R E S E A R C H
Open Access
The regularized trace formula for a fourth
order differential operator given in a finite
interval
Serpil Karayel
*and Yonca Sezer
*Correspondence:
Faculty of Arts and Science, Yildiz Technical University, Davutpas.a, ˙Istanbul, 34220, Turkey
Abstract
In this work, a regularized trace formula for a differential operator of fourth order with bounded operator coefficient is found.
MSC: 47A10; 34L20
Keywords: separable-Hilbert space; self-adjoint operator; kernel operator; spectrum; essential spectrum; resolvent
1 Introduction
Investigations into the regularized trace formulas of scalar differential operators started with the work [] firstly. After that work, regularized trace formulas for several differential operators have been studied in some works as [, ] and []. In [] a formula for the second regularized trace of the problem generated by a Sturm-Liouville operator equation with a spectral parameter dependent boundary condition is found. The list of the works on this subject is given in [] and []. The trace formulas for differential operators with operator coefficient are investigated in the works [–] and []. The boundary conditions in our work are completely different from those in [].
In this work, we find the following regularized trace formula for a self-adjoint differential operatorLof fourth order with bounded operator coefficient:
∞
m=
∞
n=
λmn–
m+
–
π
π
trQ(x)dx
=
trQ(π) –trQ() .
Here{λmn}∞n=are the eigenvalues of the operatorLwhich belong to the interval [(m+) –
Q, (m+)+Q].
1.1 Notation and preliminaries
LetHbe a separable Hilbert space with infinite dimension. Let us consider the operators
LandLin the Hilbert spaceH=L(,π;H) which are formed by the following
differen-tial expressions:
(y) =yıv(x),
(y) =yıv(x) +Q(x)y(x)
with the same boundary conditionsy() =y() =y(π) =y(π) = . Suppose that the op-erator functionQ(x) in the expression(y) satisfies the following conditions:
(Q) Q(x) :H→His a self-adjoint kernel operator for everyx∈[,π]. Moreover,Q(x)
has second order weak derivative in this interval andQ(i)(x) :H→H(i= , ) are
self-adjoint kernel operators for everyx∈[,π]. (Q) Q<.
(Q) There is an orthonormal basis{ϕn}∞n=of the spaceHsuch that
∞
n=
Q(x)ϕn<∞.
(Q) The functionsQ(i)(x)σ
(H)are bounded and measurable in the interval[,π]
(i= , , ).
Hereσ(H) is the space of kernel operators fromHtoHas in []. Moreover, we denote
the norms by·Hand·and inner products by (·,·)Hand (·,·) inHandH, respectively,
and we also denote the sum of eigenvalues of a kernel operatorAbytrA=traceA. Spectrum of the operatorLis the set
σ(L) =
,
, . . . ,
m+
, . . .
.
Every point of this set is an eigenvalue ofLwhich has infinite multiplicity. The
orthonor-mal eigenfunctions corresponding to eigenvalue (m+)are in the form
ψmn(x) =
πsin
m+
x·ϕn (n= , , . . .). (.)
2 Some relations between spectrums of operatorsL0andL
LetRλ,Rλbe resolvents of the operatorsLandL, respectively. If the operatorQ:H→H
satisfies conditions (Q) and (Q), the following can be proved:
(a) QRλ∈σ(H)for everyλ∈/σ(L).
(b) Spectrum of the operatorLis a subset of the union of pairwise disjoint intervals
Fm=
m+
–Q,
m+
+Q
(m= , , , . . .),σ(L)⊂
∞
m=
Fm.
(c) Each point of the spectrum ofL, different from(m+)inF
mis an isolated
eigenvalue which has finite multiplicity.
(d) The series∞n=[λmn– (m+)](m= , , , . . .) are absolutely convergent where {λmn}∞n=are eigenvalues of the operatorLin the intervalFm.
Letρ(L) be resolvent set of the operatorL;ρ(L) = C\σ(L). SinceQR
λ∈σ(H) for every
λ∈ρ(L), from the equationRλ=Rλ–RλQRλwe obtainRλ–Rλ∈σ(H).
On the other hand, if we consider the series
∞
n=
λmn–
m+
are absolutely convergent then we have
trRλ–Rλ
=
∞
m=
∞
n=
λmn–λ
–
(m+ )–λ
for everyλ∈ρ(L) []. If we multiply both sides of this equality withλπiand integrate this equality over the circle|λ|=bp= (p+ )(p∈N,p≥), then we find
πi
|λ|=bp
λtrRλ–Rλ
dλ
=
πi
|λ|=bp λ
p
m=
∞
n=
λmn–λ
–
(m+)–λ
dλ
+
πi
|λ|=bp λ
∞
m=p+
∞
n=
λmn–λ
–
(m+)–λ
dλ. (.)
If we consider the relations|λmn|<bp(m= , , , . . . ,p) and|λmn|>bp(m=p+ ,p+ , . . .) forn= , , , . . . , then from (.) we get
πi
|λ|=bp
λtrRλ–Rλ
dλ
= p
m=
∞
n=
πi
|λ|=bp
λdλ
λ– (m+)–
πi
|λ|=bp λdλ λ–λmn
+
∞
m=p+
∞
n=
πi
|λ|=bp
λdλ λ– (m+)–
πi
|λ|=bp λdλ λ–λmn
= p
m=
∞
n=
m+
–λmn
. (.)
Moreover, from the formulaRλ=Rλ–RλQRλ, we obtain the following equality:
Rλ–Rλ= –RλQRλ+Rλ
QRλ
–Rλ
QRλ
. (.)
If we put this equality into (.), we have
p
m=
∞
n=
λmn–
m+
=Mp+Mp+Mp. (.)
Here
Mpj= (–)j
πi
|λ|=bp λtrRλ
QRλ j
dλ (j= , ), (.)
Mp= – πi
|λ|=bp λtrRλ
QRλ
Theorem . If the operator function Q(x)satisfies condition(Q),then we have
Mpj= (–)j πij
|λ|=bp trQRλ
j
dλ.
Proof It can be shown that the operator functionQR
λis analytic with respect to the norm in the spaceσ(H) in domainρ(L) = C\σ(L) and
trQRλ j
=jtrQRλ
QRλ j–
. (.)
Considering (QR
λ)= (QRλ), we can write the formula (.) as
trQRλ j
=jtrRλ
QRλ j
. (.)
From (.) and (.), we obtain
Mpj= (–)j+
πij
|λ|=bp
λtrQRλ j
dλ.
From here, we find
Mpj= (–)j+
πij
|λ|=bp
trλQRλj –QRλjdλ
=(–) j
πij
|λ|=bp trQRλ
j
dλ+(–) j+
πij
|λ|=bp
trλQRλ j
dλ. (.)
It can be easily shown that
trλQRλ j
=trλQRλ j
.
Therefore, we have
|λ|=bp
trλQRλ j
dλ=
|λ|=bp
trλQRλ j
dλ. (.)
The integral on the right-hand side of the last equality can be written as
|λ|=bp
trλQRλj dλ=
|λ|=bp
Imλ≥
trλQRλj dλ+
|λ|=bp
Imλ≤
trλQRλj dλ. (.)
Letεbe a constant satisfying the condition <ε<bp– (p+). Consider the function
tr[λ(QRλ)j] is analytic in simple connected domains
G={λ∈C:bp–ε<λ<bp+ε,Imλ> –ε},
and
λ∈C:|λ|=bp,Imλ≥
⊂G,
λ∈C:|λ|=bp,Imλ≤
⊂G.
From (.) we obtain
|λ|=bp
trλQRλj dλ=tr–bp
QR–bpj –trbp(QRbp)
j
+trbp
QRbpj –tr–bp
QR–bpj = . (.) From (.), (.) and (.) we find
Mpj= (–)j πij
|λ|=bp
trQRλj dλ.
3 The formula of the regularized trace of the operatorL
In this section, we find a formula for the regularized trace of the operatorL. According to Theorem .,
Mp= –
πi
|λ|=bp trQR
λ dλ. (.)
Since{ψmn}∞∞m=,n=is an orthonormal basis of the spaceH, from (.) we obtain
Mp= –
πi
|λ|=bp ∞
m=
∞
n=
QRλψmn,ψmn
dλ
= – πi
|λ|=bp ∞
m=
∞
n=
(Qψmn,ψmn) (m+)–λdλ
=
∞
m=
∞
n=
(Qψmn,ψmn)· πi
|λ|=bp
λ– (m+)dλ. (.)
Considering (m+)<bp= (p+ )form≤pand (m+)>bp= (p+ )form>p, from formula (.)
Mp=
p
m=
∞
n=
(Qψmn,ψmn) πi
|λ|=bp
λ– (m+)dλ=
p
m=
∞
n=
(Qψmn,ψmn) (.)
is obtained.
From (.) and (.), we have
Mp=
p
m=
∞
n=
π
Q(x)
πsin
m+
x·ϕn,
πsin
m+
x·ϕn
H
dx
=
π
p
m=
∞
n=
π
Q(x)ϕn,ϕn
Hsin
m+
x dx
=
π
p
m=
π
∞
n=
Q(x)ϕn,ϕn
H
If we consider the formula∞n=(Q(x)ϕn,ϕn)H=trQ(x), then we get
Mp=
p+
π
π
trQ(x)dx–
π
p
m=
π
trQ(x)cos(m+ )x dx. (.)
Lemma . If the operator function Q(x)satisfies conditions(Q)and(Q),then we have
Rλ<p–
over the circle|λ|=bp.
Proof Since the operator functionQ(x) satisfies conditions (Q) and (Q), we have
{λmn}∞n=⊂
m+
–Q,
m+
+Q
(m= , , , . . .),
λmn–
m+
<Q<
(m= , , , . . . ;n= , , . . .).
If we consider this relation, we get
|λmn–λ|= λ–
m+
–
λmn–
m+
≥λ–
m+
–λmn–
m+
>|λ|–
m+
–
= (p+ )
–
m+
–
≥(p+ )–
p+
– >
p+
– >p
(.)
form≤pand
|λmn–λ|= m+
–λ–
m+
–λmn
≥
m+
–λ–
m+
–λmn ≥
m+
–|λ|–
≥
p+
– (p+ )–
> (p+ )
–
>p
(.)
form≥p+ .
On the other hand, we can write
Rλ= max
m=,,...
n=,,...
|λmn–λ|–
. (.)
From (.), (.) and (.) we get
Rλ<p– |λ|=bp
Lemma . If the operator function Q(x)satisfies condition(Q),then we have
QRλσ(H)< p
–
∞
n=
Q(x)ϕn
over the circle|λ|=bp.
Proof Let us show that the series∞m=∞n=QR
λψmnis convergent. Forλ∈/σ(L), we get
∞
m=
∞
n=
QRλψmn
=
∞
m=
∞
n=
m+
–λ
–
Qψmn
=
∞
m=
∞
n=
m+
–λ
– π
Q(x)
πsin
m+
x·ϕn
H
dx
=
π ∞
m=
∞
n=
m+
–λ
– π
Q(x)ϕnHsin
m+
x dx
<
∞
m=
∞
n=
m+
–λ
– π
Q(x)ϕn
Hdx
=
∞
m=
m+
–λ
–∞
n=
Q(x)ϕn. (.)
From this relation we obtain
∞
m=
∞
n=
QRλψmn<∞
λ∈/σ(L)
.
On the other hand, since the sequence{ψmn}∞∞m=,n=is an orthonormal basis of the space
H, we get
QRλσ
(H)≤
∞
m=
∞
n=
QRλψmn (.)
[]. From (.) and (.) we obtain
QRλσ(H)≤
∞
n=
Q(x)ϕn
∞
m=
m+
–λ
–
. (.)
Furthermore, over the circle|λ|=bpwe get
∞
m=
m+
–λ
–
= p
m=
m+
–λ
–
+
∞
m=p+
m+
–λ
< p
m=
|λ|–
m+
–
+
∞
m=p+
m+
–|λ|
–
= p
m=
(p+ )–
m+
–
+
∞
m=p+
m+
– (p+ ) –
< p
m=
(p+ )–
p+
–
+
∞
m=p+
m+
– (p+ ) –
m+
+ (p+ ) –
< (p+ )
p+
–
+ p
–
∞
m=p+
m+
– (p+ ) –
. (.)
It can be easily shown that
m+
– (p+ )>
m–p (m≥p+ ) (.)
and
∞
m=p+
m–p–< p–. (.)
From (.), (.) and (.) we get
∞
m=o m+
–λ
–
<p–+ p– < p–. (.)
From (.) and (.) we obtain
QRλσ(H)< p
–
∞
n=
Q(x)ϕn |λ|=bp
.
Theorem . If the operator function Q(x)satisfies conditions(Q), (Q), (Q),and(Q),
then we have the formula
∞
m=
∞
n=
λmn–
m+
–
π
π
trQ(x)dx
=
trQ(π) –trQ().
Proof By using Theorem ., Lemma . and Lemma ., we find
|Mp|=
π
|λ|
=bp trQRλ
dλ
< π
|λ|=bp
trQRλ
≤ π
|λ|=bp
QRλ
σ(H)|dλ|
≤ π
|λ|=bp
QRλQRλσ(H)|dλ|
≤ Q π
|λ|=bp
RλQRλσ(H)|dλ|
<c
|λ|=bp p–dλ
= π·bp·c·p–= πc(p+ )p–<cp–. (.)
Herecis a positive constant.
By using formula (.), Lemma . and Lemma ., we find
|Mp|= π
|λ|=b
p λtrRλ
QRλ
dλ
≤ bp π
|λ|=bp
Rλ
QRλ
|dλ|
≤ bp π
|λ|=bp
RλQRλσ
(H)|dλ| ≤ bp
π
|λ|=bp
RλQRλ
QRλσ(H)|dλ|
≤ bp π
|λ|=bp
RλQRλ
QRλσ(H)|dλ| < c·bpp–
=c(p+ )p–<cp–. (.)
From (.) and (.) we get
lim
p→∞Mp=plim→∞Mp= . (.)
From (.) and (.) we obtain
p
m=
∞
n=
λmn–
m+
–
π
π
trQ(x)dx
= –
π
p
m=
π
trQ(x)cos(m+ )x dx+Mp+Mp. (.)
From (.) and (.) we find
∞
m=
∞
n=
λmn–
m+
–
π
π
trQ(x)dx
= –
π ∞
m=
π
Moreover, using conditions (Q) and (Q), we get
–
π ∞
m=
π
trQ(x)cos(m+ )x dx
= – π
∞
m=
π
trQ(x)cosmx dx– (–)m π
trQ(x)cosmx dx
= –
∞
m=
π
π
trQ(x)cosmx dx
cosm +
π
π
trQ(x)dx
cos
+
∞
m=
π
π
trQ(x)cosmx dx
cosmπ+
π
π
trQ(x)dx
cosπ
=
trQ(π) –trQ(). (.)
From (.) and (.) we find
∞
m=
∞
n=
λmn–
m+
–
π
π
trQ(x)dx
=
trQ(π) –trQ() .
The theorem is proved.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Acknowledgements
The authors would like to thank Professor Ehliman Adıgüzelov for his expert assistance and for contributions in detailed editing of the last version of this manuscript.
Received: 20 April 2015 Accepted: 15 September 2015 References
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