Sturm-Liouville operators in the half axis with local perturbations

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www.elsevier.com/locate/jmaa

Sturm–Liouville operators

in the half axis with local perturbations

Rafael del Rio

a,∗,1

_{, Olga Tchebotareva}

b

a_{IIMAS–UNAM, Circuito Escolar, Ciudad Universitaria, 04510 México, D.F., Mexico} b_{Facultad de Ciencias–UNAM, Circuito Exterior, Ciudad Universitaria, 04510 México, D.F., Mexico}

Received 17 February 2006 Available online 4 August 2006

Submitted by J.A. Ball

Dedicated to Barry Simon on his 60th birthday

Abstract

We give conditions which imply equivalence of the Lebesgue measure with respect to a measureμ

generated as an average of spectral measures corresponding to Sturm–Liouville operators in the half axis. We apply this to prove that some spectral properties of these operators hold for large sets of boundary conditions if and only if they hold for large sets of positive local perturbations.

Keywords:Sturm–Liouville operators; Mixed spectra; Local perturbations

1. Introduction

A key step in important results concerning localization phenomena has been to establish ab-solute continuity of measuresμ(·):=ρλ(·) dλgenerated as averages of spectral measuresρλ

which correspond to self-adjoint operators.

Let us for example consider a family of self-adjoint operatorsLθ(q)inL2(0,∞)generated

by the differential expression

lu= −u+q(x)u

* _{Corresponding author.}

E-mail addresses:[email protected] (R. del Rio), [email protected] (O. Tchebotareva).

1 _{Partially supported by Project PAPIIT IN 111906-3.}

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and the boundary condition

u(0)cosθ−u(0)sinθ=0, θ∈ [0, π ),

and letρθ be the spectral measure corresponding toLθ. The averageμ(·):= π

0 ρθ(·) dθis not

only absolutely continuous with respect to Lebesgue measure, but is equal to it! (See e.g. [9, Theorem 1].)

Using the absolute continuity ofμthe following can be proven (see e.g. [4, Corollary 3.2]).

Theorem 1.If for almost everyE∈I,I⊂R, there exists a nontrivialL2solution oflu=Eu, then

σc(Lθ)∩I= ∅

for almost everyθ∈ [0, π ), whereσcdenotes the continuous spectrum.

For a survey of other localization results related to the absolute continuity of averages asμ

see [12].

In this paper we give conditions which imply that Lebesgue measure and a properly chosen average measureμhave the same zero sets and apply this to understand some aspects of the spec-tral theory of Sturm–Liouville operators with local perturbations. In particular, if some specspec-tral property holds for a set of positive Lebesgue measure in the boundary conditions we can show that this property holds for a set of positive measure of local perturbations leaving any boundary condition fixed. This will be a consequence of our main result Theorem 3 below. In Section 2 our main results are proven. The techniques used rely heavily on the Prüfer transform which is a very useful tool in this paper. In Section 3 we show consequences of the previous results, some applications toα-singular andα-continuous spectrum are given and an open question is presented. Some of the methods we used in this paper have their origin in [7].

2. Main results

For each(λ, θ )∈R× [0, π )let us consider the self-adjoint operatorLλθ inL2(0,∞)

gener-ated by the differential expression

lλu= − d2

dx2+V (x)+λW (x) (1)

as follows:Lλ,θu:=lλuforuin the domain ofLλθ given by

D(Lλθ)=

f∈L2(0,∞): f, flocally absolutely continuous in(0,∞), lλf ∈L2(0,∞), f (0)cosθ−f(0)sinθ=0, θ∈ [0, π )

.

We assumeLλθis regular at 0 and the limit point case holds at∞. See for these concepts [2]. The functionsV andW are real-valued and locally inL1. We assumeW (x) >0 for a.e.x∈ [0, c]

andW (x)=0 for everyx /∈ [0, c].

Denote byρθ_λthe spectral function of the operatorLλθ.

Using the same differential expression (1) we define a regular problem in the finite interval

[0, c]setting forθ∈ [0, π )the boundary conditions

u(0)cosθ−u(0)sinθ=0,

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Definition 1.FixE∈Rand let us callλ(E)an eigenvalue if there is a nontrivial solutionuof

lλu=Euwhich satisfies the conditions (2).

The following result is due to C. Sturm and J. Liouville [13]. For more recent references see for example [2, Theorem 2.1, Chapter 8, p. 212], [5, Chapter XI, p. 337], [14, Theorem 13.2], [1, Theorems 8.4.5 and 8.4.6].

Theorem 2. There exists a monotonous decreasing sequence of eigenvalues λ(0) > λ(1) > λ(2)>· · ·such thatλ(n)→ −∞ifn→ ∞.

If μandν are two measures we use the notation μν when μ is absolutely continuous with respect toν, that is ifν(A)=0 impliesμ(A)=0. Ifμνandνμwe say that the two measures are equivalent. We use| · |to denote the Lebesgue measure.

LetAbe a Borel setA⊂(E1, E2)and define a measureμθ as follows:

μθ(A):= λ2

λ1

ρ_λθ(A) dλ.

The next result gives conditions which imply equivalence betweenμθ and| · |.

Theorem 3.

(I) If λ1λ(n+1)(E1) < λ(n)(E2)λ2 whereλ(n)are eigenvalues of the regular problem in [0, c]mentioned in previous theorem then| · |μθ.

(II) Ifλ1λ2thenμθ| · |.

Before we prove the theorem let us introduce notation and recall some results. Consider real solutionsu≡0 oflλu=Eusuch that for fixeda∈R,

u(a)=sinθ,

u(a)=cosθ.

If we write the vector(u(x), u(x))in polar coordinates we obtain

u(x)=ra(x)sinφa(x),

u(x)=ra(x)cosφa(x),

whereφa(x, θ, E, λ)andra(x, θ, E, λ)are called the Prüfer phase and the Prüfer amplitude ofu.

We fix a unique value ofφaby requiringφa(a, θ, E, λ)=θ and continuity inx. These

func-tionsraandφaare jointly continuous inx, Eandλ(use arguments similar to [14, Theorem 2.1],

[2, Theorem, Chapter 2.4], [5, Theorem, Chapter V.3]). This will be very important in what fol-lows.

We shall need the next results

(a) ρθ_λ(E1, E2)= lim

b→∞

1

π E2

E1

r0(b, θ, E, λ)−2dE

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(b) (i) (∂λφa)(x, λ)= −

1

ra(x, λ)2 x

a

W (t )ra(t, λ)2sin2φa(t, λ) dt ,

(ii) (∂Eφa)(x, E)=

1

ra(x, λ)2 x

a

ra(t, E)2sin2φa(t, E) dt.

See [11, Appendix B]. (c) For anya, x, θ, E∈R,

1

π θ+π

θ

ra(x, β, E)−2dβ=1.

See [10, Corollary 12] and [11, Appendix B].

(d) For the Prüfer phaseφaof a solution oflλu=Euwe have

φa(p, λ)→ ∞ asλ→ −∞

for any pointp∈suppW (see [2, formula (2.5) in the proof of Theorem 2.1, p. 212] or [14, Theorem 13.2, part (c)]).

Proof of Theorem 3. We use an argument similar to the one used in [11, Lemma 5.10] where (II) was proven for the case of the whole line. Given anyE1< E2,the measureρλθis continuous

inE1andE2for almost anyλ.

From (a) we know that

μθ

(E1, E2)

= λ2

λ1

ρ_λθ(E1, E2) dλ= λ2 λ1 lim b→∞ 1 π E2 E1

r0(b, θ, E, λ)−2dE

dλ. (3)

Letb > c >0 wherec=supSandS= {x: W (x)=0} =suppW. Let us estimate

λ2 λ1 1 π E2 E1

r0(b, θ, E, λ)−2dE

dλ.

Using Fubini we can interchange integrals and from the equality

r0(b, θ, E, λ)−2=r0(c, θ, E, λ)−2rc

b, φ0(c, θ, E, λ), E ₋2

and the fact thatr0is jointly continuous inEandλand strictly positive we get

1

π E∈[supE1,E2] λ∈[λ1,λ2]

r0(c, θ, E, λ)−2 E2 E1 dE λ2 λ1 rc

b, φ0(c, θ, E, λ), E ₋2 dλ λ2 λ1 1 π E2 E1

r0(b, θ, E, λ)−2dE

dλ= 1 π E2 E1 λ2 λ1

r0(b, θ, E, λ)−2dλ

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1

π E∈[infE1,E2] λ∈[λ1,λ2]

r0(c, θ, E, λ)−2 E2

E1 dE

λ2

λ1

rcb, φ0(c, θ, E, λ), E−2dλ

.

Defineβ(λ):=φ0(c, θ, E, λ).

From (b)(i) we know thatβ(λ) <0. Changing variables we obtain

λ2

λ1

rcb, φ0(λ), E−2dλ= β(λ 2)

β(λ1)

rc(b, β, E)−2 β(λ) dβ=

β(λ 1)

β(λ2)

1

|β(λ)|rc(b, β, E)

−2_dβ.

Using (b)(i) and the fact thatr0is jointly continuous inλandEwe obtain β(λ)=

c

0 W (t )r

2

0(t, θ, E, λ)sin

2_{φ0(t, θ, E, λ) dt} r₀2(c, θ, E, λ)

c

0 W (t )r

2

0(t, θ, E, λ) dt r₀2(c, θ, E, λ)

sup

t∈[0,c] E∈[E1,E2]

λ_∈[λ1,λ2] _r2

0(t, θ, E, λ) r₀2(c, θ, E, λ)

c

0

W (t ) dt

supt,E,λr02(t, θ, E, λ)

infE,λr₀2(c, θ, E, λ) c

0

W (t ) dt

therefore

1

|β(λ)|

infr₀2(c, θ, E, λ)

supr₀2(t, θ, E, λ)₀cW (t ) dt =:K.

On the other hand we have that

r₀2(t, θ, E, λ)

r₀2(c, θ, E, λ)C >0

fort∈ [0, c]locally uniformly inE, λand, since sinφ0(·, θ, E, λ)has only isolated zeros, c

0

W (t )sin2φ0(t, θ, E, λ) dt >0.

By continuity inEandλwe arrive at

inf

E∈[E1,E2] λ∈[λ1,λ2]

c

0

W (t )sin2φ0(t, θ, E, λ) dt >0

and 1

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Altogether we get

C β(λ 1)

β(λ2)

rc(b, β, E)−2dβ λ2

λ1 rc

b, φ0(λ), E ₋2

dλ= β(λ 1)

β(λ2)

rc(b, β, E)−2 |β(λ)| dβ

K β(λ 1)

β(λ2)

rc(b, β, E)−2dβ

hence ˜ C E2 E1 dE

β(λ 1)

β(λ2)

rc(b, β, E)−2dβ λ2 λ1 1 π E2 E1

r0(b, θ, E, λ) ₋2 dE dλ = 1 π E2 E1 λ2 λ1

r0(b, θ, E, λ)−2dλ dE c˜ E2 E1 dE

β(λ 1)

β(λ2)

,

where

˜ c= 1

π E_∈[infE1,E2] λ_∈[λ1,λ2]

r0(b, θ, E, λ)−2K and C˜= 1

π _E_∈[sup_E₁_,E₂_] λ∈[λ1,λ2]

r0(b, θ, E, λ)−2C.

In order to handle the limit that appears in (3) we would like to get estimates for

β(λ 1)

β(λ2)

which are independent ofb. To accomplish this we look for conditions onλ1,λ2which guarantee thatβ(λ1)−β(λ2)πand then apply result (c) mentioned above.

To prove (I) we need an estimate from below. From (b)(i), (ii) and (d) we know thatβ(λ)= φ0(c, θ, E, λ)is decreasing inλ, increasing inEandβ(λ)→ ∞ifλ→ −∞orE→ ∞.

More-over,λis an eigenvalue of the problem with boundary conditions (2) if and only ifβ(λ)=mπ. For fixedEchoose two consecutive eigenvaluesλ(n),λ(n+1)of this problem and take

λ2> λ(n)> λ(n+1)> λ1

then we will have β(λ1)−β(λ2)π. Now since β is increasing in E we can choose λ2λ(n)(E2) > λ(n+1)(E1) > λ1and we get

β(λ1, E)−β(λ2, E)π for everyE∈ [E1, E2],

then we can conclude

λ2 λ1 1 π E2 E1

r0(b, θ, E, λ)−2dE

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To get the estimate from above that we need to prove (II) it is not necessary to impose extra conditions onλ1,λ2. We split the interval[β(λ2), β(λ1)]into intervals of length at mostπand

apply (c),

β(λ 1)

β(λ2)

rc(b, β, E)−2dβπ

β(λ1)−β(λ2) π +1

.

Remember thatβ(λ)=φ0(c, θ, E, λ)is bounded locally uniformly inE, then

λ2

λ1

1

π E2

E1

r0(b, θ, E, λ) ₋2

dE

dλπC(Eˆ 2−E1),

where

ˆ

C=β(λ1)−β(λ2)+π _˜

C(E2−E1).

Since _π1E2

E1(r0(b, θ, E, λ))

−2_dE_{is bounded uniformly for every}_{b > c}_and_λ_{∈ [}_{λ1, λ2]}_(use

[8, Lemma 1.1, Chapter 2] and joint continuity) we can apply the Lebesgue theorem of dominated convergence to obtain

πC(Eˆ 2−E1)μθ

(E1, E2)

πc(E˜ 2−E1).

Using countable additivity the theorem follows for general Borel setA. 2

The following result is a consequence of Theorem 3.

Corollary 1.LetI:=(E1, E2)⊂Rbe open and defineLλθ as above. For anyλ∈R, the op-eratorLλθ has singular continuous spectrum inI for a set of positive Lebesgue measure ofθ’s if and only if for anyθ∈ [0, π ), Lλθ has singular continuous spectrum inI for a setB ofλ’s

of positive Lebesgue measure. Moreover, if λ(n)(E)are the eigenvalues of (1) with boundary conditions(2), then

B∩λ(n+1)(E1), λ(n)(E2)>0 for everyn0.

Proof. (⇒) LetSbe the set of pointsEfor which there are subordinate solutions oflλu=Eu

which are not inL2. It is known that this set is a support of the singular continuous part ofLλθ

and it does not depend onλandθ(see [6]). Sinceρ_λθ(S∩I ) >0 for a set of positive measure in

θby hypothesis using equality (see e.g. [9])

|S∩I| = π

0

ρ_λθ(S∩I ) dθ

we deduce|S∩I|>0. This implies using Theorem 3(I) thatμθ(S∩I ) >0 for anyθ. From here

we knowρ_λθ(S∩I ) >0 forλ∈Band

B∩λ(n+1)(E1), λ(n)(E2)>0.

(⇐) AssumeLλθ has singular continuous spectrum inIfor a setBofλ’s of positive Lebesgue measure. Thenρ_λθ(S∩I ) >0 forλ∈Band

B

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Therefore there exists an intervalJsuch that

J

ρθ_λ(S∩I ) dλ

B∩J

ρ_λθ(S∩I ) dλ >0.

Using Theorem 3(II) we obtain|S∩I|>0 and therefore

π

0

ρθ_λ(S∩I ) dθ= |S∩I|>0

for every fixed λ. Therefore Lλθ has singular continuous spectrum in I for a set of positive

Lebesgue measure inθ. 2

If instead of taking the supportS as above we take the setP of subordinate solutions which are inL2we get the same result for the pure point part and takingP ∪Swe obtain the result for

the singular part ofLλθ.

3. Consequences and remarks

For the pure point part of the spectrum the following result can be proven. DefineLλθ(U )as it was done in Section 2 but using instead the differential expression

lλ(U )= − d2

dx2 +V (x)+U (x)+λW (x),

where we assumeV is bounded from below(Lλθ =Lλθ(U )ifU≡0). Using Corollary 1.8 of

[6] the following version of Corollary 1 above can be proven:

Proposition 1.For anyλ∈Rthe operatorLλθ has point spectrum inI for a set of positive

Lebesgue measure inθ’s if and only if for anyθthe operatorLλθ(U )has point spectrum in I

for a set ofλ’s of positive Lebesgue measure. We assume the functionU (x)is locally inL1and U (x)eA|x|dx <∞

for allA >0.

We recall the following related result which was proven in [4].

Corollary 2.LetI ⊂Rbe open. IfLθ(q)has singular continuous spectrum inI for a set of

positive measure ofθ’s, the same is true forq+vwherevis a perturbation of compact support

(vdoes not need to be positive).

The next proposition states that, in a way, the roles ofθandvin Corollary 2 can be exchanged. The proof follows directly from Corollary 1.

Proposition 2.Letθ0, θ1∈ [0, π ). The operatorLλθ0 has singular continuous spectrum inI for

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Similar results can be proven for theα-continuous andα-singular spectrum. Recall that for

α∈(0,1)theα-dimensional Hausdorff measure is defined for Borel setsAby

hα(A)≡lim

δ→0δ-coversinf

∞

v₌1 |bv|α,

where aδ-cover is a countable collection of intervals each of length at mostδsoA⊂∞_v₌₁bv. Givenα∈ [0,1]we define a measureμto beα-continuous(αc)ifμ(S)=0 for any setSwith

hα(S)=0 andα-singular(αs)if it is supported on a setSwithhα(S)=0. For every suchαand any measureμ, one can uniquely decomposeμ=μαc₊_μαs _with_μαc_α_{-continuous and}_μαs_, α-singular.

Denoteρ:=ρ_λθ. It is possible to find setsAα andBαsuch that

dραc=dρ(Aα∩.),

dρsc=dρ(Bα∩.)

and it happens thatAα andBαare independent ofθandλ. See [6] and references therein.

Using the same reasoning as above one can prove Proposition 2 and Corollary 1 forα-singular andα-continuous instead of singular continuous spectrum.

Remark.DefineB to be the set ofλ’s such thatLλθ has singular continuous (point, singular, α-continuous,α-singular) spectrum in an intervalI=(E1, E2). From the results above, it

fol-lows in particular that it is not possible to have|B|>0 andB⊂J whereJ is a finite interval, since|B∩ [λ(n+1)(E1), λ(n)(E2)]|>0 for everyn0 andλ(n)→ −∞ifn→ ∞. It remains to

answer this forθ. We believe it is possible to have a setT ofθ’s such that|T|>0 andT ⊂J

whereJ is an interval properly contained in[0, π ). Our belief is based on the following result [3, Theorem 1.2] with regard to abstract rank one perturbations. LetAbe a self-adjoint operator with a cyclic vectorϕ. Denote byAλthe perturbations ofA:

Aλ=A+λ(·, ϕ)ϕ, λ∈R.

Theorem 4. For any measurable set B⊂R there exists a family of rank one perturbations

{Aλ}λ∈R such thatAλ has dense absolutely continuous and dense singular spectrum for almost everyλ∈Band dense absolutely continuous(but no singular)spectrum for almost everyλ /∈B.

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