Semiclassical bounds in magnetic bottles

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DOI: 10.1142/S0129055X16500021 · Source: arXiv
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Abstract
The aim of the paper is to derive spectral estimates into several classes of magnetic systems. They include three-dimensional regions with Dirichlet boundary as well as a particle in $\mathbb{R}^3$ confined by a local change of the magnetic field. We establish two-dimensional Berezin-Li-Yau and Lieb-Thirring-type bounds in the presence of magnetic fields and, using them, get three-dimensional estimates for the eigenvalue moments of the corresponding magnetic Laplacians.
SEMICLASSICAL BOUNDS
IN MAGNETIC BOTTLES
Diana Barseghyana,b, Pavel Exnera,c, Hynek Kovaˇr´ıkd,
Timo Weidle
a) Department of Theoretical Physics, Nuclear Physics Institute ASCR,
25068 ˇ
Reˇz near Prague, Czech Republic
b) Department of Mathematics, Faculty of Science, University
of Ostrava, 30. dubna 22, 70103 Ostrava, Czech Republic
c) Doppler Institute for Mathematical Physics and Applied
Mathematics, Czech Technical University, Bˇrehov´a 7, 11519 Prague
d) Dicatam, Sezione di Matematica, Universit`a degli Studi
di Brescia, via Branze 38, 25123 Brescia, Italy
e) Fakult¨at f¨ur Mathematik und Physik Institut f¨ur Analysis,
Dynamik und Modellierung, Universit¨at Stuttgart, Pfaffenwaldring 57,
70569 Stuttgart, Germany
dianabar@ujf.cas.cz, diana.barseghyan@osu.cz, exner@ujf.cas.cz,
hynek.kovarik@unibs.it, weidl@mathematik.uni-stuttgart.de
Abstract. The aim of the paper is to derive spectral estimates into several
classes of magnetic systems. They include three-dimensional regions with
Dirichlet boundary as well as a particle in R3confined by a local change of
the magnetic field. We establish two-dimensional Berezin-Li-Yau and Lieb-
Thirring-type bounds in the presence of magnetic fields and, using them, get
three-dimensional estimates for the eigenvalue moments of the corresponding
magnetic Laplacians.
1 Introduction
Let be the Dirichlet Laplacian corresponding to an open bounded do-
main Ω Rd, defined in the quadratic form sense on H1
0(Ω). The operator
is obviously non-negative and since the embedding H1
0L2(Ω) is compact,
its spectrum is purely discrete accumulating at infinity only. It is well known
1
that for d= 3, up to a choice of the scale, the eigenvalues describe energies
of a spinless quantum particle confined to such a hard-wall ‘bottle’.
Motivated by this physical problem, we consider in the present work a
magnetic version of the mentioned Dirichlet Laplacian, that is, the operator
H(A)=(i+A(x))2associated with the closed quadratic form
k(i+A)uk2
L2(Ω) , u ∈ H1
0(Ω) ,
where the real-valued and sufficiently smooth function Ais a vector potential.
The magnetic Sobolev norm on the bounded domain Ω is equivalent to the
non-magnetic one and the operator H(A) has a purely discrete spectrum as
well. We shall denote the eigenvalues by λk=λk(Ω, A), assuming that they
repeat according to their multiplicities.
One of the objects of our interest in this paper will be bounds of the
eigenvalue moments of such operators. For starters, recall that for non-
magnetic Dirichlet Laplacians the following bound was proved in the work
of Berezin, Li and Yau [Be72a, Be72b, LY83],
X
k
λk(Ω,0))σ
+Lcl
σ,d ||Λσ+d
2for any σ1 and Λ >0,(1.1)
where ||is the volume of Ω and the constant on the right-hand side,
Lcl
σ,d =Γ(σ+ 1)
(4π)d
2Γ(σ+1+d/2) ,
is optimal. Furthermore, the bound (1.1) holds true for 0 σ < 1 as well,
but with another, probably non-sharp constant on the right-hand side,
X
k
λk(Ω,0))σ
+2σ
σ+ 1σ
Lcl
σ,d ||Λσ+d
2,0σ < 1.(1.2)
see [La97]. In the particular case σ= 1 the inequality (1.1) is equivalent, via
Legendre transformation, to the lower bound
N
X
j=1
λj(Ω,0) Cd||2
dN1+ 2
d, Cd=4πd
d+ 2Γ(d/2 + 1) 2
d.(1.3)
Turning next to the magnetic case, we note first that the pointwise dia-
magnetic inequality [LL01], namely
|∇|u(x)|| ≤ |(i+A)u(x)|for a.a. x,
2
implies λ1(Ω, A)λ1(Ω,0), however, the estimate λj(Ω, A)λj(Ω,0) fails
in general if j2. Nevertheless, momentum estimates are still valid for
some values of the parameters. In particular, it was shown [LW00] that the
sharp bound (1.1) holds true for arbitrary magnetic fields provided σ3
2,
and the same sharp bound holds true for constant magnetic fields if σ1,
see [ELV00]. Furthermore, in the dimension d= 2 the bound (1.2) holds true
for constant magnetic fields if 0 σ < 1 and the constant on its right-hand
side cannot be improved [FLW09].
Our main aim in the present work is to derive sufficiently precise two-
dimensional Berezin-type estimates for quantum systems exposed to a mag-
netic field and to apply them to the three-dimensional case. We are going
to address two questions, one concerning eigenvalue moments estimates for
magnetic Laplacians on three dimensional domains having a bounded cross
section in a fixed direction, and the other about similar estimates for mag-
netic Laplacians defined on whole R3.
Let us review the paper content in more details. In Sec. 2 we will describe
the dimensional-reduction technique [LW00] which allows us to derive the
sought spectral estimates for three-dimensional magnetic ‘bottles’ using two-
dimensional ones. Our next aim is to derive a two-dimensional version of
the Li-Yau inequality (1.3) in presence of a constant magnetic field giving
rise to an extra term on the right-hand side. The result will be stated and
proved in first part of Sec. 3. This in turn will imply, by means of Legendre
transformation, a magnetic version of the Berezin inequality which we are
going to present in second part of Sec. 3. It has to be added that the question
of semiclassical spectral bounds for such systems has been addressed before,
in particular, another version of the magnetic Berezin inequality was derived
by two of us [KW13]. In final part of Sec. 3 we are going to compare the
two results and show that the one derived here becomes substantially better
when the magnetic field is strong.
In some cases the eigenvalues of the magnetic Dirichlet Laplacian with a
constant magnetic field can be computed exactly in terms of suitable special
functions. In the first part of Sec. 4 we are present such an example con-
sidering the magnetic Dirichlet Laplacian on a two-dimensional disc with a
constant magnetic field. Its eigenvalues will be expressed in terms of Kummer
function zeros. Next, in the second part, we are going to consider again the
magnetic Dirichlet Laplacian on a two-dimensional disc, now in a more gen-
eral situation when the magnetic field is no longer homogeneous but retains
the radial symmetry; we will derive the Berezin inequality for the eigenvalue
3
moments. In Sec. 5 we shall return to our original motivation and use the
mentioned reduction technique to derive Berezin-type spectral estimates for
a class of three-dimensional magnetic ‘bottles’ characterized by a bounded
cross section in the x3direction.
Turning to the second one of the indicated questions, from Sec. 6 on, we
shall be concerned with magnetic Laplacians in L2(R3) associated with the
magnetic field B:R3R3which is as a local perturbation of a constant
magnetic field of intensity B0>0. Again, as before, we first derive suitable
two-dimensional estimates; this will be done in Sec. 6. In the last two sections
we apply this result to the three-dimensional case. In Sec. 7 we show that the
essential spectrum of the magnetic Laplacian with corresponding perturbed
magnetic field coincides with [B0,). The Sec. 7.1 we then prove Lieb-
Thirring-type inequalities for the moments of eigenvalues below the threshold
of the essential spectrum for several types of magnetic ‘holes’.
2 Dimensional reduction
As indicated our question concerns estimating eigenvalues due to confine-
ment in a three-dimensional ‘bottle’ by using two-dimensional Berezin type
estimates. In such situation one can use the dimension-reduction tech-
nique [LW00]. In particular, let be the Dirichlet Laplacian on an open
domain Ω R3, then for any σ3
2the inequality
tr (Λ ())σ
+Lcl
1ZR
tr Λ(ω(x3))σ+1
2
+dx3(2.1)
is valid, where ω(x3)is the Dirichlet Laplacian on the section
ω(x3) = x0= (x1, x2)R2|x= (x0, x3) = (x1, x2, x3),
see [LW00], and also [ELM04, Wei08]. The integral at the right-hand side of
(2.1), in fact restricted to those x3for which inf spec(ω(x3))<Λ, yields the
classical phase space volume. Note that in this way one can obtain estimates
also in some unbounded domains [GW11] as well as remainder terms [Wei08].
A similar technique can be used also in the magnetic case. To describe
it, consider a sufficiently smooth magnetic vector potential A(·) : Ω R3
generating the magnetic field
B(x) = (B1(x), B2(x), B3(x)) = rot A(x).
4
For the sake of definiteness, the shall use the gauge with A3(x) = 0. Fur-
thermore, we consider the magnetic Dirichlet Laplacians
H(A) = (ixA(x))2on L2(Ω)
and e
Hω(x3)(e
A)=(ix0e
A(x))2on L2(ω(x3)) ,
where e
A(x) := (A1(x), A2(x)). Note that for the fixed x3the two-dimensional
vector potential e
A(x0, x3) corresponds to the magnetic field
˜
B(x0, x3) = B3(x) = ∂A2
∂x1A1
∂x2
.
Referring to [LW00, Sec. 3.2] one can then claim that for a σ3
2we have
tr(Λ − H(A))σ
+Lcl
1ZR
tr(Λ e
Hω(x3)(e
A))σ+1/2
+dx3.(2.2)
3 Berezin-Li-Yau inequality with a constant
magnetic field
Suppose that the motion is confined to a planar domain ωbeing exposed to
influence of a constant magnetic field of intensity B0perpendicular to the
plane, and let A:R2R2be a vector potential generating this field. We
denote by Hω(A) the corresponding magnetic Dirichlet Laplacian on ωand
µj(A) will be its eigenvalues arranged in the ascending with repetition ac-
cording to their multiplicity. Our first aim is to extend the Li-Yau inequality
(1.3) to this situation with an additional term on the right-hand side depend-
ing on B0only. This will be then used to derive the Berezin-type inequality.
Conventionally we denote by Nthe set of natural numbers, while the set of
integers will be denoted by Z.
The following result is not new. Indeed, it can be recovered from [ELV00,
Sec. 2], however, for the sake of completeness we include a proof.
3.1 Li-Yau estimate
Theorem 3.1. Assume that ωR2is open and finite. Then the inequality
X
jN
µj(A)2πN 2
|ω|+B2
0
2π|ω|m(1 m) (3.1)
5
holds, where m:= n2πN
B0|ω|ois the fractional part of 2πN
B0|ω|.
Proof. Without loss of generality we may assume that B0>0. Let Pkbe the
orthogonal projection onto the k-th Landau level, B0(2k1), of the Landau
Hamiltonian (i+A(x))2in L2(R2) which is an integral operator with the
kernel Pk(x, y) – see [KW13]. Note that we have
Pk(x, x) = 1
2πB0,(3.2)
ZR2Zω|Pk(y, x)|2dxdy=ZωZR2
Pk(y, x)Pk(x, y) dydx
=Zω
Pk(x, x) dx=B0
2π|ω|.(3.3)
Let φjbe a normalized eigenfunction corresponding to the eigenvalue µj(A).
We put fk,j(y) := RωPk(y, x)φj(x) dx, where yR2, and furthermore
FN(k) := X
jNkfk,j k2
L2(R2).
We have the following identity,
X
jN
µj(A) = X
jNk(i∇ − A)φjk2
L2(ω)
=X
jNX
kNk(i∇ − A)fk,j k2
L2(R2)
=X
kN
B0(2k1) X
jNkfk,j k2
L2(R2)
=X
kN
B0(2k1)FN(k) =: J[FN].
Moreover, the normalization of the functions φjimplies
X
kN
FN(k) = X
jNX
kNkfk,j k2
L2(R2)=X
jNkφjk2
L2(ω)=N . (3.4)
6
Finally, in view of Bessel’s inequality the following estimate holds true,
FN(k) = X
jNkfk,j k2
L2(R2)=ZR2X
jNZω
Pk(y, x)φj(x) dx
2
dy
ZR2Zω|Pk(y, x)|2dxdy=B0
2π|ω|.(3.5)
Let us now minimize the functional J[FN] under the constraints (3.4) and
(3.5). To this aim, recall first the bathtub principle [LL01]:
Given a σ-finite measure space (Ω,Σ, µ), let fbe a real-valued measurable
function on Ω such that µ{x:f(x)< t}is finite for all tR. Fix further a
number G > 0 and define a class of measurable functions on Ω by
C=g: 0 g(x)1 for all xand Z
g(x)µ(dx) = G.
Then the minimization problem of the functional
I= inf
g∈C Z
f(x)g(x)µ(dx)
is solved by
g(x) = χ{f<s}(x) + {f=s}(x),(3.6)
giving rise to the minimum value
I=Z{f<s}
f(x)µ(dx) + csµ{x:f(x) = s},
where
s= sup{t:µ{x:f(x)< t} ≤ G}
and
{x:f(x) = s}=Gµ{x:f(x)< s}.
Moreover, the minimizer given by (3.6) is unique if G=µ{x:f(x)< s}or
if G=µ{x:f(x)s}.
Applying this result to the functional J[FN] with the constraints (3.4) and
(3.5) we find that the corresponding minimizers are
FN(k) = B0
2π|ω|, k = 1,2, . . . , M ,
7
FN(M+ 1) = B0
2π|ω|m ,
FN(k) = 0, k > M + 1,
where M=h2πN
B0|ω|iis the entire part and m=n2πN
B0|ω|o, so that M+m=2πN
B0|ω|.
Consequently, we have the lower bound
J[FN]B0
2π|ω|
M
X
k=1
(2k1)B0+B0
2π|ω|m(2M+ 1)B0
=B0
2π|ω|(M2+ 2Mm +m)
=B2
0
2π|ω|(M+m)2+B2
0
2π|ω|(mm2)
which implies X
jN
µj(A)2πN 2
|ω|+B2
0
2π|ω|m(1 m).
This is the claim we have set out to prove.
Since 0 m < 1 by definition the last term can regarded as a non-
negative remainder term, which is periodic with respect to N
Φ, where Φ = B0|ω|
2π
is the magnetic flux, i.e. the number of flux quanta through ω. Note that for
N < Φ the right-hand side equals NB and for large enough B0this estimate
is better than the lower bound in terms of the phase-space volume.
3.2 A magnetic Berezin-type inequality
The result obtained in the previous subsection allows us to derive an exten-
sion of the Berezin inequality to the magnetic case. We keep the notation
introduced above, in particular, Hω(A) is the magnetic Dirichlet Laplacian
on ωcorresponding to a constant magnetic field B0and µj(A) are the respec-
tive eigenvalues. Without loss of generality we assume again that B0>0.
Then we can make the following claim.
Theorem 3.2. Let ωR2be open and finite, then for any Λ> B0we have
N
X
j=1
µj(A)) 2B2
0)|ω|
8π+B0)B0|ω|
4πΛ + B0
2B0.(3.1)
8
Proof. Subtracting NΛ from both sides of inequality (3.1), we get
N
X
j=1
µj(A)) NΛ2πN 2
|ω|B2
0
2π|ω|m(1 m),(3.2)
and consequently
N
X
j=1
µj(A))+NΛ2πN 2
|ω|B2
0
2π|ω|m(1 m)+
.
We are going to investigate the function f:R+R,
f(z) := zΛ2πz2
|ω|B2
0|ω|
2π2πz
B0|ω|12πz
B0|ω|,
on the intervals
B0|ω|k
2πz < B0|ω|(k+ 1)
2π, k = 0,1,2,...,
looking for an upper bound. It is easy to check that
f0(z) = Λ 4π
|ω|zB2
0|ω|
2π
2π
B0|ω|+2B2
0|ω|
2π2πz
B0|ω|2π
B0|ω|
= Λ 4π
|ω|zB0+ 2B02πz
B0|ω|,
thus the extremum of fis achieved at the point z0such that
ΛB04π
|ω|z0+ 2B02πz0
B0|ω|= 0 .(3.3)
Denoting x0:= 2πz0
B0|ω|, the condition reads Λ 2B0x0B0+ 2B0{x0}= 0
giving
x0=ΛB0+ 2B0{x0}
2B0
.
9
It yields the value of function fat z0, namely
f(z0) = ΛB0|ω|
2π
B0+ 2B0{x0})
2B0B2
0|ω|
2πΛB0+ 2B0{x0}
2B02
B2
0|ω|
2π{x0}(1 − {x0})
=Λ|ω|
4πB0+ 2B0{x0})|ω|
8πB0+ 2B0{x0})2
B2
0|ω|
2π{x0}(1 − {x0})
=|ω|
4πΛ(Λ B0+ 2B0{x0})B0+ 2B0{x0})2
2
2B2
0{x0}(1 − {x0})
=|ω|
4πΛ2ΛB0+ 2ΛB0{x0} − Λ2
2+ ΛB0B2
0
2B0{x0}
+2B2
0{x0} − 2B2
0{x0}22B2
0{x0}+ 2B2
0{x0}2
=|ω|2B2
0)
8π.(3.4)
Furthermore, the values of fat the endpoints B0k|ω|
2π, k = 0,1,2, . . . , equal
fB0k|ω|
2π=B0Λk|ω|
2π2π
|ω|
B2
0k2|ω|2
4π2=B0k|ω|
2πkB0).
10
Consider now an integer msatisfying 1 mhΛ+B0
2B0i, then
fB0|ω|
2πΛ + B0
2B0m
=B0|ω|
2πΛ + B0
2B0mΛΛ + B0
2B0mB0
B0|ω|
2πΛ + B0
2B0mΛΛ + B0
2B0mB0+Λ + B0
2B0B0
=(2m1)B0)|ω|
4πΛ + (2m1)B0
2+Λ + B0
2B0B0
=2(2m1)2B2
0)|ω|
8π+(2m1)B0)B0|ω|
4πΛ + B0
2B0
2B2
0)|ω|
8π+B0)B0|ω|
4πΛ + B0
2B0.(3.5)
On the other hand, for integers satisfying khΛ+B0
2B0ione can check easily
that
4B2
0k24B0Λk+ Λ2B2
00,
which means B0k|ω|
2πB0k)2B2
0)|ω|
8π.(3.6)
Combining inequalities (3.5) and (3.6) we conclude that at the interval end-
points, z=B0k|ω|
2π, k = 0,1,2, . . . , the value of function fdoes not exceed
2B2
0)|ω|
8π+B0)B0|ω|
4πnΛ+B0
2B0o. Hence in view of (3.4) we have
f(z)2B2
0)|ω|
8π+B0)B0|ω|
4πΛ + B0
2B0
for any z0. Combining this inequality above with the bound (3.2), we
arrive at the desired conclusion.
Remark 3.3. Using the Aizenman-Lieb procedure [AL78] and the fact that
inf σ(Hω(A)) B0we can get also bound for other eigenvalue moments.
11
Specifically, for any σ3/2 Theorem 3.2 implies
N
X
j=1
µj(A))σ+1/2
+=Γ(σ+ 3/2)
Γ(σ1/2)Γ(2) Z
0
t)σ3/2
+
N
X
j=1
(tµj(A))+dt
Γ(σ+ 3/2)
Γ(σ1/2) Z
0
t)σ3/2
+(t2B2
0)+|ω|
8π
+(tB0)+B0|ω|
4πΛ + B0
2B0dt
Γ(σ+ 3/2)|ω|
Γ(σ1/2) 2B2
0)+
8π
+B0)+B0
4πΛ + B0
2B0Z
0
t)σ3/2
+dt
=Γ(σ+ 3/2)Λσ1/2|ω|
Γ(σ1/2)(2σ1) 2B2
0)+
4π+B0)+B0
2πΛ + B0
2B0.
3.3 Comparison to earlier results
Given a set ωR2and a point xω, we denote by
δ(x) = dist(x, ∂ω) = min
y∂ω |xy|
the distance of xto the boundary, then
R(ω) = sup
xω
δ(x)
is the in-radius of ω. Furthermore, given a β > 0 we introduce
ωβ={xω:δ(x)< β}, β > 0,
and define the quantity
σ(ω) := inf
0<β<R(ω)|ωβ|
β.(3.7)
Using these notions and the symbols introduced above we can state the fol-
lowing result obtained in the work of two of us [KW13]:
12
Theorem 3.4. Let ωR2be an open convex domain, then for any Λ> B0
we have
N
X
j=1
µj(A)) Λ2|ω|
8π1
512π
σ2(ω)
|ω|Λ (3.8)
B2
01
2Λ + B0
2B02|ω|
2π1
128π
σ2(ω)
|ω|Λ.
To make a comparison to the conclusions of the previous section, let us make
both B0and Λ large keeping their ratio fixed. Specifically, we choose a Λ
from the interval (B0,2B0) writing it as Λ = B0(1 + α) with an α(0,1).
The second term on the right-hand side of (3.1) is then α2B2
0|ω|
8π, and we want
to show that the difference between the bounds (3.8) and (3.1) tends to plus
infinity as B0→ ∞. To this aim, we write Λ = B0(1 + α) with an α(0,1),
then
2B2
0)|ω|
8π+B0)B0|ω|
4πΛ + B0
2B0=B2
0|ω|
4πα(1 + α).(3.9)
On the other hand, a short calculation shows that for our choice of B0and
Λ the right-hand side of the bound (3.8) becomes
=Λ2|ω|
8πB2
0|ω|
2π1
2α
22
+Λ
512π
σ2(ω)
|ω|1 + (1 α)2
(1 + α)2,
in particular, after another easy manipulation we find that for large B0this
expression behaves as B2
0|ω|
2πα+O(B0). Comparing the two bounds we see
that
rhs of (3.8) rhs of (3.1) = B2
0|ω|
4πα(1 α) + O(B0) (3.10)
tending to plus infinity as B0→ ∞. At the same time,
rhs of (3.8)
rhs of (3.1) =2
1 + α+O(B1
0) (3.11)
illustrating that the improvement represented by Theorem 3.2 is most pro-
nounced for eigenvalues near the spectral threshold.
13
4 Examples: a two-dimensional disc
Spectral analysis simplifies if the domain ωallows for a separation of vari-
ables. In this section we will discuss two such situations.
4.1 Constant magnetic field
We suppose that ωis a disc and the applied magnetic field is homogeneous.
As usual in cases of a radial symmetry, the problem can be reduced to de-
generate hypergeometric functions. Specifically, we will employ the Kummer
equation
rd2ω
dr2+ (br)dω
dr= 0 (4.1)
with real valued parameters aand bwhich has two linearly independent
solutions M(a, b, r) and U(a, b, r), the second one of which has a singularity
at zero [AS64].
Given an α > 0, we denote by ak
|m|kNthe set of the first parameter
values such that M(ak
|m|,|m|+ 1, α) = 0. Since for any a, b 0 the function
M(a, b, r) has no positive zeros [AS64], all the ak
|m|are negative. Then the
following claim is valid.
Theorem 4.1. Let Hω(A)be the magnetic Dirichlet Laplacian corresponding
to a constant magnetic field B0and ωbeing the two dimensional disc with
center at the origin and radius r0>0. The eigenvalues of Hω(A)coincides
with nB0+B0|m| − m2ak
|m|,B0r0/2omZ, kN.
Proof. We employ the standard partial wave decomposition – see, e.g., [Er96]
L2(ω) =
M
m=−∞
L2((0, r0),2πr dr),(4.2)
and Hω(A) = L
m=−∞ hm, where
hm:= d2
dr21
r
d
dr+m
rB0r
22
.(4.3)
The last named operator differs by mB0from the operator
˜
hm=d2
dr21
r
d
dr+m2
r2+B2
0r2
4(4.4)
14
on the interval (0, r0) with Dirichlet boundary condition at the endpoint r0.
Looking for solutions to the eigenvalue equation
˜
hmu=λu (4.5)
we employ the Ansatz
u(r) = r|m|eB0r2/4v(r),
where vL2((0, r0), rdr). Computing the first two derivatives we get
˜
hmu=v00 2|m|+ 1
rv0+B0(|m|+ 1)v(r) + B0rv0r|m|eB0r2/4,
hence the equation (4.5) can rewritten as
v00 +2|m|+ 1
rB0rv0(B0(|m|+ 1) λ)v= 0 .(4.6)
Using the standard substitution we pass to the function g(r) = v2r
B0
belonging to L2(0, B0r2
0/2). Expressing the derivatives of vin terms of those
of g, one can rewrite equation (4.6) as
rg00(r)+(|m|+ 1 r)g0((|m|+ 1)B0λ)
2B0
g(r) = 0 ,
which is the Kummer equation with b=|m|+ 1 and a=(|m|+1)B0λ
2B0. The
mentioned singularity of its solution U(a, b, r) for small r, namely [AS64]
U(a, b, r) = Γ(b1)
Γ(a)r1b+O(rb2) for b > 2
and
U(a, 2, z) = 1
Γ(a)
1
r+O(ln r), U(a, 1, r) = 1
Γ(a)ln r+O(1)
means that u(r) = r|m|eB0r2/4U(|m|+1)B0λ
2B0,|m|+ 1,B0r2
2does not belong
to H1
0((0, r0), rdr). Consequently, the sought solution of (4.5) has the form
r|m|eB0r2/4M(|m|+ 1)B0λ
2B0
,|m|+ 1,B0r2
2,
15
and in view of the Dirichlet boundary conditions at r0we arrive at the spec-
tral condition
M(|m|+ 1)B0λ
2B0
,|m|+ 1,B0r2
0
2= 0 .
which gives n(|m|+ 1)B02B0ak
|m|,B0r0/2omZ, kNas the eigenvalue set;
returning to the original operator hmwe get the claim of the theorem.
4.2 Radial magnetic field
If the magnetic field is non-constant but still radially symmetric, in general
one cannot find the eigenvalues explicitly but it possible to find a bound to
the eigenvalue moments in terms of an appropriate radial two-dimensional
Schr¨odinger operator.
Theorem 4.2. Let Hω(A)be the magnetic Dirichlet Laplacian Hω(A)on a
disc ωof radius r0>0centered at the origin with a radial magnetic field
B(x) = B(|x|). Assume that
α:= Zr0
0
sB(s) ds < 1
2.(4.7)
Then for any Λ, σ 0, the following inequality holds true
tr(Λ Hω(A))σ
+1
12α+ sup
nNn
12α (4.8)
×tr
Λ
ω
D+1
x2+y2 Zx2+y2
0
sB(s) ds!2
σ
+
.
In particular, the estimate (4.8) implies
inf σ(Hω(A)) inf σ
ω
D+1
x2+y2 Zx2+y2
0
sB(s) ds!2
.
Proof. Let us again employ the partial-wave decomposition (4.2), with the
angular component (4.3) replaced by
hm:= d2
dr21
r
d
dr+m
r1
rZr
0
sB(s) ds2
,(4.9)
16
and inspect the eigenvalues of this operator. Obviously, for m0 we have
hm≥ − d2
dr21
r
d
dr+m2
r2+1
r2Zr
0
sB(s) ds2
,(4.10)
while for any m > 0 we can use the inequality
2|m|
r2Zr
0
sB(s) ds2m2
r2Zr
0
sB(s) ds
which in view of the assumption (4.7) yields
hm≥ − d2
dr21
r
d
dr+ (1 2α)m2
r2+1
r2Zr
0
sB(s) ds2
.
Next we divide the set of natural numbers into groups such that for all the
elements of any fixed group the entire part 12α mis the same, and we
estimate the operator hmfrom below by
hm≥ − d2
dr21
r
d
dr+12α m2
r2+1
r2Zr
0
sB(s) ds2
.(4.11)
Since the number of elements in each group is bounded from above by the
sum 1
12α+ supnNnn
12αo, using (4.10) and (4.11) one infers that
tr(Λ Hω(A))σ
+1
12α+ sup
nNn
12α
×
X
m=−∞
tr Λ d2
dr21
r
d
dr+m2
r2+1
r2Zr
0
sB(s) ds2!!σ
+
=1
12α+ sup
nNn
12α
×tr Λ
M
m=−∞ d2
dr21
r
d
dr+m2
r2+1
r2Zr
0
sB(s) ds2!!σ
+
with any σ, Λ0. However, the direct sum in the last expression is nothing
else than a partial-wave decomposition of the two-dimensional Schr¨odinger
operator with the radial potential V(r) = 1
r2Rr
0sB(s) ds2and the Dirichlet
condition at the boundary of the disc; this yields the desired claim.
17
5 Application to the three-dimensional case
Let us return now to our original motivation of estimating eigenvalues due to
confinement in a three-dimensional ‘bottle’. One can employ inequality (2.2)
in combination with the results of the previous sections to improve in some
cases the spectral bound by taking the magnetic field into account instead
of just dropping it.
Let Ω R3with the bounded x3cross sections. The class of fields to
consider are those of the form B(x) = (B1(x), B2(x), B3(x3)), that is, those
for which the component B3perpendicular to the cross section depends on
the variable x3only. Such fields certainly exist, for instance, one can think of
the situation when the ‘bottle’ is placed into a homogeneous magnetic field.
The field is induced by an appropriate vector potential A(·):ΩR3,
B(x) = (B1(x), B2(x), B3(x3)) = rot A(x),
and we consider the magnetic Dirichlet Laplacians
H(A) = (ixA(x))2on L2(Ω).
We use the notion introduced in Sec. 2. In view of the variational principle
we know that the ground-state eigenvalue of e
Hω(x3)(e
A) cannot fall below the
first Landau level B3(x3). Consequently, integrating with respect to x3in the
formula (2.2) one can drop for all the x3for which B3(x3)Λ. Combining
this observation with Remark 3.3 we get
tr(Λ − H(A))σ
+Γ(σ+ 3/2)Λσ1/2
4π(2σ1)Γ(σ1/2) Lcl
1Z{x3:B3(x3)<Λ}|ω(x3)|
×Λ2B3(x3)2+ 2B3ΛB3(x3)Λ + B3
2B3dx3
for any σ3/2.
Example 5.1. (circular cross section) Let Ω be a three-dimensional cusp
with a circular cross section ω(x3) of radius r(x3) such that r(x3)0 as
x3→ ∞. Then the above formula in combination with Theorem 4.1 yields
tr(Λ − H(A))σ
+Lcl
1X
mZ, kNZRΛB3(x3)
B3(x3)|m| − m2ak
|m|,B3(x3)r0(x3)/2σ+1/2
+
dx3
18
for any σ3/2. The particular case B(x) = {0,0, B}applies to a cusp-
shaped region placed to a homogeneous field parallel to the cusp axis.
Example 5.2. (radial magnetic field) Consider the same cusp-shaped region
Ω in the more general situation when the third field component can depend
on the radial variable, B(x)=(B1(x), B2(x), B3(x2
1+x2
2, x3)), assuming that
sup
x3R
α(x3) = sup
x3RZr0(x3)
0
sB3(s, x3) ds < 1
2.
Then the dimensional reduction in view of Theorem 4.2 gives
tr(Λ − H(A))σ
+Lcl
1ZR 1
p12α(x3)+ sup
nN(n
p12α(x3))!
×tr
Λ
ω(x3)
D+1
x2
1+x2
2 Zx2
1+x2
2
0
sB3(s, x3) ds!2
σ+1/2
+
for any σ3/2.
6 Spectral estimates for eigenvalues from
perturbed magnetic field
Now we change the topic and consider situations when the discrete spectrum
comes from the magnetic field alone. We are going to demonstrate a Berezin-
type estimate for the magnetic Laplacian on R2with the field which is a radial
and local perturbation of a homogeneous one. We consider the operator
H(B) in L2(R2) defined as follows,
H(B) = 2
x+ (i∂y+A2)2, A =0, B0xf(x, y),(6.1)
with fgiven by
f(x, y) = Z
x
g(pt2+y2) dt .
with g:R+R+; the operator H(B) is then associated with the magnetic
field
B=B(x, y) = B0g(px2+y2).
19
Since have chosen the vector potential in such a way that the unperturbed
part corresponds to the Landau gauge, we have
H(B0) = 2
x+ (i∂y+B0x)2.
Using a partial Fourier transformation, it is easy to conclude from here that
the corresponding spectrum consists of identically spaced eigenvalues of infi-
nite multiplicity, the Landau levels,
σ(H(B0)) = {(2n1)B0, n N}.(6.2)
It is well known that inf σess(H(B)B) = 0, hence the relative compactness
of B0Bwith respect to H(B)B0in L2(R2) implies
inf σess(H(B)) = B0.
We have to specify the sense in which the magnetic perturbation is local. In
the following we will suppose that
(i) the function gL(R+) is non-negative and such that both fand
x2fbelong to L(R2), and
lim
x2
1+x2
2→∞ |x2f(x1, x2)|+|f(x1, x2)|= 0 .
(ii) kgkB0.
Let us next rewrite the vector potentials A0and Aassociated to B0and B
in the polar coordinates. Passing to the circular gauge we obtain
A0= (0, a0(r)) , A = (0, a(r)) ,(6.3)
with
a0(r) = B0r
2, a(r) = B0r
21
rZr
0
g(s)sds . (6.4)
Hence the operators H(B0) and H(B) are associated with the closures of the
quadratic forms in L2(R+, rdr) with the values
Q(B0)[u] = Z2π
0Z
0|ru|2+|ir1θu+a0(r)u|2rdrdθ(6.5)
20
and
Q(B)[u] = Z2π
0Z
0|ru|2+|ir1θu+a(r)u|2rdrdθ , (6.6)
respectively, both defined on C
0(R+). Furthermore, for every kN0we
introduce the following auxiliary potential,
Vk(r) := 2k
r(a0(r)a(r)) + a2(r)a2
0(r),(6.7)
and the functions
ψk(r) = sB0
Γ(k+ 1) B0
2k/2
rkexp B0r2
4.(6.8)
Finally let us denote by
α=Z
0
g(r)rdr(6.9)
the flux associated with the perturbation; recall that in the rational units we
employ the flux quantum value is 2π. Now we are ready to state the result.
Theorem 6.1. Let the assumptions (i) and (ii) be satisfied, and suppose
moreover that α1. Put
Λk=ψk,Vk(·)ψkL2(R+,rdr).(6.10)
Then the inequality
tr(H(B)B0)γ
2γ
X
k=0
Λγ
k, γ 0,(6.11)
holds true whenever the right-hand side is finite.
Remark 6.2. For a detailed discussion of the asymptotic distribution of
eigenvalue of the operator H(B) we refer to [RT08].
Proof. We are going to employ the fact that both A0and Aare radial func-
tions, see (6.3), and note that by the partial-wave decomposition
tr (H(B)B0)γ
=X
kZ
tr (hk(B)B0)γ
,(6.12)
21
where the operators hk(B) in L2(R+, rdr) are associated with the closures of
the quadratic forms
Qk[u] = Z
0 |ru|2+
k
rua(r)u
2!rdr ,
defined originally on C
0(R+), and acting on their domain as
hk(B) = 2
r1
rr+k
ra(r)2
.
In view of (6.7) it follows that
hk(B) = hk(B0) + Vk(r),
where
hk(B0) = 2
r1
rr+k
ra0(r)2
.
To proceed we need to recall some spectral properties of the two-dimensional
harmonic oscillator,
Hosc =∆ + B2
0
4(x2+y2) in L2(R2).
It is well known that the spectrum of Hosc consists of identically spaced
eigenvalues of a finite multiplicity,
σHosc={nB0, n N},(6.13)
where the first eigenvalue B0is simple and has a radially symmetric eigen-
function. The latter corresponds to the term with k= 0 in the partial-wave
decomposition of Hosc, which implies
σHosc=[
kZ
σ2
r1
rr+k2
r2+B2
0r2
4,
where the operators in the brackets at the right-hand side act in L2(R+, rdr).
Hence in view of (6.13) we have
inf
k6=0 σ2
r1
rr+k2
r2+B2
0r2
42B0.(6.14)
22
On the other hand, for k < 0 it follows from (ii), (6.7) and (6.9) that
Vk(r) = 2k
rZr
0
g(s)sdsB0Zr
0
g(s)sds+1
r2Zr
0
g(s)sds2
kB0B0.
By (6.14) we thus obtain the following inequality which holds in the sense of
quadratic forms on C
0(R+) for any k < 0,
hk(B) = hk(B0) + Vk(r) = 2
r1
rr+k2
r2+B2
0r2
4kB0+Vk(r)
≥ −2
r1
rr+k2
r2+B2
0r2
4α B0
(2 α)B0.
Since α1 holds by hypothesis, this implies that
tr (H(B)B0)γ
=X
kZ
tr (hk(B)B0)γ
=X
k0
tr (hk(B)B0)γ
,(6.15)
see (6.12). In order to estimate tr (hk(B)B0)γ
for k0 we employ
Πk= (·, ψk)L2(R+,rdr)ψk,
the projection onto the subspace spanned by ψk, and note that
ψkker(hk(B0)B0),kψkkL2(R+,rdr)= 1 kN∪ {0}.(6.16)
Let Qk= 1 Πk. From the positivity of Vk(·)it follows that for any
uC
0(R+) it holds
u, ΠkVk(·)Qk+QkVk(·)Πku
u, ΠkVk(·)Πku+u, QkVk(·)Qku,(6.17)
where the scalar products are taken in L2(R+, rdr). From (6.17) we infer
that
hk(B)B0= (Πk+Qk) (hk(B0)B0+Vk(·)) (Πk+Qk)
k+Qk)hk(B0)B0Vk(·)k+Qk)
Πkhk(B0)B02Vk(·)Πk
+Qkhk(B0)B02Vk(·)Qk.(6.18)
23
The operator hk(B0) has for each kN0discrete spectrum which consists
of simple eigenvalues. Moreover, from the partial-wave decomposition of the
operator H(B0) we obtain
σ(H(B0)) = {(2n1)B0, n N}=[
kZ
σ(hk(B0)) ,
see (6.2). It means that
kZ:σ(hk(B0)) ⊂ {(2n1)B0, n N},
and since ψkis an eigenfunction of hk(B0) associated to the simple eigenvalue
B0, see (6.16), it follows that
Qk(hk(B0)B0)Qk2B0Qk,kN∪ {0}.(6.19)
On the other hand, by (6.7) and (6.9) we infer
sup
r>0Vk(r)α B0kN∪ {0}.
The last two estimates thus imply that
Qkhk(B0)B02Vk(·)QkQk(2 B0(1 α)) Qk0,
where we have used the assumption α1. With the help of (6.18) and the
variational principle we then conclude that
tr (hk(B)B0)γ
tr Πkhk(B0)B02Vk(·)Πkγ
= tr 2 ΠkVk(·)Πkγ
= 2γtr ΠkVk(·)Πkγ
= 2γψk,Vk(·)ψkγ
L2(R+,rdr)= 2γΛγ
k,
see (6.10). To complete the proof it now remains to apply equation (6.15).
7 Three dimensions: a magnetic ‘hole’
Let us return to the three-dimensional situation and consider a magnetic
Hamiltonian H(B) in L2(R3) associated to the magnetic field B:R3R3re-
garded as a perturbation of a homogeneous magnetic field of intensity B0>0
pointing in the x3-direction,
B(x1, x2, x3) = (0,0, B0)b(x1, x2, x3),(7.1)
24
with the perturbation bof the form
b(x1, x2, x3) = ω0(x3)f(x1, x2),0, ω(x3)gqx2
1+x2
2.
Here ω:RR+,g:R+R+and
f(x1, x2) = Z
x1
gqt2+x2
2dt . (7.2)
The resulting field Bthus has the component in the x3-direction given the
B0plus a perturbation which is a radial field in the x1, x2plane with a
x3dependent amplitude ω(x3). The first component of Bthen ensures that
∇ · B= 0, which is required by the Maxwell equations which include no
magnetic monopoles; it vanishes if the field is x3-independent.
A vector potential generating this field can be chosen in the form
A(x1, x2, x3) = (0, B0x1ω(x3)f(x1, x2),0) ,
which reduces to Landau gauge in the unperturbed case, and consequently,
the operator H(B) acts on its domain as
H(B) = 2
x1+ (i∂x2+B0x1ω(x3)f(x1, x2))22
x3.(7.3)
We have again to specify the local character of the perturbation: we will
suppose that
(i) the function gL(R+) is non-negative, such that fand x2fbelong
to L(R2), and
lim
x2
1+x2
2→∞ |x2f(x1, x2)|+|f(x1, x2)|= 0 ,
(ii) ω0, ωL2(R)L(R), and
kωkkgkB0,lim
|x3|→∞ ω(x3) = 0 .
Lemma 7.1. The assumptions (i) and (ii) imply σess (H(B)) = [B0,).
25
Proof. We will show that the essential spectrum of H(B) coincides with the
essential spectrum of the operator
H(B0) = 2
x1+ (i∂x2+B0x1)22
x3,
which is easy to be found, we have σ(H(B0)) = σess (H(B0)) = [B0,). Let
T=H(B)− H(B0) = 2ωf (i∂x2+B0x1)iω ∂x2f+ω2f2.
From assumption (i) in combination with [Da, Thm. 5.7.1] it follows that the
operator (ω ∂x2f+ω2f2)(∆ + 1)1is compact on L2(R3). The diamagnetic
inequality and [Pi79] thus imply that the sum iω ∂x2f+ω2f2is relatively
compact with respect to H(B0).
As for the first term of the perturbation T, we note that since (i∂x2+B0x1)
commutes with H(B0), it holds
ωf (i∂x2+B0x1) (H(B0) + 1)1
=ωf (H(B0) + 1)1/2(i∂x2+B0x1) (H(B0) + 1)1/2.(7.4)
In the same way as above, with the help of [Da, Thm.5.7.1], diamagnetic
inequality, and [Pi79], we conclude that ω f (H(B0) + 1)1/2is compact on
L2(R3). On the other hand, (i∂x2+B0x1) (H(B0) + 1)1/2is bounded on
L2(R3). As their product the operator (7.4) is compact; by Weyl’s theorem
we then have σess (H(B)) = σess(H(B0)) = [B0,).
7.1 Lieb-Thirring-type inequalities for H(B)
Now we are going to formulate Lieb-Thirring-type inequalities for the nega-
tive eigenvalues of H(B)B0in three different cases corresponding to differ-
ent types of decay conditions on the function g. Let us start from a general
result. We denote by
α(x3) = ω(x3)Z
0
g(r)rdr
the magnetic flux (up to the sign) through the plane {(x1, x2, x3):(x1, x2)
R2}associated with the perturbation. From Theorem 6.1 and inequality
(2.2) we make the following conclusion.
26
Theorem 7.2. Let assumptions (i) and (ii) be satisfied. Suppose, moreover,
that supx3α(x3)1and put
Λk(x3) = ψk,Vk(·;x3)ψkL2(R+,rdr).(7.5)
Then the inequality
tr (H(B)B0)σ
Lcl
σ,12σ+1
2ZR
X
k=0
Λk(x3)σ+1
2dx3, σ 3
2,(7.6)
holds true whenever the right-hand side is finite.
7.1.1 Perturbations with a power-like decay
Now we come to the three cases mentioned above, stating first the results
and then presenting the proofs. We start from magnetic fields (7.1) with the
perturbation gwhich decays in a powerlike way. Specifically, we shall assume
that
0g(r)B0(1 + pB0r)2β, β > 1.(7.7)
We have included the factor B0on the right hand side of (7.7) having
in mind that B1/2
0is the Landau magnetic length which defines a natural
length unit in our model.
For any β > 1 and γ > max n1
β1,2owe define the number
K(β, γ ) = 2γ+
X
k=1 Γ ((k+ 1 β)+)
Γ(k)+1
22πk γ
,(7.8)
and recall also the classical Lieb-Thirring constants in one dimension,
Lcl
1=Γ(σ+ 1)
2πΓ(σ+ 3/2) , σ > 0.(7.9)
Theorem 7.3. Assume that gsatisfies (7.7) and that kωk2(β1).
Then
tr (H(B)B0)σ
Lcl
1Kβ, σ +1
22B0
β1σ+1
2ZR
ω(x3)σ+1
2dx3
holds true for all
σ > max 3
2,3β
2β2.(7.10)
27
Remark 7.4. Since ωL(R)L2(R), it follows that ωLσ+1
2(R) for
any σ3/2. Note also that by the Stirling formula we have
Γ (k+ 1 β)
Γ(k)k1βas k→ ∞.
Hence the constant Kβ, σ +1
2is finite for any σsatisfying (7.10).
7.1.2 Gaussian decay
Next we assume that the perturbation ghas a Gaussian decay, in other words
0g(r)B0eεB0r2, ε > 0.(7.11)
Theorem 7.5. Assume that gsatisfies (7.11) and that kωk2ε. Then
for any σ > 3/2it holds
tr (H(B)B0)σ
Lcl
σ,1B0
εσ+1
2
G(ε, σ)ZR
ω(x3)σ+1
2dx3,
where
G(ε, σ) = 1 +
X
k=1 (1 + 2ε)k+1
22πk σ+1
2
.(7.12)
7.1.3 Perturbations with a compact support
Let Dbe a circle of radius Rcentered at the origin and put
g(r) = B0rR
0r > R .(7.13)
Theorem 7.6. Assume that gsatisfies (7.13) with Rsuch that B0R22.
Suppose moreover that kωk1. Then for any σ > 3/2it holds
tr (H(B)B0)σ
Lcl
σ,1JB0, σBσ+1
2
0ZR
ω(x3)σ+1
2dx3,(7.14)
where
J(B0, σ) = B0R2σ+1
2
1 +
X
k=1 B0R2
2k+1 1
k!+1
22πk !σ+1
2
.
(7.15)
28
7.2 The proofs
Note that the assumptions of these theorems ensure that supx3α(x3)1,
hence in all the three cases we may apply Theorem 6.1 and, in particular,
the estimate (7.6). To this note it is useful to realize that by (6.4), (6.7) and
(6.9) we have
Vk(r;x3) = α(x3)B0+2α(x3)k
r22k ω(x3)
r2Z
r
g(s)sds
+B0ω(x3)Z
r
g(s)sds+ω2(x3)
r2Zr
0
g(s)sds2
.(7.16)
Consequently, we obtain a simple upper bound on the negative part of Vk,
Vk(r;x3)2k ω(x3)
r2Z
r
g(s)sds+α(x3)B02k
r2+
(7.17)
for all kN∪ {0}. For k= 0 we clearly we have
Λ0(x3)α(x3)B0,(7.18)
by (6.16). In order to estimate Λk(x3) with k1 we denote by λk(x3) the
contribution to Λk(x3) coming from the first term on the right-hand side of
(7.17), i.e.
λk(x3) = 2 ω(x3)kZ
0
ψ2
k(r)Z
r
g(s)sdsr1dr . (7.19)
Before coming to the proofs we need an auxiliary result.
Lemma 7.7. For any kNit holds
Λk(x3)λk(x3) + α(x3)B0
2πk .
Proof. In view of (7.5), (7.17), and (7.19) the claim will follow if we show
that Z
0
ψ2
k(r)B02k
r2+
rdrB0
2πk .(7.20)
29
Let rk=q2k
B0. Using (6.8) and the substitution s=B0r2
2we then find
Z
0
ψ2
k(r)B02k
r2+
rdr=B0Z
rk
ψ2
k(r)rdr2kZ
rk
ψ2
k(r)r1dr
=B0
Γ(k+ 1) Z
k
esskdsB0
Γ(k)Z
k
essk1ds .
Integration by parts gives
Z
k
esskds=ekkk+kZ
k
essk1ds ,
hence Z
0
ψ2
k(r)B02k
r2+
rdr=ekkkB0
Γ(k+ 1) ,
and inequality (7.20) follows from the Stirling-type estimate [AS64, Eq. 6.1.38]
Γ(k+ 1) = k!2π k k+1
2ek, k N;
this concludes the proof.
Proof of Theorem 7.3. In view of (7.18) and Lemma 7.7 it suffices to
estimate λk(x3) in a suitable way from above for k1. Using (7.7) we find
Z
0
g(r)rdrB0Z
0
(1 + pB0r)2βrdrB0Z
0
(1 + pB0r)12βdr
=Z
0
(1 + s)12βds=1
2(β1) ,
which implies
α(x3)ω(x3)
2(β1) .(7.21)
Moreover, by virtue of (7.7)
Z
r
g(s)sdspB0Z
r
(1 + pB0s)12βds=1
2β2(1 + pB0r)22β.
30
Assume first that 1 kβ1. In this case a combination of (6.8) and the
last equation gives
λk(x3)ω(x3)B0
(β1) Γ(k)B0
2kZ
0
eB0r2
2r2k1(1 + pB0r)22βdr
=ω(x3)B0
(β1) Γ(k)Z
0
essk1(1 + 2s)22βds
ω(x3)B0
(β1) Γ(k)Z
0
esds=ω(x3)B0
(β1) Γ(k),(7.22)
where we have used again the substitution s=B0r2
2.
On the other hand, for k > β 1 we have
λk(x3)ω(x3)B0
(β1) Γ(k)B0
2kZ
0
eB0r2
2r2k1(1 + pB0r)22βdr
ω(x3)B0
(β1) Γ(k)B0
2kZ
0
eB0r2
2r2k1(B0r2)1βdr
ω(x3)B0
(β1) Γ(k)Z
0
esskβds=ω(x3)B0Γ(k+ 1 β)
(β1) Γ(k).
This together with equations (7.21), (7.18), (7.22) and Lemma 7.7 shows that
X
k=0
Λγ
k(x3)K(β, γ )B0
β1γ
ω(x3)γ,
with the constant K(β, γ) given by (7.8). The claim now follows from (7.6)
upon setting γ=σ+1
2.
Proof of Theorem 7.5. We proceed as in the proof of Theorem 7.3 and
use equation (7.18) and Lemma 7.7. Since
α(x3)ω(x3)B0Z
0
B0eεB0r2rdr=ω(x3)
2ε(7.23)
holds in view of (7.11), for k= 0 we get
Λ0(x3)α(x3)B0ω(x3)B0
2ε.
31
On the other hand,
Z
r
g(s)sdsB0Z
r
eεB0s2sds=1
2εeεB0r2.
Hence using the substitution s=B0r2
2(1 + 2ε), we obtain
λk(z)ω(x3)B0
εΓ(k)B0
2kZ
0
eB0r2
2(1+2ε)r2k1dr
=ω(x3)B0
2ε
(1 + 2ε)k
Γ(k)Z
0
essk1ds=ω(x3)B0
2ε(1 + 2ε)k
for any k1. Summing up gives
X
k=0
Λγ
k(x3)ω(x3)B0
2εγ 1 +
X
k=1 (1 + 2ε)k+1
22πk γ!.
Theorem 6.1 applied with γ=σ+1
2then completes the proof.
Proof of Theorem 7.6. In this case we have
α(x3) = ω(x3)B0R2
2.
Inequality (7.18) thus implies
Λ0(z)ω(z)B2
0R2
2.
For k1 we note that in view of (7.13)
Z
r
g(s)sds=
1
2(R2r2)rR
0r > R
Hence from (6.8) and (7.19) we conclude that
λk(z)B2
0R2ω(x3)
Γ(k)B0
2kZR
0
eB0r2
2r2k1dr
B2
0R2ω(x3)
2Γ(k)ZB0R2
2
0
essk1ds
B2
0R2ω(z)
2kΓ(k)B0R2
2k
=B0ω(x3)
Γ(k+ 1) B0R2
2k+1
, k N.
32
This in combination with the above estimate on Λ0(x3) and Lemma 7.7 im-
plies
X
k=0
Λγ
k(x3)ω(x3)γBγ
0B0R2
2γ 1 +
X
k=1 B0R2
2k1
k!+1
2πk !γ!,
and the claim follows again by applying Theorem 6.1 with γ=σ+1
2.
Acknowledgements
The research was supported by the Czech Science Foundation (GAˇ
CR) within
the project 14-06818S. D.B. acknowledges the support of the University of Os-
trava and the project “Support of Research in the Moravian-Silesian Region
2013”. H. K. was supported by the Gruppo Nazionale per Analisi Matemat-
ica, la Probabilit`a e le loro Applicazioni (GNAMPA) of the Istituto Nazionale
di Alta Matematica (INdAM). The support of MIUR-PRIN2010-11 grant for
the project “Calcolo delle variazioni” (H. K.) is also gratefully acknowledged.
T.W. was in part supported by the DFG project WE 1964/4-1 and the DFG
GRK 1838.
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