) are bounded uniformly in L∞ by a constant depending
only on Ω. The ellipticity constant of A does not change and the Lipschitz constant of
A only improves (since λ ≥ 1), so we also have bounds on them depending only on Ω.
A simple change of variables yields that for y ∈ B(0, r1 ) and r ≤ r1 − |y|,
Hvx0 ,λ (y, r) = λn−1 Hv (x0 + y/λ, r/λ),
Dvx0 ,λ (y, r) = λn−2 Dv (x0 + y/λ, r/λ),
Ivx0 ,λ (y, r) = λn−2 Iv (x0 + y/λ, r/λ),
Nvx0 ,λ (y, r) = Nv (x0 + y/λ, r/λ).
Remark 1.3.7. Notice that for x ∈ Ω ∩ D, the principal part of equation (1.46) is the
Laplacian, i.e. A = I. Hence, in points from Ω ∩ D, the metric g associated with A is
trivial, and in these points we have
Z
Hv (r) =
v2,
Z ∂Br
Dv (r) =
|∇ v|2 ,
Br
Z
Iv (r) =
(|∇ v|2 + ub · ∇ v + cv 2 ).
Br
23
In particular, we can skip the argument about H(r) and
they are equal in this case.
R
∂Br
v 2 being comparable –
Next, we apply Theorem 4, Proposition 1.2.11, Corollary 1.2.12, Theorem 5 and
Lemma 1.2.13 to the function vx0 ,λ , and using the scaling above we rewrite the results
in terms of v. We immediately obtain the following results.
Proposition 1.3.8. Let B(x0 , r1 /λ) ⊂ D. Then there exist constants c1 , c2 > 0, r0 ∈
(0, r1 ) depending only on r1 and Ω such that
Nv (x0 , R1 ) ≤ c1 + c2 Nv (x0 , R2 )
for any 0 < R1 < R2 ≤ r0 /λ.
(1.52)
Proposition 1.3.9. Let B(x0 , r1 /λ) ⊂ D. Then there exist constants C, c1 , c2 > 0,
r0 ∈ (0, r1 ) depending only on r1 and Ω such that for any 0 < R1 < R2 < r0 /λ,
Z
R2 c1 +c2 Nv (x0 ,R2 )
v ≤C
v2,
−
R1
∂B(x0 ,R2 )
∂B(x0 ,R1 )
c1 +c2 Nv (x0 ,R2 ) Z
Z
R2
v2 ≤ C
−
v2.
−
R
1
B(x0 ,R2 )
B(x0 ,R1 )
Z
−
2
Lemma 1.3.10. Let B(x0 , r1 /λ) ⊂ D. Assume
Z
Z
2
v ≥ κ−
−
v2
(1.53)
(1.54)
(1.55)
B(x0 ,r)
B(x0 ,ζr)
for some κ, ζ ∈ (0, 1) and r ∈ (0, r0 /λ], where r0 depends on r1 and Ω and is chosen so
that Propositions 1.3.8 and 1.3.9, and the analogies of Proposition 1.2.11 and Corollary
1.2.12 hold. Then there exists a constant Cζ > 0 depending on r1 , Ω and ζ such that
Nv (x0 , ζr) ≤ Cζ (1 − log κ).
(1.56)
In particular, for any β < ζ, there exist constants C1 , C2 depending on r1 , Ω and ζ, β
such that
Z
Z
1 C2
v2 ≥
v2.
(1.57)
−
κ −
C
1
B(x0 ,βr)
B(x0 ,ζr)
1.3.3
Proof of Theorem 7
We will use an argument of the type as in Proposition 1.2.7 in a chain. For that, we
will need to start in a point where the frequency is reasonably bounded/where we have
a doubling condition with a reasonable constant. We will require even somewhat more:
the integral of u2 over a small ball with center in this special starting point y∗ ∈ ∂Ω
will control the global integral of u2 . Recall the Notation 1.3.2: δ > 0 is a constant
depending on Ω such that each point in Ωδ = {x ∈ Rn | dist(x, ∂Ω) < δ} ∩ Ω has a single
closest point on ∂Ω.
24
Lemma 1.3.11. For any ρ < δ, there exists a point y∗ ∈ ∂Ω such that for some constant
C depending only on Ω, there holds
Z
Z
2
−2n+1
u2 .
u ≤ Cρ
B(y∗ ,ρ)∩Ω
Ω
In the proof we crucially use this interior estimate for harmonic functions:
Proposition 1.3.12. Let w be a harmonic function in B1 . Then
sup |w| ≤ C
B1/2
Z
w
2
B1
1/2
,
where C is a constant depending only on n.
Proof. This can be found e.g. in [24], Remark 1.19.
Proof of Lemma 1.3.11. Choose a ρ/2-net of points y1 , y2 , . . . , ym ∈ ∂Ω, i.e. for each y ∈
∂Ω there exists 1 ≤ i ≤ m such that |y − yi | ≤ ρ/2. We can always make m ≤ Cρ−(n−1) ,
where C depends only on Ω. Then the balls {B(yi , ρ)}m
i=1 cover Ωρ/2 , and therefore there
exists y∗ ∈ {y1 , y2 , . . . , ym } such that
Z
Z
u2 .
(1.58)
u2 ≤ Cρ−(n−1)
Ωρ/2
Now we just need to bound
R
Ωu
B(y∗ ,ρ)∩Ω
2
in terms of
R
Ωρ/2
u2 . Let
n
ρo
= Ω \ Ωρ .
Ωcρ/4 = x ∈ Ω : dist(x, ∂Ω) ≥
4
4
Then ∂Ωcρ/4 = {x ∈ Ω : dist(x, ∂Ω) = ρ/4} and for any x ∈ ∂Ωcρ/4 , by the scaling of the
interior estimate in Proposition 1.3.12 we obtain
Z
Z
sup u2 ≤ Cρ−n
u2 ≤ Cρ−n
u2 .
B(x,ρ/4)
B(x,ρ/8)
Ωρ/2
Hence, by the maximum principle for the harmonic function u inside Ωcρ/4 ,
2
2
sup u ≤ sup u ≤
Ωcρ/4
Z
∂Ωcρ/4
2
u =
Ω
Z
sup
u +
Ω\Ωρ/2
≤ Cρ−n
sup u ≤ Cρ
−n
x∈∂Ωcρ/4 B(x,ρ/8)
2
Z
2
Z
2
Ωρ/2
2
u ≤ C sup u +
Ωcρ/4
u2 .
Ωρ/2
Combining this with (1.58), we obtain the desired estimate.
25
Z
Z
u2 ,
Ωρ/2
u2
Ωρ/2
Proof of Theorem 7. Fix r1 := δλ1 , where λ1 is the smallest positive Steklov eigenvalue
for Ω. Then B(x0 , r1 /λ) ⊂ D for every x0 ∈ ∂Ω and every Steklov eigenvalue λ. Fix
r0 < r1 so that Propositions 1.3.8 and 1.3.9 and Lemma 1.3.10 hold with this r0 . Note
that r1 and r0 depend only on Ω.
To prove the Theorem, it is enough to prove that there exists a constant C depending
only on Ω and n such that for every y ∈ ∂Ω
Z
Z
2
Cλ5
v2.
(1.59)
v ≤2
B(y,r0 /2λ)
B(y,r0 /λ)
Indeed, this will prove (1.50) for r = r0 /2λ. Once we know it for this r, from Lemma
1.3.10 we obtain a bound for the frequency N (r0 /2λ) ≤ Cλ5 , and from Propositions
1.3.8 and 1.3.9 we obtain (1.50) for any r ≤ r0 /2λ.
The rest of the proof is organized as follows. First, we prove a doubling condition
for v in a special point y = y∗ ∈ ∂Ω, move to a close-by point inside Ω, and switch to u.
As the next step, we show a doubling condition for u in a point near y∗ for a ball with
fixed radius (independent of λ), and propagate the doubling condition estimate through
Ω into a neighborhood of any other given point y0 ∈ ∂Ω. In step 3, we gradually pass
from the ball with fixed radius to a small ball in ∼ 1/λ-neighborhood of ∂Ω, still for u
and within Ω. In the last step, we switch back to v and deduce the doubling condition
(1.59) in the point y0 ∈ ∂Ω.
Step 1:
We start in a point y = y∗ ∈ ∂Ω, which we obtain from Lemma 1.3.11 for ρ = r0 /2λ,
i.e. there holds
Z
r −2n+1 Z
0
2
u ≤C
u2 .
2λ
Ω
B(y∗ ,r0 /2λ)∩Ω
In Ω, we have |v(x)| = |u(x)eλd(x) | ≥ |u(x)|. Hence
Z
Z
Z
1 r0 2n−1
2
2
u2 .
u ≥
v ≥
C 2λ
Ω
B(y∗ ,r0 /2λ)∩Ω
B(y∗ ,r0 /2λ)
(1.60)
R
R
On the other hand, we show that the integral B(y,r0 /λ) v 2 is controlled by Ω u2 for
any y ∈ ∂Ω. Fix y ∈ ∂Ω. Notice that r0 /λ < δ. Recall that the reflection map Ψ
introduced in Subsection 1.3.1 maps any x ∈ Ωδ , written as x = z + tν(z), where z ∈ ∂Ω,
into Ψ(x) = z − tν(z), and this is a 1-to-1 map from Ωδ onto Ω′δ = D \ Ω. Since
B(y, r0 /λ) \ Ω ⊂ Ω′r0 /λ ⊂ Ω′δ , the map Ψ−1 maps B(y, r0 /λ) \ Ω onto a subset of Ωr0 /λ ,
and by change of variables, using the boundedness of the Jacobian of Ψ−1 (the bound
depends only on Ω), we obtain
Z
Z
2
v ≤C
v2.
Ωr0 /λ
B(y,r0 /λ)\Ω
On Ωr0 /λ we have
|v(x)| = |u(x)eλd(x) | ≤ |u(x)|er0 ≤ C|u(x)|,
26
(1.61)
with C depending only on Ω. Hence,
Z
Z
2
v =
B(y,r0 /λ)
2
v +
B(y,r0 /λ)∩Ω
≤C
Z
2
Ωr0 /λ
u ≤C
Z
Z
2
B(y,r0 /λ)\Ω
v ≤C
Z
v2
Ωr0 /λ
(1.62)
2
u .
Ω
Now we just need to propagate the estimate (1.60) into any point y ∈ ∂Ω – combined
with (1.62), we will obtain a doubling condition for v.
Using (1.60) and (1.62) in y = y∗ , we know that
Z
Z
1 1
2
v2.
v ≥
C λ2n−1 B(y∗ ,r0 /λ)
B(y∗ ,r0 /2λ)
Hence, from Lemma 1.3.10 used for r = rλ0 , ζ = 21 , κ = C1 λ−2n+1 , β = 41 , we get
Z
Z
1 1
2
v ≥
v2,
(1.63)
C1 λC2 B(y∗ ,r0 /2λ)
B(y∗ ,r0 /4λ)
where C1 , C2 depend only on Ω.
Now fix y = y0 ∈ ∂Ω in which we want to prove (1.59). Recalling Notation 1.3.2,
we know that there exist points z1 , z2 ∈ Ω such that dist(z1 , y∗ ) = dist(z2 , y0 ) = δ,
B(z1 , δ), B(z2 , δ) ⊂ Ω, and there is a curve Γ in Ω with endpoints z1 , z2 such that if we
look at Γ as a set of points in Ω, then {x ∈ Rn : dist(x, Γ) < δ} ⊂ Ω. We will propagate
the doubling condition estimate along this curve. First choose x1 on the segment y∗ z1
such that dist(y∗ , x1 ) = r0 /4λ. Then B(x1 , r0 /4λ) lies inside Ω and touches ∂Ω in y∗ .
We also have
B(y∗ , r0 /4λ) ⊂ B(x1 , r0 /2λ) ⊂ B(x1 , 3r0 /4λ) ⊂ B(y∗ , r0 /λ).
Hence, using (1.63) and (1.60),
Z
B(x1 ,r0 /2λ)
2
v ≥
Z
v2
B(y ,r /4λ)
∗ 0
Z
1 1
v2
≥
C1 λC2 B(y∗ ,r0 /2λ)
Z
1 1
≥
u2 .
C 1 λC 2 Ω
(1.64)
On the other hand, from B(x1 , 3r0 /4λ) ⊂ B(y∗ , r0 /λ) and (1.62) (used for y = y∗ ), we
obtain
Z
Z
Z
u2 ,
v2 ≤ C
v2 ≤
B(y∗ ,r0 /λ)
B(x1 ,3r0 /4λ)
so together with (1.64) we have
Z
1 1
v ≥
C 1 λC 2
B(x1 ,r0 /2λ)
2
27
Z
Ω
v2.
B(x1 ,3r0 /4λ)
Hence it follows from Lemma 1.3.10 used in the center x1 for r = 3r0 /4λ, ζ = 2/3,
κ = C11 λC1 2 , β = 1/6, that
Z
B(x1 ,r0 /8λ)
v2 ≥
1 1
f1 λCf2
C
Z
B(x1 ,r0 /2λ)
v2 ≥
1 1
C 1 λC 2
Z
u2 ,
Ω
where in the last inequality we used (1.64). Notice that starting from the estimate (1.60)
for a ball with center y∗ ∈ ∂Ω, we moved to an estimate for a ball B(x1 , r0 /8λ) inside
Ω. Since B(x1 , r0 /8λ) ⊂ Ωr0 /λ , using (1.61) we can move from v towards the harmonic
function u:
Z
Z
Z
1
1 1
u2 ≥
v2 ≥
u2 .
(1.65)
C2
C
C
λ
1
B(x1 ,r0 /8λ)
B(x1 ,r0 /8λ)
Ω
Step 2:
Now we are ready to move towards balls with fixed radii which will not depend on
λ. Take the point z1 for which B(z1 , δ) lies inside Ω and touches ∂Ω in y∗ . Since x1 lies
on the segment y∗ z1 in distance r0 /4λ from y∗ , we have
B(x1 , r0 /8λ) ⊂ B(z1 , δ − r0 /8λ) ⊂ B(z1 , δ) ⊂ Ω.
Hence, by (1.65) we obtain
Z
Z
2
u ≥
1 1
u ≥
C
C
1λ 2
B(x1 ,r0 /8λ)
B(z1 ,δ−r0 /8λ)
2
Z
1 1
u ≥
C
C
1λ 2
Ω
2
Z
u2 .
(1.66)
B(z1 ,δ)
Now we can use Lemma 1.2.6 for the harmonic function u on the ball B(z1 , δ) and
r0
r0
α = 1 − 8λδ
, γ = 1 − 16λδ
, κ = C11 λC1 2 , β = 1/2. For this choice, we have
r0
r0 n
1
≤1−
γn = 1 −
=1−
,
16λδ
16λδ
Cλ
1
1 − γn ≥
,
Cλ
r0
r0
1 − 16λδ
r0
γ
=
> e 32λδ ,
r0 ≥ 1 +
α
1 − 8λδ
16λδ
1
log(γ/α) >
,
Cλ
log(α/β) ≤ log 2,
and Lemma 1.2.6 together with (1.66) yield
Z
z1
2
B(z1 ,δ/2)
u ≥
1 1
C 1 λC 2
Cλ
e Z
Cλ log λ Z
1
u ≥
u2 .
2
B(z1 ,δ−r0 /8λ)
Ω
2
(1.67)
Now we propagate the estimate (1.67) along Γ into the point z2 . We choose points
= z1 , z 2 , . . . , z k = z2 on Γ such that
dist(z i , z i+1 ) ≤ δ/4
for i = 1, 2, . . . , k − 1.
28
Notice that k is bounded by a constant depending only on Ω. Using induction, assume
Z
Ci λ log λ Z
1
u ≥
u2
2
i
B(z ,δ/2)
Ω
2
(1.68)
for some constant Ci depending only on Ω – for i = 1, this is (1.67). Then
Z
Ci λ log λ Z
1
u2
u ≥
2
i
i
B(z ,δ)
B(z ,δ/2)
2
and the inclusion B(z i+1 , δ/2) ⊃ B(z i , δ/4), Lemma 1.2.6 and (1.68) imply
Z
Z
2
u2
u ≥
B(z i ,δ/4)
B(z i+1 ,δ/2)
Cλ log λ Z
Ci+1 λ log λ Z
1
1
2
≥
u ≥
u2
2
2
i
B(z ,δ/2)
Ω
(here C comes from Lemma 1.2.6 and depends on Ci ), which is (1.68) for i + 1 instead
of i. After k − 1 steps, the constant Ck will be bounded by a constant depending only
on Ω, and we obtain
Cλ log λ Z
Z
1
2
u2 .
(1.69)
u ≥
2
Ω
B(z2 ,δ/2)
Step 3:
Now we need to get from the ball B(z2 , δ/2) closer to the boundary, where we will
be able to switch back to the function v. In each step, we multiply the distance from y0
by a factor of 5/6. Hence, we will need ∼ log λ steps to get to a neighborhood of ∂Ω of
order 1/λ.
Recall that B(z2 , δ) lies inside Ω and touches ∂Ω in y0 . Consider the sequence
i
0
x , x1 , . . . , xl of points on the segment z2 y0 such that dist(xi , y0 ) = δ 65 (i.e. x0 = z2 )
r0
and l is chosen to be the smallest integer such that dist(xl , y0 ) ≤ 4λ
, i.e.
l=
log λ
+ C,
log 6/5
4δ−log r0
4δ−log r0
≤ C < log log
+ 1. Denote ri := δ
where log log
6/5
6/5
lies in Ω and touches ∂Ω in y0 .
By induction, we will show that
Z
5 i
6 ,
so that the ball B(xi , ri )
C2i λ log λ
Z
1
2i −1
u ≥
u2 ,
τ
2
B(xi ,r i /6)
Ω
2
(1.70)
25 n
where τ = 24
− 1. For i = 0, this follows from (1.69) and Lemma 1.2.6 used for r = δ,
α = 1/2, β = 1/6.
29
Now assume that (1.70) holds for some i ∈ {0, 1, . . . , l − 1}, and we will show it for
i+1. Since dist(xi , xi+1 ) = ri /6, we have B(xi+1 , 2ri+1 /5) = B(xi+1 , ri /3) ⊃ B(xi , ri /6).
On the other hand, we have B(xi+1 , ri+1 ) ⊂ Ω. Hence, (1.70) implies
Z
C2i λ log λ
Z
1
2i −1
u ≥
u ≥
τ
u2
2
i
i
i+1
i+1
B(x ,r /6)
B(x
,2r
/5)
Ω
C2i λ log λ
Z
1
i
τ 2 −1
u2 .
≥
2
B(xi+1 ,r i+1 )
2
Z
2
(1.71)
Apply Lemma 1.2.6 in point xi+1 for r = ri+1 , α = 2/5, β = 1/6, γ = 24/25, κ =
i
5 n 1 C2 λ log λ 2i −1
τ
. This is chosen so that the exponent in (1.19) is one:
2
2
log(α/β)
= 1,
log(γ/α)
and our computation becomes much easier. We obtain
Z
u2
B(xi+1 ,r i+1 /6)
!
n Z
n C2i λ log λ
24
5
1
(1/6)n
i
2 −1
u2
τ
1−
≥
n
(2/5)
2
2
25
B(xi+1 ,2r i+1 /5)
C2i+1 λ log λ
Z
1
2i+1 −1
≥
τ
u2
2
i+1
i+1
B(x
,r
)
(the last inequality follows from (1.71)), which is (1.70) for i + 1.
Using (1.70) for i = l, we obtain
Z
C2l λ log λ
Z
1
2l −1
τ
u2
u ≥
2
l
l
Ω
B(x ,r /6)
2
log 2
Cλ1+ log(6/5) log λ Z
1
u2
≥
2
Ω
Cλ5 Z
1
u2 .
≥
2
Ω
(1.72)
r0
Step 4: Recall that l was chosen so that rl = dist(xl , y0 ) ≤ 4λ
. Hence, B(y0 , r0 /2λ)∩
l
l
Ω ⊃ B(x , r /6), and using that |v(x)| ≥ |u(x)| in Ωδ and (1.72), we obtain
Cλ5 Z
1
u2 .
u ≥
u ≥
v ≥
2
Ω
B(xl ,r l /6)
B(y0 ,r0 /2λ)∩Ω
B(y0 ,r0 /2λ)
R
Together with (1.62) used for y = y0 which bounds B(y0 ,r0 /λ) v 2 by a constant times
R 2
Ω u , we obtain the desired doubling condition (1.59) in y = y0 , what we wanted.
Z
2
Z
2
30
Z
2
Remark 1.3.13. In Step 3, instead of our explicit choice for decreasing the distance to
the boundary by the multiplicative factor of 5/6, the radius of the smaller ball in control
being 1/6 of the distance to the boundary, and the auxiliary constant γ = 24/25, we
can do the calculation with general constants. While the computation becomes more
log 2
complicated, we do not gain too much: instead of our exponent 1 + log(6/5)
≈ 4.8, a
numerical calculation suggests that the optimal choice of the constants gives an exponent
of roughly 4.6.
1.3.4
Doubling condition on ∂Ω
In this section, we will use Theorem 7 to prove Theorem 2:
Theorem 2. There exist constants r0 , C depending only on Ω and n such that for any
r ≤ r0 /λ and x ∈ ∂Ω,
Z
Z
2
Cλ5
u2 .
(1.40)
u ≤2
B(x,r)∩∂Ω
B(x,2r)∩∂Ω
As a connection between the boundary integrals and integrals over solid balls, we will
use a quantitative Cauchy uniqueness theorem which was proved in [29]. In [29] it is
formulated for solutions of general elliptic equations under the same assumptions as used
in Subsection 1.2.2: consider the equation
n
X
(aij (x)wxi )xj +
in B1 (0) ⊂
bi (x)wxi + c(x)w = 0
(1.73)
i=1
i,j=1
Rn
n
X
with coefficients satisfying the following assumptions:
aij (x)ξi ξj ≥ α|ξ|2 ,
∀ξ ∈ Rn , x ∈ B1 (0);
X
X
(ii)
|aij (x)| +
|bi (x)| + |c(x)| ≤ K,
∀x ∈ B1 (0);
(i)
i,j
(iii)
X
i,j
i
|aij (x) − aij (y)| ≤ γ|x − y|,
(1.74)
∀x, y ∈ B1 (0);
for some positive constants α, K, γ.
Lemma 1.3.14 ([29], Lemma 4.3). Let w be a solution of (1.73) in B1+ ⊂ Rn with the
coefficients of the equation satisfying (1.74) and ||w||L2 (B + ) ≤ 1. Suppose that
1
||w||H 1 (Γ) + ||(∂w/∂xn )||L2 (Γ) ≤ ǫ ≪ 1,
where Γ = {(x′ , 0) ∈ Rn : |x′ | < 3/4}. Then ||w||L2 (B +
1/2
constants C, β which depend only on n, α, K, γ.
)
≤ Cǫβ for some positive
Notice that Lemma 1.3.14 relates the H 1 -norm on a hypersurface to the L2 -norm
in the solid ball. For a doubling condition, we need the L2 -norm also on the hypersurface. The following lemma gives us the missing connection between the H 1 -norm on
31
the hypersurface and the L2 -norm on the hypersurface; the additional H 2 -norm on the
solid will be bounded using interior elliptic estimates. Technically it is a combination of a
J. L. Lions-type lemma estimating the H 1 (Rn−1 )-norm in terms of the H 3/2 (Rn−1 )-norm
and the L2 (Rn−1 )-norm, and the standard trace theorem.
Lemma 1.3.15. Let w ∈ H 2 (Rn ) and consider the trace of w onto
{x ∈ Rn : xn = 0} = Rn−1 ,
which we denote by w. Then there exists a constant C depending only on n such that
for any η > 0,
C
||∇w||L2 (Rn−1 ) ≤ η||w||H 2 (Rn ) + 2 ||w||L2 (Rn−1 ) .
(1.75)
η
Proof. First we will show that for any w ∈ H 3/2 (Rn−1 ), η > 0, there holds
||∇w||L2 (Rn−1 ) ≤ η||w||Ḣ 3/2 (Rn−1 ) +
1
||w||L2 (Rn−1 ) .
η2
(1.76)
Using the Fourier transform
ŵ(ξ) =
1
(2π)(n−1)/2
Z
w(x)e−ix·ξ dx,
Rn−1
Plancherel identity and Young’s inequality |ξ|2 ≤ 23 η 2 |ξ|3 + 31 η14 , we have
Z
2
|iξ ŵ|2 dξ
||∇w||L2 (Rn−1 ) =
n−1
R Z
Z
2 2
1 1
3
2
≤ η
|ŵ|2 dξ
|ξ| |ŵ| dξ +
3
3 η 4 Rn−1
Rn−1
1 1
2
||w||2L2 (Rn−1 ) ,
= η 2 ||w||2Ḣ 3/2 (Rn−1 ) +
3
3 η4
and (1.76) follows. Using the trace theorem H 3/2 (Rn−1 ) ⊂ H 2 (Rn ) (continuously), for
any w ∈ H 2 (Rn ) we obtain from (1.76)
||∇w||L2 (Rn−1 ) ≤ Cη||w||H 2 (Rn ) +
1
||w||L2 (Rn−1 ) .
η2
Using η/C instead of η, we obtain (1.75).
Proof of Theorem 2. Since u = v on ∂Ω, we can rewrite (1.40) (which we want to prove
for r small enough) as
Z
Z
2
Cλ5
v2.
(1.77)
v ≤2
B(x,r)∩∂Ω
B(x,2r)∩∂Ω
From Theorem 7, we know that there exist constants r0 , C depending only on Ω and n
such that for any r ≤ r0 /2λ and x ∈ ∂Ω,
Z
Z
2
Cλ5
v2.
(1.78)
v ≤2
B(x,r)
B(x,2r)
32
Fix this r0 and also fix the point x = x0 ∈ ∂Ω in which we want to prove (1.77). In order
to have coefficients of the underlying equation bounded independently of λ as in (1.74),
we use the scaling from Section 1.3.2: vx0 ,λ (x) = v(x0 + x/λ), x ∈ B(0, r0 ). Then vx0 ,λ
satisfies (1.51) with the coefficients Ax0 ,λ (y), bx0 ,λ (y) and cx0 ,λ (y) bounded uniformly in
L∞ by a constant depending only on Ω, i.e. they satisfy (1.74) with α, K, γ independent
of λ (the Lipschitz constant γ only improves with the scaling, ellipticity α stays the
same). Proving Theorem 2 is equivalent to proving, for any r < r1 (where r1 is to be
determined, depending on Ω),
Z
Z
2
Cλ5
vx0 ,λ ≤ 2
vx20 ,λ ,
(1.79)
B(0,2r)∩∂Ωx0 ,λ
B(0,r)∩∂Ωx0 ,λ
where Ωx0 ,λ = {x : x0 + x/λ ∈ Ω}. From (1.78), we know that for any r ≤ r0 /2,
Z
Z
2
Cλ5
vx20 ,λ .
(1.80)
vx0 ,λ ≤ 2
B(0,r)
B(0,2r)
Since we want to use the lemmas above, we need to flatten out the hypersurface ∂Ω.
Since Ω is a C 2 -domain, we can assume that in B(x0 , r0 /λ) it is a graph of a C 2 -function
whose C 2 -norm is bounded by a constant M independent of x0 and λ – otherwise we
just diminish r0 . After scaling by λ, this bound only improves. Hence, we can assume
that
B(0, r0 ) ∩ ∂Ωx0 ,λ = {x = (x′ , xn ) ∈ B(0, r0 ) : xn = Φ(x′ )},
where Φ ∈ C 2 (B n−1 (0, r0 )), Φ(0) = 0, ∇Φ(0) = 0, ||Φ||C 2 ≤ M , and hence (by possibly
diminishing r0 )
||Φ||C 1 (B n−1 (0,r0 )) ≤ ǫ,
(1.81)
where ǫ ≤ 0.1 will be chosen later. Now define the injective map F : B n−1 (0, r0 )×R → Rn
by
F (x′ , xn ) := (x′ , xn + Φ(x′ )).
Using (1.81) one can easily compute that
F (B(0, r)) ⊂ B(0, (1 + ǫ)r)
for any r ≤ r0 .
(1.82)
The inverse F −1 (x′ , xn ) = (x′ , xn − Φ(x′ )) is well-defined on B(0, r0 ) and similarly,
F (B(0, r)) ⊃ B(0, r/(1 + ǫ))
for any r ≤ r0 .
(1.83)
Hence we have for any r < r0 /(1 + ǫ) that
F (B n−1 (0, r/(1 + ǫ)) × {0}) ⊂ B(0, r) ∩ ∂Ωx0 ,λ ⊂ F (B n−1 (0, (1 + ǫ)r) × {0}). (1.84)
Therefore, to prove (1.79), it is enough to show
Z
Z
5
vx20 ,λ ≤ 2Cλ
F (B n−1 (0,2r)×{0})
33
F (B n−1 (0,r)×{0})
vx20 ,λ
(1.85)
for all r ≤ r1 for some r1 ≤ r0 /2(1 + ǫ) depending only on Ω (we can use this doubling
2
2
2
≥ 1.12 2 >
condition twice to account for the loss of the factor (1+ǫ)2 in radii: (1+ǫ)
2
2).
Now define
w(x) := vx0 ,λ (F (x)) for x ∈ B(0, r0 /(1 + ǫ)).
We apply area formula to both sides of (1.85) under the map F |B n−1 (0,2r)×{0} . The
(generalized) Jacobian of this map will be
q
q
p
T
J = det((In−1 , ∇Φ)(In−1 , ∇Φ) ) = det(In−1 + (∇Φ)(∇Φ)T ) = 1 + |∇Φ|2 ,
so using (1.81), we get 1 ≤ J ≤ 1 + ǫ ≤ 1.1. Hence, to prove (1.85), it is enough to prove
Z
Z
2
Cλ5
w2
(1.86)
w ≤2
B n−1 (0,r)×{0}
B n−1 (0,2r)×{0}
for all r ≤ r1 for some r1 ≤ r0 /2(1 + ǫ) depending only on Ω. Using the same arguments
((1.82), (1.83), using the doubling condition twice and the Jacobian of F being 1), it
follows from (1.80) that
Z
Z
B(0,2r)
w2 ≤ 2Cλ
5
w2
(1.87)
B(0,r)
for any r ≤ r0 /2(1 + ǫ).
Using that vx0 ,λ solves the equation (1.51), we obtain that w = vx0 ,λ ◦ F solves
div(Ã∇w) + b̃ · ∇w + c̃w = 0,
(1.88)
where
Ã(x) = Ax0 ,λ (F (x)) + P (F (x)),
b̃(x) = bx0 ,λ (F (x)) + Q(F (x)),
c̃(x) = cx0 ,λ (F (x)).
The perturbation P is symmetric and for x = (x′ , xn ),
P ij (x) = 0
P in (x) = −
P
nn
n−1
X
for i, j = 1, . . . , n − 1,
aij (x)Φxj (x′ )
for i = 1, . . . , n − 1,
j=1
(x) = −2
n−1
X
in
′
a (x)Φxi (x ) +
n−1
X
aij (x)Φxi (x′ )Φxj (x′ ).
i,j=1
i=1
For the perturbation Q we have
Qi (x) =
n−1
X
′
aij
xn (x)Φxj (x )
j=1
Qn (x) = −
n−1
X
i=1
bi (x)Φxi (x′ ) +
for i = 1, . . . , n − 1,
n−1
X
i=1
′
ain
xn (x)Φxi (x ) −
34
n−1
X
i,j=1
′
′
aij
xn (x)Φxi (x )Φxj (x ).
Hence, using the bound (1.81) on the derivatives of Φ, we have
||P (x)||L∞ ≤ CKǫ,
||Q(x)||L∞ ≤ C(K + γ)ǫ,
and the coefficients of equation (1.88) will satisfy L∞ -bounds depending only on Ω, the
Lipschitz constant of à as well, and if we choose ǫ small enough, we also get the ellipticity
of à depending only on Ω (or even an absolute bound like α̃ ≥ 1/2).
Now we are ready to prove (1.86) for r small enough. Fix r ≤ r0 /2(1 + ǫ), so that
(1.87) holds. We use scaling again so that we move to balls of fixed radii: denote
w̃(x) := c̃w(rx),
where the constant c̃ is chosen so that
Z
x ∈ B(0, 2),
(1.89)
w̃2 = 1.
(1.90)
B(0,2)
Using the doubling condition (1.87) twice, we obtain
Z
5
w̃2 ≥ 2−Cλ .
(1.91)
B(0,1/2)
Note that w̃ satisfies an equation of type (1.73) with the coefficients satisfying (1.74)
with α, K,γ depending only on Ω: compared to the equation (1.88) satisfied by w, after
the scaling, as we have seen before, the L∞ bounds of the lower-order coefficients only
improve, while the leading order coefficients keep the L∞ -bounds and ellipticity and
improve the Lipschitz constant. Using Lemma 1.3.14, we will show that
||w̃||H 1 (B n−1 (0,3/4)) + ||(∂ w̃/∂xn )||L2 (B n−1 (0,3/4)) ≥ 2−Cλ
5
(1.92)
with a different constant C, which still depends only on Ω. Take the constants C = C1 , β
from Lemma 1.3.14 used for w̃ which only depend on n and Ω. Then if we have
||w̃||H 1 (B n−1 (0,3/4)) + ||(∂ w̃/∂xn )||L2 (B n−1 (0,3/4)) = ǫ̃ ≪ 1,
then Lemma 1.3.14 and (1.90) imply
||w̃||L2 (B(0,1/2)) ≤ C1 ǫ̃β .
Since we have the lower bound (1.91), it follows that
−1/β −Cλ5 /2β
ǫ̃ ≥ C1
2
,
so we get (1.92) with the constant (C + 2 log2 C1 )/2β.
Now we will show that
||∇w̃||L2 (B n−1 (0,3/4)) ≥
1
||(∂ w̃/∂xn )||L2 (B n−1 (0,3/4)) ,
C
35
(1.93)
and hence from (1.92) we will get a lower bound for ||w̃||H 1 (B n−1 (0,3/4)) . After that,
we will show that ||w̃||L2 (B n−1 (0,7/8)) controls ||w̃||H 1 (B n−1 (0,3/4)) , and hence we will get
a lower bound for it, which will be a significant part of the doubling condition on the
hypersurface.
Recall that w̃(x) = c̃w(rx) = c̃vx0 ,λ (F (rx)), F (B n−1 (0, 2r) × {0}) ⊂ ∂Ωx0 ,λ , and the
normal derivative of vx0 ,λ on ∂Ωx0 ,λ is zero. After using the area formula as above, we
obtain (∇T denotes the tangential gradient):
||∇w̃||L2 (B n−1 (0,3/4)) = c̃r−
≥ c̃r
n−3
2
− n−3
2
= c̃r−
n−3
2
||(∂ w̃/∂xn )||L2 (B n−1 (0,3/4)) = c̃r−
n−3
2
≤ c̃r
− n−3
2
||∇w||L2 (B n−1 (0,3r/4))
C −1 ||∇T vx0 ,λ ||L2 (F (B n−1 (0,3r/4)))
C −1 ||∇vx0 ,λ ||L2 (F (B n−1 (0,3r/4))) ,
||(∂w/∂xn )||L2 (B n−1 (0,3r/4))
C||∇vx0 ,λ ||L2 (F (B n−1 (0,3r/4))) ,
and (1.93) follows. Together with (1.92), we hence obtain (with a different constant than
in (1.92))
5
||w̃||H 1 (B n−1 (0,3/4)) ≥ 2−Cλ .
(1.94)
Now we will show that
||w̃||H 1 (B n−1 (0,3/4)) ≥ ǫ̃ ⇒ ||w̃||L2 (B n−1 (0,7/8)) ≥ ǫ̃3 /C.
(1.95)
We will use Lemma 1.3.15. Since it involves functions defined in the whole Rn , introduce
a cut-off function ϕ ∈ C ∞ (Rn ) such that
ϕ=1
on B(0, 3/4),
ϕ=0
on Rn \ B(0, 7/8),
||ϕ||C 2 (Rn ) ≤ C.
Then using Lemma 1.3.15 for the function w̃′ = w̃ · ϕ (extended by 0 beyond Rn \
B(0, 7r/8)), we get that for any η > 0,
||∇w̃||L2 (B n−1 (0,3/4)) ≤ ||∇w̃′ ||L2 (Rn−1 )
C
||w̃′ ||L2 (Rn−1 )
η2
C
≤ Cη||w̃||H 2 (B(0,7/8)) + 2 ||w̃||L2 (B n−1 (0,7/8))
η
C
≤ Cη||w̃||L2 (B(0,1)) + 2 ||w̃||L2 (B n−1 (0,7/8))
η
C
≤ Cη + 2 ||w̃||L2 (B n−1 (0,7/8)) ,
η
≤ η||w̃′ ||H 2 (Rn ) +
36
(1.96)
where in the last two lines we used the interior elliptic estimate for strong solutions
||w̃||H 2 (B(0,7/8)) ≤ C||w̃||L2 (B(0,1))
(see [19, Theorem 9.11]) and (1.90). Choosing η = ǫ̃/2C in (1.96), where
||w̃||H 1 (B n−1 (0,3/4)) ≥ ǫ̃,
we obtain
ǫ̃ − ||w̃||L2 (B n−1 (0,3/4)) ≤ ||∇w̃||L2 (B n−1 (0,3/4))
C
≤ ǫ̃/2 + 2 ||w̃||L2 (B n−1 (0,7/8)) ,
ǫ̃
and (1.95) follows. Together with (1.94), we obtain (with a different constant)
5
||w̃||L2 (B n−1 (0,7/8)) ≥ 2−Cλ .
(1.97)
On the other hand, the trace theorem, the interior elliptic estimate [19, Theorem 9.11],
and (1.90) imply
||w̃||L2 (B n−1 (0,7/4)) ≤ C||w̃||H 1 (B n (0,7/4)) ≤ C̃||w̃||L2 (B(0,2)) = C̃,
so
5
||w̃||L2 (B n−1 (0,7/8)) ≥ 2−Cλ ||w̃||L2 (B n−1 (0,7/4)) ,
which translates into
5
||w||L2 (B n−1 (0,7r/8)) ≥ 2−Cλ ||w||L2 (B n−1 (0,7r/4)) .
Since r ≤ r0 /2(1+ǫ) was arbitrary, we proved (1.86) for any r ≤ r1 , r1 = 7r0 /(16·1.1).
In the process of the above proof, we showed that the average integral of u2 over a
small ball on the boundary controls the average integral of u2 over a small solid ball with
the same center. Since we will use this result in the next chapter (independently of the
doubling condition), let us formulate it precisely.
Corollary 1.3.16. There exist C, r0 > 0 depending only on Ω such that for all r ≤ r0
and x0 ∈ ∂Ω, there holds
Z
Z
2
−Cλ5 λ
u ≥2
u2 .
(1.98)
r
r B (x0 , 2r )∩Ω
B ( x0 , λ
)∩∂Ω
λ
Proof. This follows from (1.97). Looking back at the definition of c̃ by (1.89) and (1.90),
we have
Z
Z
1
1
2
w
(rx)dx
=
vx20 ,λ (F (x))
=
n
c̃2
r
B(0,2)
B(0,2r)
Z
n Z
λ
λn
2
u2 .
≥ n
v ≥ n
r B x0 , 2r
r B x0 , 2r ∩Ω
(1+ǫ)λ
(1+ǫ)λ
37
We used a change of variables by the map F , whose Jacobian is 1 and for which (1.82)
holds. On the other hand, from (1.97), we obtain
Z
Z
1 −Cλ5
1
1
2
w̃ = n−1
w2
2
≤ 2
c̃2
c̃ B n−1 (0,7/8)
r
B n−1 (0,7/8r)
Z
1
vx20 ,λ (F (x))
= n−1
r
n−1
B
(0,7/8r)
Z
Z
n−1
λ
λn−1
2
≤ n−1
v = n−1
u2 .
7(1+ǫ)
7(1+ǫ)
r
r
B x0 ,
B x0 ,
r ∩∂Ω
r ∩∂Ω
8
8
Here we used a change of variables by the map F |B n−1 (0,r)×{0} , whose Jacobian is larger
or equal to 1 and for which (1.84) holds.
Combining the above two estimates and using ǫ such that (1 + ǫ)2 < 8/7 (this gives
us r0 > 0 depending only on Ω such that the estimates hold for r ≤ r0 ), we obtain
(1.98).
38
Chapter 2
Size of the Nodal Sets of Steklov
Eigenfunctions
2.1
Nodal Sets of Steklov Eigenfunctions on Analytic Domains
In this section we prove Theorem 1:
Theorem 1. Let Ω ⊂ Rn be an analytic domain. Then there exists a constant C
depending only on Ω and n such that for any λ > 0 and u which is a (classical) solution
to (1) there holds
Hn−2 ({x ∈ ∂Ω : u(x) = 0}) ≤ Cλ6 .
(2.1)
The proof relies on the doubling condition on the boundary (Theorem 2) and the
approach used in [29] for analytic solutions of elliptic equations on a solid domain.
For the rest of this section, assume the assumptions of Theorem 1: Ω ⊂ Rn is an
analytic domain and u is a solution to (1) with eigenvalue λ > 0.
2.1.1
Analyticity of the eigenfunctions
First we will see that u is analytic up to the boundary ∂Ω, and get an estimate on its
derivatives which we will need in a complexification argument.
Proposition 2.1.1. The eigenfunction u is real-analytic on a neighborhood of Ω. More
precisely, there exists r0 > 0 depending only on Ω such that u can be harmonically
extended onto an rλ0 -neighborhood of Ω. Moreover, there exists a constant C such that
for every x0 ∈ ∂Ω and r ≤ r0 ,
α!
|Dα u(x0 )| ≤ C
(r/2neλ)|α|
39
Z
−
B(x0 ,r/λ)
u2
!1/2
.
(2.2)
Proof. By the theorem on elliptic iterates of Lions and Magenes ([30, Theorem VIII.1.2]),
u is analytic on a neighborhood of Ω. To determine the size of this neighborhood, for a
fixed x0 ∈ ∂Ω, in a similar way as in Section 1.3.1, denote
ux0 ,λ (x) := u(x0 + x/λ), x ∈ Ωx0 ,λ ,
where Ωx0 ,λ = {x : x0 + x/λ ∈ Ω}. By (1) we have
∆ux0 ,λ = 0
∂u
=u
∂ν
in Ωx0 ,λ ,
on ∂Ωx0 ,λ .
It follows from sections VIII.1 and VIII.2 in [30] that ux0 ,λ can be analytically extended
onto B(0, r0 ), where r0 depends only on Ω, not on x0 or λ (the domain Ωx0 ,λ changes, but
only becomes flatter, i.e. better with increasing λ). This means that u can be analytically
extended onto B(x0 , r0 /λ). Then ∆u is also analytic on B(x0 , r0 /λ) and is equal to zero
on the open set B(x0 , r0 /λ) ∩ Ω, hence it is zero in B(x0 , r0 /λ) and u is harmonic in
B(x0 , r0 /λ). Hence, from Proposition 1.13 and Remark 1.19 in [24], we obtain
|Dα u(x0 )| ≤ C
α!
(r/2neλ)|α|
α!
≤C
(r/2neλ)|α|
max
B(x0 ,r/2λ)
|u|
Z
−
u2
B(x0 ,r/λ)
!1/2
.
Corollary 2.1.2. Let r0 be as in Proposition 2.1.1, x0 ∈ ∂Ω. Then u can be extended
n
into the complex ball B C (x0 , r0 /6nλ) such that for any r ≤ r0 ,
sup
n
z∈B C (x0 ,r/6nλ)
Z
|u(z)| ≤ C −
B(x0 ,r/λ)
u
2
!1/2
.
(2.3)
Proof. Write the real Taylor polynomial of u in x0 . By the estimate (2.2) on its con
efficients |Dα u(x0 )|/α!, it converges for all complex z ∈ B C (x0 , r0 /2neλ), and for
n
z ∈ B C (x0 , r/6nλ) we also obtain the estimate (2.3).
2.1.2
Proof of Theorem 1
Flattening the boundary and complexification:
As in the proof of Theorem 2 in Section 1.3.4, we cover ∂Ω by finitely many pieces
Γi ⊂ ∂Ω, i = 1, 2, . . . , k, such that k depends only on Ω and each Γi is a graph of a
(this time) analytic function (in some coordinate system). We require some additional
properties of this cover. For each Γi = Γ, after choosing the right coordinate system, we
want
Γ = {x = (x′ , xn ) ∈ Rn : x′ ∈ Γ̃, xn = Φ(x′ )},
40
where Γ̃ ⊂ Rn−1 is a compact set, Φ is analytic on the larger set
2r0
′
n−1
Γ̃ = x ∈ R
: dist(x, Γ̃) <
,
λ1
where r0 > 0 is a constant depending on Ω,
{(x′ , Φ(x′ )) : x′ ∈ Γ̃′ } ⊂ ∂Ω,
∇Φ(x0 ) = 0 for some point x0 ∈ Γ̃, and Φ can be extended to a complex analytic function
on
−r0 r0 n−1
′
,
Γ̃ ×
λ1 λ1
with ∇z Φ(x0 ) = 0,
′
|∇z Φ(z)| ≤ 0.1 for z ∈ Γ̃ ×
−r0 r0
,
λ1 λ1
n−1
.
(2.4)
This can be achieved by considering a neighborhood of each x0 ∈ ∂Ω which can be
parametrized this way and then choosing a finite cover of ∂Ω by these neighborhoods.
Denote
r0
′
n−1
Γ̃m = x ∈ R
: dist(x, Γ̃) <
λ1
an “intermediate” set (Γ̃ ⊂ Γ̃′m ⊂ Γ̃′ ).
In a similar way as in the proof of Theorem 2, denote
w(z ′ ) := u(z ′ , Φ(z ′ )).
By the definition of Φ and Corollary 2.1.2, this is well-defined if
−r0 r0 n−1
r0
′
′
z ∈ Γ̃ ×
,
,
and
dist((z ′ , Φ(z ′ )), ∂Ω) ≤
λ1 λ1
6nλ
where r0 is the minimum of r0 ’s coming up in Corollary 2.1.2 and in the discussion above
about properties of Φ. If z ′ = x′ + iy ′ , x′ ∈ Γ̃′ , |y ′ | ≤ λr01 , then (x′ , Φ(x′ )) ∈ ∂Ω, and we
have
dist((z ′ , Φ(z ′ )), ∂Ω) ≤ |(x′ + iy ′ , Φ(x′ + iy ′ )) − (x′ , Φ(x′ ))|
≤ |y ′ | + |y ′ | sup |∇z Φ(x′ + ity ′ )| ≤ 1.1|y ′ |
t∈[0,1]
by (2.4). Hence, w is well-defined and analytic for
z ′ = x′ + iy ′ , x′ ∈ Γ̃′ , |y ′ | ≤
r0
,
6.6nλ
and by Corollary 2.1.2, for all r ≤ r0 , x′ ∈ Γ̃′m we also have
sup
r
)
(x′ , 6.6nλ
n−1
BC
Z
|w| ≤ C −
B((x′ ,Φ(x′ )),r/λ)
41
u2
!1/2
.
(2.5)
Instead of the bound on the right hand side which involves an integral over Rn -balls,
using Corollary 1.3.16 we can bound the left hand side by an integral over balls on ∂Ω.
Denote x0 = (x′ , Φ(x′ )). Although Corollary 1.3.16 gives a bound only on an integral
over B(x0 , r/λ) ∩ Ω, in our case u is harmonic also on B(x0 , r/λ) ∩ Ωc and satisfies the
c
almost identical boundary condition ∂u
∂ν = −λu on ∂Ω (the normal has opposite sign).
Since all arguments in the proof of Corollary 1.3.16 are local and the sign of λ never
mattered, applying it to B(x0 , r/λ) ∩ Ωc gives us
Z
Z
2
−Cλ5 λ
u ≥2
u2
r
r B (x0 , r )∩Ωc
B (x0 , 2λ
)∩∂Ω
λ
and together with Corollary 1.3.16 for regular Ω and (2.5), we obtain
!1/2
n−1 Z
λ
2
u
r
B (x0 , r )∩∂Ω
Cλ5
BC
sup
|w| ≤ 2
r
(x′ , 3.3nλ
)
n−1
λ
for all x′ ∈ Γ̃′m , x0 = (x′ , Φ(x′ )), and r ≤ r0 , where r0 depends only on Ω (it is different
than above). By a change of variables as in the proof of Theorem 2, using (2.4) for
bounding the Jacobian, this translates into
Cλ5
sup
|w| ≤ 2
r
(x′ , 3.3nλ
)
B n−1 (x′ , 1.1r
λ )
n−1
BC
(since x′ ∈ Γ̃′m , we have B n−1 x′ , 1.1r
λ
Hence we proved
Z
−
w2
!1/2
⊂ Γ̃′ and the right hand side is well-defined).
Cλ5
|w| ≤ 2
sup
r
)
(x′ , 4nλ
Z
−
r
B n−1 (x′ , λ
)
n−1
BC
w2
!1/2
(2.6)
for all x′ ∈ Γ̃′m and r ≤ r0 , where r0 > 0 depends only on Ω (different from r0 above).
Also note that the doubling condition, Theorem 2, translates into the following for
w:
Z
Z
2
Cλ5
w ≤2
w2
(2.7)
r
B n−1 (x′ , λ
)
r
B n−1 (x′ , 2λ
)
for all x′ ∈ Γ̃′m and r ≤ r0 , where r0 > 0 depends only on Ω (different from r0 above). We
actually proved this in the process of proving Theorem 2 as (1.86) (the w|B n−1 (0,r0 )×{0}
in there was defined exactly as our w shifted to 0 and scaled by λ).
By the change of variable as in the proof of Theorem 2, using (2.4), to prove (2.1),
it is enough to prove
Hn−2 ({x′ ∈ Γ̃ : w(x′ ) = 0}) ≤ Cλ6 .
(2.8)
To estimate the nodal set, we use an estimate on the number of zero points for analytic
functions.
42
Lemma 2.1.3. Suppose f : B1 ⊂ C → C is analytic with
|f (0)| = 1
sup |f | ≤ 2N ,
and
B1
for some positive constant N . Then for any r ∈ (0, 1) there holds
#{z ∈ Br : f (z) = 0} ≤ cN,
where c is a positive constant depending only on r. For r = 1/2, we have
#{z ∈ B1/2 : f (z) = 0} ≤ N.
Proof. A similar version of this lemma was first proved in [10]. This version can be found
in [22] as Lemma 2.3.2.
Proof of Theorem 1. It remains to prove (2.8). Take r0 such that both (2.6) and (2.7)
r0
hold for all r ≤ r0 and x′ ∈ Γ̃′m . Choose a cover of Γ̃ by balls of radii R = (16n+1)λ
.
n−1
We need Cλ
balls in the cover, where C depends only on Ω. Take one of them and
denote it B(p, R), i.e. the center is p ∈ Γ̃. Choose xp ∈ B(p, R) ⊂ Γ̃′m such that
Z
|w(xp )| ≥ −
B(p,R)
|w|
2
!1/2
.
(2.9)
Then B(p, R) ⊂ B(xp , 2R), so the Cλn−1 balls B(xp , 2R) cover Γ̃. Next,
B(xp , 16nR) ⊂ B(p, (16n + 1)R),
and since 4R < r0 /4nλ, by (2.6) we have
Cλ5
sup
BC
n−1
(xp ,4R)
|w| ≤ 2
Z
−
w
2
B n−1 (xp ,16nR)
Cλ5
≤2
Z
−
!1/2
w
B n−1 (p,(16n+1)R)
2
!1/2
.
Now use the doubling condition (2.7) with center p, ⌈log2 (16n + 1)⌉-times, and obtain
sup
B
Cn−1
(xp ,4R)
Cλ5
|w| ≤ 2
Z
−
w
B n−1 (p,R)
2
!1/2
5
≤ 2Cλ |w(xp )|
(2.10)
(we used (2.9) in the last inequality). For each direction ω ∈ Rn−1 , |ω| = 1, consider
the function fω (z) = w(xp + 4Rzω) of one complex variable z defined on B C (0, 1). From
(2.10) and Lemma 2.1.3, we obtain
#{x ∈ B n−1 (xp , 2R) : x − xp ||ω, w(x) = 0} ≤ #{z ∈ B1/2 : fω (z) = 0}
= N (ω) ≤ Cλ5 .
43
By the integral geometry estimate [14, 3.2.22], we obtain
Z
n−2
n−1
n−2
H
{x ∈ B
(xp , 2R) : w(x) = 0} ≤ c(n)R
Sn−2
N (ω)dω ≤ C
1
λn−2
λ5 .
Since there were Cλn−1 balls B n−1 (xp , 2R) which cover Γ̃, summing up these estimates
gives (2.8).
2.2
Nodal Sets of Steklov Eigenfunctions on
Smooth Domains
In this section we briefly discuss the possible directions of research to obtain an estimate
for the size of the nodal set of Steklov eigenfunctions in the case that Ω is not analytic,
using the doubling condition which we proved also in this setting.
Following an argument first employed in [25], the following bound was proved for
solutions of general elliptic equations with non-analytic coefficients in [22] as Corollary
5.3.8:
Theorem 10. Suppose that w is a nonzero solution of (1.20) in B1 , with the coefficients
of the equation satisfying (1.21), (1.22) and (1.23). Assume that for each B(p, r) ⊂ B1 ,
there holds
R
r B(p,r) |Dw|2
R
≤ N.
(2.11)
2
∂B(p,r) w
Then
Hn−1 ({x ∈ B1/2 : w(x) = 0}) ≤ (c1 N )c2 N ,
where c1 and c2 are positive constants depending only on n, α, Γ and K.
The proof of this theorem splits the nodal set of w into a good part, where the
gradient of w is large, and a bad part, where the gradient is small. The good part can
be well approximated by the nodal set of a harmonic function and estimated that way.
For the bad part, the fact that the singular set where both w and Dw vanish has a
lower dimension, is used. Unfortunately, that part does not translate into our case of
Steklov eigenfunctions: since we consider them on the hypersurface ∂Ω, the singular set
can have the same dimension as the nodal set. The Hn−2 -measure of the singular set
can be estimated, but the known results in [21] require the coefficients of the equation
to be smooth and use their C M -norms, where M depends (among others) on N from
(2.11), which in our case translates to requiring C M -smoothness of the domain Ω with
M depending on λ. The estimate on the size of the singular set is also not explicit.
Another way how to estimate the size of the nodal set of Steklov eigenfunctions on
non-analytic Ω is to follow the approach used in [23] to estimate the measure of slices of
nodal sets. However, this approach also uses the C M -norms of coefficients, M depending
on N much like above, and the estimate is also not explicit.
Using either approach, one can hope to prove that for C ∞ -smooth Ω, the Hn−2 measure of the nodal set of Steklov eigenfunctions is finite for any λ.
44
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