Miskolc Mathematical Notes HU e-ISSN 1787-2413 Vol. 20 (2019), No. 1, pp. 17–32 DOI: 10.18514/MMN.2019.2621
THE SECOND REGULARIZED TRACE FORMULA FOR THE STURM-LIOUVILLE OPERATOR
F. AYDIN AKGUN, M. BAYRAMOGLU, AND A. BAYRAMOV Received 02 May, 2018
Abstract. In this paper, the second regularized trace formula for the differential operator with antiperiodic boundary conditions is obtained.
2010Mathematics Subject Classification: 34L20; 35R10 Keywords: eigenvalue, eigenfunction, resolvent, regularized trace
1. INTRODUCTION
In the Hilbert spaceH DL2Œ0; , we consider operator L generated by the dif- ferential expression
l.y/D y00Cq.x/y;
with the antiperiodic boundary conditions
y.0/D y./; y0.0/D y0./;
whereq.x/2C2Œ0; is real function and satisfies the condition
q.0/Dq./: (1.1)
It is well known from [13] that the eigenvalues of the operator L form double series n;jD
2n 1C c0
2n 1Co 1
n 2
; wherec0D21 R
0 q.x/dx; j D1; 2,o n1
includej andnD1; 2; ::::LetL0denote the operatorLwithq.x/0. The eigenvalues of operatorL0are
nD.2n 1/2; nD1; 2; ::::
The orthonormal eigenfunctions corresponding to the eigenvaluesnare
1 nD
r2
cos.2n 1/x; n2D r2
sin.2n 1/x; nD1; 2; :::: (1.2)
c 2019 Miskolc University Press
It is natural to define second regularized trace of operatorLas
1
X
nD1
0 2n;1C2n;2 2.2n 1/4
; (1.3)
where the symbol ”0” means that something in this sum is discarded to provide its convergence. The sum in (1.3) is the main interest of this article.
The regularized trace formula
1
X
nD0
n n2 1
Z 0
q.x/dx
Dq.0/Cq./
4
1 2
Z 0
q.x/dx;
of the Stum-Liouvile operator
y00Cq.x/yDy; y0.0/D0; y0./D0; (1.4) with q.x/2C1Œ0; was first studied by Gelfand and Levitan ([6]), where the n
are the eigenvalues of the operator in (1.4). Afterwards, trace formulas for different differential operators are studied by several mathematicians(see [1–5,8,10–12,14,15]
and references therein).
Note that the first regularized trace formula
1
X
nD0
n;1Cn;2 2.2n 1/2 2c0
D0;
was obtained for operatorLwith a real potentialq.x/2L2.0; /, by [10] and with an arbitrary complexq.x/2L2.0; /, by [15].The similar formula was obtained by [12] for the operatorLwith operator functionq.x/.
Trace formulas are used in inverse problems of spectral analysis of differential equa- tions(see [14]) and for approximate calculation of the first eigenvalues of the related operator [1,4,5,7,9,14].
2. SOME FORMULAS ABOUT THE OPERATORqR0
LetR0andRbe the resolvents of the operatorL0andL, respectively. Then, for any2.L/and2.L0/where.:/is the resolvent set of an operator,RWH! H andR0WH!H are trace class operators, that is,RR021.H /. Therefore,
t r.R R0/D
1
X
nD1
1
n;1 C 1 n;2
2 .2n 1/2
:
Multiplying both sides of the above equality by 2=2 i then integrating over the circlejj Dbp D.2p 1/2C4p,
1 2 i
Z
jjDbp
2t r.R R0/d D
p
X
nD1
2.2n 1/4 2n;1 2n;2
(2.1)
is obtained. Taking in the account thatR R0D RqR0, from equation (2.1),
p
X
nD1
2n;1C2n;2 2.2n 1/4 D
N
X
jD1
MpjCMpN (2.2)
is obtained. Here
Mpj D. 1/jC1 2 i
Z
jjDbp
2t rh
R0.qR0/ji d ; and
MpN D . 1/N 2 i
Z
jjDbp
2t rh
R.qR0/NC1i
d ; (2.3)
whereqDq.x/andN is an integer.
The formula
Mpj D. 1/j 2 ij
Z
jjDbp
t r.qR0/jd ; (2.4) can be proved similarly as in Theorem2in [12].
From equations (1.2) and (2.4), we write Mp1D 1
i Z
jjDbp
(
1
X
nD1
Œ.qR0 n1; n1/C.qR0 n2; n2/
) d
D2
1
X
nD1
Œ.q n1; n1/C.q n2; n2/ 1 2 i
Z
jjDbp
n
d
D 4
p
X
nD1
.2n 1/2 Z
0
q.x/Œcos2.2n 1/xCsin2.2n 1/xdx
D 4
p
X
nD1
.2n 1/2 Z
0
q.x/dx: (2.5)
Now, we shall computeMp2. From equation (2.4), Mp2D 1
2 i Z
jjDbp
( 1
X
nD1
.qR0/2 n1; n1
C .qR0/2 n2; n2 )
d
D 1 2 i
Z
jjDbp
( 1
X
nD1
1
n
qR0q n1; n1
C qR0q n2; n2 )
d
D 1 2 i
Z
jjDbp
( 1
X
nD1 1
X
rD1
1
.n /.r /
q n1; r1
q r1; n1 C q n1; r2
q r2; n1
C q n2; r1
q r1; n2
C q n2; r2
q r2; n2 d :
For convenience, let
qnrD j.q n1; r1/j2C j.q n1; r2/j2C j.q n2; r1/j2C j.q n2; r2/j2: (2.6) Then,
Mp2D
1
X
nD1 1
X
rD1
qnr
1 2 i
Z
jjDbp
. n/. r/d D
p
X
nD1 p
X
rD1
qnr
1 2 i
Z
jjDbp
. n/. r/d C
p
X
nD1 1
X
rDpC1
qnr
1 2 i
Z
jjDbp
. n/. r/d C
1
X
nDpC1 p
X
rD1
qnr
1 2 i
Z
jjDbp
. n/. r/d C
1
X
nDpC1 1
X
rDpC1
qnr
1 2 i
Z
jjDbp
. n/. r/d D
p
X
nD1 p
X
rD1
qnrC2
p
X
nD1 1
X
rDpC1
qnr
n
r n
D
p
X
nD1 1
X
rD1
qnr p
X
nD1 1
X
rDpC1
rCn
r n
qnr: Thus,
Mp2D
p
X
nD1 1
X
rD1
qnr ˛p; (2.7)
where
˛p D
p
X
nD1 1
X
rDpC1
rCn
r n
qnr: By using equations (1.2) and (2.6), we have
˛pD˛p1C˛p2C˛p3; (2.8)
where
˛p1D2 2
p
X
nD1 1
X
rD1
.2r 1/2C.2n 1/2 .2r 1/2 .2n 1/2
(ˇ ˇ ˇ ˇ
Z 0
q.x/cos2.n r/xdx ˇ ˇ ˇ ˇ
2
C
C ˇ ˇ ˇ ˇ
Z 0
q.x/sin2.n r/xdx ˇ ˇ ˇ ˇ
2)
;
˛p2D2 2
p
X
nD1 1
X
rD1
.2r 1/2C.2n 1/2 .2r 1/2 .2n 1/2 ˇ ˇ ˇ ˇ
Z 0
q.x/cos2.nCr 1/xdx ˇ ˇ ˇ ˇ
2
;
˛p3D2 2
p
X
nD1 1
X
rD1
.2r 1/2C.2n 1/2 .2r 1/2 .2n 1/2 ˇ ˇ ˇ ˇ
Z 0
q.x/sin2.nCr 1/xdx ˇ ˇ ˇ ˇ
2
: The formula of˛p1can be written as
˛p1D2 2
1
X
iD1
X r nDi
np r > p
1C 2.2n 1/2 .2r 1/2 .2n 1/2
(2.9)
(ˇ ˇ ˇ ˇ
Z 0
q.x/cos2ixdx ˇ ˇ ˇ ˇ
2
C ˇ ˇ ˇ ˇ
Z 0
q.x/sin2ixdx ˇ ˇ ˇ ˇ
2)
: (2.10)
Forip X
cr nDi np r>p
.2n 1/2
.2r 1/2 .2n 1/2 D
i 1
X
jD0
.2.p j / 1/2
.2.p jCi / 1/2 .2.p j / 1/2
D
i 1
X
jD0
.2.p j / 1Ci /2 2.2.p j / 1Ci /iCi2 4i.2.p j / 1Ci /
D2p 1
4 C1 2i
4 C
i 1
X
jD0
i
4.2p 1 2jCi /:
(2.11)
For any integerspandi, let
ED f.r; n/Wr; n2NIr nDiInpIr > pg: Then using (2.11), we write
X
n;r2E
1C 2.2n 1/2 .2r 1/2 .2n 1/2
DiC2 X
cr nDi np r>p
.2n 1/2 .2r 1/2 .2n 1/2
DpCi 2
p
X
jDp iC1
1
2j 1Ci; .ip/:
(2.12)
It is easy to see that
i 2
p
X
jDp iC1
1
2j 1Ci <i2 p: Using this inequality and equation (2.12),
X
n;r2E
1C 2.2n 1/2 .2r 1/2 .2n 1/2
DpCi2O.p 1/; .ip/ (2.13) is obtained.
HereO.p 1/depends onpandi, and satisfies the inequality jO.p 1/j< const:p 1: In a similar form, forip, it can be shown that
X
n;r2E
1C 2.2n 1/2 .2r 1/2 .2n 1/2
DO.p/: (2.14)
HereO.p/depends onpandi, and satisfies the inequality jO.p/j< const:p:
From (2.9), (2.13) and (2.14), we obtain
˛p1D2 2p
1
X
iD1
(ˇ ˇ ˇ ˇ
Z 0
q.x/cos2ixdx ˇ ˇ ˇ ˇ
2
C ˇ ˇ ˇ ˇ
Z 0
q.x/sin2ixdx ˇ ˇ ˇ ˇ
2)
C
p
X
iD1
i2O.p 1/ (ˇ
ˇ ˇ ˇ
Z 0
q.x/cos2ixdx ˇ ˇ ˇ ˇ
2
C ˇ ˇ ˇ ˇ
Z 0
q.x/sin2ixdx ˇ ˇ ˇ ˇ
2)
C
1
X
iDpC1
O.p/
(ˇ ˇ ˇ ˇ
Z 0
q.x/cos2ixdx ˇ ˇ ˇ ˇ
2
C ˇ ˇ ˇ ˇ
Z 0
q.x/sin2ixdx ˇ ˇ ˇ ˇ
2)
Dp
Z
0 jq.x/j2dx p 2
ˇ ˇ ˇ ˇ
Z 0
q.x/dx ˇ ˇ ˇ ˇ
2
C˛p1.1/C˛p1.2/:
(2.15)
Here, sinceq.0/Dq./, j˛p1.1/j D
ˇ ˇ ˇ ˇ ˇ
p
X
iD1
i2O.p 1/ (ˇ
ˇ ˇ ˇ
Z 0
q.x/cos2ixdx ˇ ˇ ˇ ˇ
2
C ˇ ˇ ˇ ˇ
Z 0
q.x/sin2ixdx ˇ ˇ ˇ ˇ
2)ˇ ˇ ˇ ˇ ˇ const:p 1
Z
0 jq0.x/j2dx;
(2.16)
and j˛p1.2/j D
ˇ ˇ ˇ ˇ ˇ ˇ
1
X
iDpC1
O.p/
(ˇ ˇ ˇ ˇ
Z 0
q.x/cos2ixdx ˇ ˇ ˇ ˇ
2
C ˇ ˇ ˇ ˇ
Z 0
q.x/sin2ixdx ˇ ˇ ˇ ˇ
2)ˇ ˇ ˇ ˇ ˇ ˇ const:p 1
Z
0 jq0.x/j2dx
(2.17)
is obtained. From (2.15)-(2.17), we get
˛p1D p
Z 0
q2.x/dx p 2
ˇ ˇ ˇ ˇ
Z 0
q.x/dx ˇ ˇ ˇ ˇ
2
CO.p 1/: (2.18) Sinceq.x/satisfies the condition in equation (1.1), it can be shown that
j˛pjj const:.p 1/; .j D2; 3/: (2.19) From equation (2.6), we have
p
X
nD1 1
X
rD1
qnrD
p
X
nD1
˚kq n1k2C kq n2k2 D2p
Z 0
q2.x/dx: (2.20) From equations (2.7), (2.8) and (2.18)-(2.20),
Mp2Dp
Z 0
q.x/2dxC p 2
Z 0
q.x/dx 2
CO.p 1/;
is obtained.
3. THE SECOND REGULARIZED TRACE FORMULA
In this section we obtain the second regularized trace formula for the operatorL.
To do this, we will first show that the formulas
plim!1Mpj D0; j3; (3.1)
plim!1MpN D0; N 6; (3.2)
are satisfied.
The inequalities
kqR0k1.H /< C;kR0k< Cp 1;kRk< Cp 1; .jj Dbp/ (3.3) are true(see [12]). HereC > 0is a constant. From (2.4) and (3.3), we have
jMpjj D 1 j
ˇ ˇ ˇ ˇ ˇ Z
jjDbp
t r.qR0/jd ˇ ˇ ˇ ˇ ˇ bp
j Z
jjDbp
.qR0/j
1.H /jd j
bp j
Z
jjDbp
.qR0/
1.H /
.qR0/j 1 jd j C bp
j Z
jjDbp
kqkj 1 .R0/
j 1jd j
< C1p5 j.C1> 0/:
This implies that
plim!1Mpj D0; .j 6/;
but we claim that it is also true forj D3; 4; 5:Now let we prove the formula (3.1) forj D3.
Mp3D 1 3 i
1
X
nD1 1
X
rD1 1
X
kD1
Z
jjDbp
d
.n /.r /.k / ˚
..q n1; r1/Œ.q r1; k1/ k1C.q r1; k2/ k2 C.q n1; r2/Œ.q r2; k1/ k1C.q r2; k2/ k2; n1/ C..q n2; r1/Œ.q r1; k1/ k1C.q r1; k2/ k2
C.q n2; r2/Œ.q r2; k1/ k1C.q r2; k2/ k2; n2/ D 1
3 i
1
X
nD1 1
X
rD1 1
X
kD1
Z
jjDbp
d
.n /.r /.k /F .n; r; k/;
(3.4)
where
F .n; r; k/D.q n1; r1/.q r1; k1/.q k1; n1/C.q n1; r1/.q r1; k2/.q k2; n1/ C.q n1; r2/.q r2; k1/.q k1; n1/C.q n1; r2/.q r2; k2/.q k2; n1/ C.q n2; r1/.q r1; k1/.q k1; n2/C.q n2; r1/.q r1; k2/.q k2; n2/ C.q n2; r2/.q r2; k1/.q k1; n2/C.q n2; r2/.q r2; k2/. k2; n2/:
SinceF .n; r; k/DF .r; n; k/DF .k; n; r/DF .k; r; n/DF .n; k; r/DF .r; k; n/, from equation (3.4)
Mp3D 1 i
p
X
nD1 1
X
rDpC1 1
X
kDpC1
Z
jjDbp
d
. n/. r/. k/F .n; r; k/
C 1 i
p
X
nD1 p
X
rD1 n¤r
1
X
kDpC1
Z
jjDbp
d
. n/. r/. k/F .n; r; k/
C 1 i
p
X
nD1 1
X
kDpC1
Z
jjDbp
d
. n/. k/F .n; n; k/
D2
p
X
nD1 1
X
rDpC1 1
X
kDpC1
n
.r n/.k n/F .n; r; k/
C4
p
X
nD1 p
X
crD1 n¤r
1
X
kDpC1
n
.n r/.n k/F .n; r; k/
2
p
X
nD1 1
X
kDpC1
k
.k n/2F .n; n; k/
(3.5)
is obtained. Let
F1.n; r; k/D 3 Z
0
q.x/cos2.n r/xdx Z
0
q.x/cos2.r k/xdx Z
0
q.x/cos2.k n/xdx;
(3.6)
F2.n; r; k/DF .n; r; k/ F1.n; r; k/; (3.7) Api D
p
X
nD1 1
X
rDpC1 1
X
kDpC1
n
.r n/.k n/Fi.n; r; k/; (3.8) BpiD
p
X
nD1 p
X rD1 n¤r
1
X
kDpC1
n
.n r/.n k/Fi.n; r; k/; (3.9)
Cpi D
p
X
nD1 1
X
kDpC1
k
.k n/2Fi.n; n; k/; .i D1; 2/: (3.10) From equation (3.6), we obtain
F1.n; r; k/DF1.n; k; r/; F1.n; n; k/DF1.n; k; k/:
Using these equalities and equation (3.8), we have Ap1D
p
X
nD1 1
X
crDpC1 r>k
1
X
kDpC1
n
.r n/.k n/F1.n; r; k/
C
p
X
nD1 1
X
kDpC1
n
.k n/2F1.n; n; k/:
(3.11)
In similar way, it can be shown that, Bp1D
p
X
nD1 p
X
crD1 n<r
1
X
kDpC1
k
.r n/.k n/F1.n; r; k/: (3.12) Let
Ap1D
p
X
nD1 1
X
crDpC1 r>k
1
X
kDpC1
n
.r n/.k n/F1.n; r; k/; (3.13)
Bp1D Bp1; (3.14)
Cp1D
p
X
nD1 1
X
kDpC1
1 k n
F1.n; n; k/: (3.15) Hence, by using equations (3.5)-(3.15), we obtain
Mp3D4Ap1 4Bp1 2Cp1C2Ap2C4Bp2 2Cp2: (3.16) Let us find a formula forAp1.
Let
E1D f.n; r; k/Wn; r; k2NIr nDiIk nDjInpIr; k > pg; wherep; i andj are integers such thatpj,i j, then
Ap1D
p
X
nD1 1
X
rDpC1 r>k
1
X
kDpC1 k np
n
.r n/.k n/F1.n; r; k/
C
p
X
nD1 1
X
rDpC1 r>k
1
X
kDpC1 k n>p
n
.r n/.k n/F1.n; r; k/
D 3
1
X
iD2 i >j
p
X
jD1
2 4
X
n;r;k2E1
n
.r n/.k n/ Z
0
q.x/cos2ixdx
Z
0
q.x/cos2.i j /xdx Z
0
q.x/cos2jxdx
C
p
X
nD1 1
X
rDpC1 r>k
1
X
kDpC1 k n>p
n
.r n/.k n/F1.n; r; k/:
Let
ˇij D 3 Z
0
q.x/cos2ixdx Z
0
q.x/cos2.i j /xdx Z
0
q.x/cos2jxdx;
(3.17) and
Ap11D
1
X
iD2 p
X
jD1 i >j
2 4
0
@ X
n;r;k2E1
n
.r n/.k n/ 1 Aˇij
3 5;
Ap12D
p
X
nD1 1
X
rDpC1 r>k
1
X
kDpC1 k n>p
n
.r n/.k n/F1.n; r; k/; (3.18) then we write
Ap1DAp11CAp12: (3.19)
By similar proof of (2.14), it can be shown that X
n;r;k2E1
n
.r n/.k n/D D X
n;r;k2E1
.2n 1/2 .2r 1/2 .2n 1/2
.2k 1/2 .2n 1/2 D 1
16iCj pO.1/;
(3.20)
whereO.1/satisfies the condition
jO.1/j< const:
and depends onp; i andj. Moreover, ifq.x/has a continuous derivative of second order atŒ0; and satisfies the condition in (1.1), then it can be shown that
ˇij Dconst:
i2j2 : (3.21)
From (3.18), (3.20) and (3.21), Ap11D
1
X
iD2 p
X
jD1 i >j
ˇij
16i CO.1/
pi2j
;
is obtained.
Since ˇ ˇ ˇ ˇ ˇ ˇ
1
X
iD2 p
X
jD1
O.1/
pi2j ˇ ˇ ˇ ˇ ˇ ˇ
const:p 1
1
X
iD1
i 2
! p X
jD1
j 1< const:p 1lnp;
we find
Ap11D
1
X
iD2 p
X
jD1 i >j
ˇij
16i Co.1/: (3.22)
Hereo.1/is an expression which satisfies the condition
plim!1o.1/D0;
and depends onp. From3.19and3.22, we obtain Ap1D
1
X
iD2 p
X
jD1 i >j
ˇij
16i CAp12Co.1/: (3.23)
Now, to find the formula forBp1, let Bp11D
p
X
nD1 p
X
rD1 1
X
kDpC1 n<r;k np
k
.k n/.k r/F1.n; r; k/;
and
Bp12D
p
X
nD1 p
X
rD1 1
X
kDpC1 n<r;k n>p
k
.k n/.k r/F1.n; r; k/;
then from (3.12) and (3.14), we have
Bp1DBp11CBp12: (3.24)
By using equations in (3.6) and (3.7),Bp11can be written as Bp11D
p
X
jD2 p 1
X
iD1 j >i
0
@ X
n;r;k2E2
k
.k n/.k r/ 1
Aˇij; (3.25)
whereE2is a set defined by
E2D f.n; r; k/Wn; r; k2NIr nDiIk nDjIn; rpIk > pg
fori < j p. Moreover it can be shown that X
n;r;k2E2
k
.k n/.k r/D 1 16j C i
pO.1/: (3.26)
From (3.21), (3.25) and (3.26), we obtain Bp11D
p
X
jD2 p 1
X
iD1 j >i
ˇij
16j CO.1/
pj2i
:
and, sinceˇij Dˇj i, we write Bp11D
p
X
jD2 p 1
X
iD1 j >i
ˇij
16j CO.1/:
By using above equation and3.24, we have Bp1D
p
X
iD2 p 1
X
jD1 i >j
ˇij
16i CBp12Co.1/: (3.27) From (3.24), we get
ˇ ˇ ˇ ˇ ˇ
p
X
iD2
ˇip
i ˇ ˇ ˇ ˇ ˇ
p
X
iD2
1
i3p2 Do.1/;
ˇ ˇ ˇ ˇ ˇ ˇ
1
X
iDpC1 p
X
jD2
ˇij i
ˇ ˇ ˇ ˇ ˇ ˇ
1
X
iDpC1 p
X
jD2
1 i3j2 D
0
@
1
X
iDpC1
1 i3
1 A
0
@
p
X
jD1
1 j2
1
ADo.1/:
Therefore, by (3.27), we have Bp1D
1
X
iD2 p
X
jD1 i >j
ˇij
16iCBp12Co.1/: (3.28) From (3.16), (3.23) and (3.28), we obtain
Mp3D4Ap12 4Bp12 2Cp1C2Ap2C4Bp2 2Cp2Co.1/: (3.29) Here, it can be easily seen that,
plim!1Ap12D lim
p!1Bp12D lim
p!1Cp1D lim
p!1Ap2D lim
p!1Bp2D lim
p!1Cp2D0:
(3.30)
From (3.29) and (3.30), we obtain
plim!1Mp3D0:
In a similar way, it can be proved that
plim!1Mp4D lim
p!1Mp5D0:
Now, let us prove the expression given in (3.2). By employing (2.2) and (3.3), we obtain
jMpNj D 1 2
ˇ ˇ ˇ ˇ ˇ ˇ ˇ
Z
jjDbp
2t rh
R.qR0/NC1i d
ˇ ˇ ˇ ˇ ˇ ˇ ˇ bp2
2 Z
jjDbp
R.qR0/NC1
1.H /jd j bp2
Z
jjDbp
kRk
.qR0/NC1
1.H /jd j C bp2p 1
Z
jjDbp
qR0
N .qR0/
1.H /jd j const:p5 N:
This shows that
plim!1MpN D0; N 6:
By using the equations (2.2), (2.5), (3.1) and (3.2), we find
p
X
nD1
2n;1C2n;2 2.2n 1/4 D 4
p
X
nD1
.2n 1/2 Z
0
q.x/dx Cp
Z
0
q2.x/dxC p 2
Z 0
q.x/dx 2
Co.1/:
As a result, we get
p
X
nD1
2n;1C2n;2 2.2n 1/4 4
.2n 1/2 Z
0
q.x/dx 1
Z
0
q2.x/dx 1 2
Z 0
q.x/dx 2#
D0:
(3.31)
The left hand-side of this equality is called the second regularized trace of the oper- ator L. Thus we have proven the main result of this article given by the following theorem.
Theorem 1. If q.x/2 C2Œ0; is a real function which satisfies the condition (1.1), then the equality obtained in (3.31) holds for the second regularized trace of the operatorL.
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Authors’ addresses
F. Aydin Akgun
Yildiz Technical University, Department of Mathematical Engineering, Istanbul, Turkey E-mail address:fakgun@yildiz.edu.tr
M. Bayramoglu
Institute of Academy of Sciences of Azerbaijan, Baku, Azerbaijan E-mail address:mamed.bayramoglu@yahoo.com
A. Bayramov
Institute of Academy of Sciences of Azerbaijan, Baku, Azerbaijan E-mail address:azadbay@gmail.com
