On identity for eigenvalues of second order differential operator equation

Mathematical and Computer Modelling 49 (2009) 403–412

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Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm

On identity for eigenvalues of second order differential operator equation Azad Bayramov a , Zerrin Oer b,∗ , Oya Baykal b a

Department of Statistics, Faculty of Arts and Science, Yıldız Technical University, (34210), Davutpas . a, İstanbul, Turkey

b

Department of Mathematics, Faculty of Arts and Science, Yıldız Technical University, (34210), Davutpas . a, İstanbul, Turkey

article

a b s t r a c t

info

Article history: Received 5 October 2006 Received in revised form 10 June 2008 Accepted 25 September 2008

In this study we investigate the spectrum and regularized trace of the second order differential operator equation with periodic boundary conditions. © 2008 Elsevier Ltd. All rights reserved.

Keywords: Hilbert space Self-adjoint operator Kernel operator Spectrum Regularized trace

1. Introduction Let H be a separable Hilbert space. In the Hilbert space H1 = L2 ([0, π], H ) we investigate the regularized trace of the self-adjoint operator L generated by the expression l(y) = −y00 (x) + Q (x)y(x) with the semi-periodic boundary conditions y(0) = −y(π ),

y0 (0) = −y0 (π ).

(1)

Let us assume that the operator function Q (x) satisfies the following conditions: (1) For ∀x ∈ [0, π], Q (x) : H → H is a self-adjoint kernel operator. (2) sup[0,π] kQ (x)kH < 4. (3) There exists an orthonormal basis {ϕn }∞ n=1 of the space H such that ∞ X

kQ (x)ϕn kH1 < ∞.

n =1

(4) The function kQ (x)kσ1 (H ) is bounded and measurable in the interval [0, π]. Here σ1 (H ) is the space of kernel operators from H to H as in [1].

∗

Corresponding author. E-mail addresses: [email protected] (A. Bayramov), [email protected] (Z. Oer), [email protected] (O. Baykal).

0895-7177/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2008.09.005

404

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

00 Let L0 be the operator  generated by the expression l0 (y) = −y (x) with the boundary conditions (1). The spectrum of 2 ∞ this operator is the set (2m − 1) m=1 and every point of this set is eigenvalue of infinite multiplicity. The orthonormal

eigenfunctions corresponding to the eigenvalue (2m − 1)2 have the form

r

2

π

r cos(2m − 1)x · ϕn ,

2

π

sin(2m − 1)x · ϕn ,

n = 1, 2, . . . .

For convenience we denote

r ψ

0 1n

=

2

π r

0 ψ3n =

µ1 = 1,

2

π

r cos x · ϕn ,

ψ

cos 3x · ϕn ,

0 2n

=

2

π r

0 ψ4n =

µ2 = 1,

sin x · ϕn ,

2

π

µ3 = 9,

(2)

sin 3x · ϕn , . . .

µ4 = 9, . . . .

The formula of regularized trace of Sturm–Liouville operator is firstly obtained by Gelfand and Levitan [2]. After this work numerous investigations on calculation of regularized trace of various differential and differential operator equations appeared (see, for example [3–13]). We note that formula of regularized traces can be used approximately in calculation of first eigenvalues of differential operators (see [3,4]). 2. Investigation of the spectrum Let R0λ and Rλ be the resolvents of the operators L0 and L respectively. Lemma 1. If the operator function Q (x) satisfies the condition (1) and (3) and λ ∈ ρ(L0 ) then QR0λ : H1 → H1 is the nuclear operator: QR0λ ∈ σ1 (H1 ) Proof. The system (2) of eigenfunctions of operator L0 is the orthonormal basis of the space H1 . Then as shown in [1], it is P∞ P ∞ 0 sufficient to show that the series m=1 n=1 kQR0λ ψmn kH1 is a convergent series in order to prove QR0λ ∈ σ1 (H1 ). By means of (2) we have ∞ X ∞ X

∞ X ∞ X

0 kQR0λ ψmn kH 1 =

m=1 n=1

=

∞ X ∞ X

|(2m − 1)2 − λ|−1

π

"Z +

"  Z 

m=1 n=1

r (Q (x)

0

≤2

0 |µm − λ|−1 kQ ψmn kH 1

m=1 n=1

∞ X ∞ X

2

π

π

r ( Q ( x)

0

r sin(2m − 1)xϕn , Q (x)

2

π 2

π

r cos(2m − 1)xϕn , Q (x)

sin(2m − 1)xϕn )dx

2

π  #1/2 

#1/2 cos(2m − 1)xϕn )dx



|(2m − 1)2 − λ|−1 kQ (x)ϕn k2H1 .

(3)

m=1 n=1

By virtue of the condition (3) we obtain, from (3), that ∞ X ∞ X

0 kQR0λ ψmn kH1 < ∞.

m=1 n=1

Hence lemma is proved.



Theorem 1. If the operator function Q (x) satisfies the conditions (1) and (3), then the spectrum of the operator L is the subset of the union of the intervals

Ωm = [(2m − 1)2 − kQ kH1 , (2m − 1)2 + kQ kH1 ],

m = 1, 2, . . .

which are pairwise disjoint and (a) Every point different from (2m − 1)2 , belonging to the interval Ωm of the spectrum of the operator L is a discrete eigenvalue, whose multiplicity is finite. (b) (2m − 1)2 may be the eigenvalue, whose multiplicity is finite or infinite of the operator L. (c) The equality limn→∞ λmn = (2m − 1)2 holds. Here {λmn }∞ n=1 are the eigenvalues, belonging to the interval Ωm of the operator L and each eigenvalue has been repeated according to its multiplicity.

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

405

Proof. The resolvent Rλ of the operator L satisfies the equation R0λ − Rλ QR0λ = Rλ . If λ ∈ R \ 1

S∞

m=1

(4)

Ωm then

|λ − (2m − 1)2 | > kQ kH1 m = 1, 2, . . . .

(5)

For self-adjoint operator R0λ = (L0 − λI )−1 , we have

kR0λ kH1 = max |λ − (2m − 1)2 |−1 .

(6)

m

From this and (5), we obtain 1 kR0λ kH1 < kQ k− H1 .

Hence

kQR0λ kH1 ≤ kQ kH1 kR0λ kH1 < 1. Thus A(B) = R0λ − BQR0λ is constrait operator from £(H1 ) to £(H1 ). Here £(H1 ) is the linear bounded operators S space from H1 to H1 . This gives ∞ A(Rλ ) = Rλ , that is, the Eq. (4) has a single solution Rλ ∈ £(H1 ). Thus, every point λ 6∈ m=1 Ωm is the regular point of S∞ the self adjoint operator L. So, the spectrum of the operator L is σ (L) ⊂ m=1 Ωm . From the formula (4) and the Lemma 1, T for every λ ∈ ρ(L) ρ(L0 ), Rλ − R0λ belongs to σ1 (H1 ). This means that Rλ − R0λ is the nuclear operator. In this case, as it is proved in [14], the Continuous parts of the spectra of the operators L0 and L coincide. According to this and since the spectrum of the operator L0 is Continuous, the Continuous part of the spectrum operator L is the set {(2m − 1)2 }∞ m=1 . This also means that the assertions (a), (b) and (c) of Theorem 1 are satisfied.  3. A formula for the regularized trace ∞ Let{ψmn }∞ m,n=1 be the orthonormal eigenfunctions corresponding to the eigenvalues {λmn }m,n=1 of operator L and 0 0 B0mn = (·, ψmn )H1 ψmn ,

Bmn = (·, ψmn )H1 ψmn .

The difference Rλ − R0λ satisfies the following formula Rλ − R0λ =

∞ X ∞ X

Bmn

m=1 n=1

λmn − λ

−

∞ X ∞ X

B0mn

m=1 n=1

µm − λ

.

(7)

Since the operator function Rλ − R0λ belongs to σ1 (H1 ) for every λ ∈ ρ(λ), from the formula (7) we have tr (Rλ − R0λ ) =

∞ X ∞  X m=1 n=1

1

−

λmn − λ



2

(2m − 1)2 − λ

.

By multiplying with 2λπ i both sides of this equality and integrating over the circle |λ| = bp = (2p − 1)2 + 4p we obtain 1

Z

2π i |λ|=bp

1

λtr (Rλ − R0λ )dλ =

Z

2π i |λ|=bp

+

1

λ

Z

2π i |λ|=bp

p X ∞  X

λmn − λ ∞ X ∞  X 1

m=1 n=1

λ

1

m=p+1 n=1

−



2

(2m − 1)2 − λ

λmn − λ

−

dλ

2

(2m − 1)2 − λ



dλ.

(8)

For m ≤ p, p ≥ 1 by condition (2) we have

λmn ≤ (2m − 1)2 + kQ kH1 ≤ (2p − 1)2 + kQ kH1 < (2p − 1)2 + 4p = bp and moreover, for m > p we obtain

λmn > (2m − 1)2 − kQ kH1 ≥ [(2p + 1) − 1]2 − kQ kH1 > (2p − 1)2 + 4p. So,



|λmn | < bp |λmn | > bp

if m ≤ p; p ≥ 1, if m > p; p ≥ 1,

n = 1, 2, . . . n = 1, 2, . . . .

(9)

406

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

From (8) and (9) we have

Z

1

2π i |λ|=bp

λtr (Rλ − R0λ )dλ =

p X ∞ X

2(2m − 1)2 − λmn .




(10)

m=1 n=1

On the other hand, from the formula Rλ = R0 − Rλ QR0λ the equality Rλ − R0λ =

N X

(−1)j R0λ (QR0λ )j + (−1)N +1 Rλ (QR0λ )N +1

j =1

is obtained for every natural number N. From last equality and (10), we write p X ∞ X

N
X λmn − 2(2m − 1)2 = Mpj + MpN

m=1 n=1

(11)

j =1

where, Mpj

(−1)j+1 = 2π i

Z |λ|=bp

λtr [R0λ (QR0λ )j ]dλ

(12)

and MpN

(−1)N = 2π i

Z |λ|=bp

λtr [Rλ (QR0λ )N +1 ]dλ.

(13)

Theorem 2. Mpj =

(−1)j 2π ij

Z |λ|=bp

tr (QR0λ )j dλ.

(14)

In order to prove this theorem we prove the following three theorems. Theorem 3. Operator function QR0λ in the domain ρ(L0 ) = C \ {(2m − 1)2 }∞ m=1 is analytic with respect to the norm in σ1 (H1 ) and

(QR0λ )0 = Q (R0λ )2 .

(15)

Proof. ∀λ ∈ ρ(λ0 )R0λ is a nuclear operator and R0λ+1λ − R0λ = 1λR0λ+1λ R0λ . From here we have



QR0

0

λ+1λ − QRλ 0 2 − Q (Rλ )

1λ

= kQR0λ+1λ R0λ − Q (R0λ )2 kσ1 (H1 ) σ1 (H1 )

= kQR0λ (R0λ+1λ − R0λ )kσ1 (H1 ) ≤ kQR0λ kσ1 (H1 ) k(R0λ+1λ − R0λ )k1 .

(16)

Clearly lim kR0λ+1λ − R0λ k1 = 0.

(17)

1λ→0

From (16) and (17) we have



QR0

0

λ+1λ − QRλ 0 2 lim − Q (Rλ ) 1λ→0

1λ Formula (15) is proved.

= 0. σ1 (H1 )



Theorem 4. Let ∀λ ∈ G ⊂ C , operator function B(λ) ∈ σ1 (H ) be analytic with respect to the norm in σ1 (H ). Then for every n∈N tr [(B (λ)) ] = tr n

is valid.

0



dBn (λ) dλ



= ntr [B0 (λ)B(n−1) (λ)

(18)

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

407

Proof. Firstly we by the inductive method prove:

(Bn (λ))0 =

n −1 X

Bi (λ)B0 (λ)Bn−1−i (λ).

(19)

i=0

For n = 1 formula (19) is clear. For n = 2,

(B2 (λ))0 = B0 (λ)B(λ) + B(λ)B0 (λ) = (B2 (λ))0 =

1 X

Bi (λ)B0 (λ)B1−i (λ).

i =0

We suppose formula (19) is true for n = m. For n = m + 1 we have

(Bm+1 (λ))0 = (Bm (λ))0 B(λ) + Bm (λ)B0 (λ).

(20)

From (19) and (20) we have

"

#

m−1

(B

m+1

X

(λ)) = 0

B (λ)B (λ)B i

0

m−1−i

(λ) B(λ) + Bm (λ)B0 (λ)

i=0 m−1

=

X

Bi (λ)B0 (λ)Bm−i (λ) + Bm (λ)B0 (λ)

i =0

=

m X

Bi (λ)B0 (λ)Bm−i (λ).

i=0

Formula (19) is proved. From the condition of Theorem 4 and (19) we have tr [(Bn (λ))0 ] =

n −1 X

tr [Bi (λ)B0 (λ)Bn−1−i (λ)].

(21)

i =0

From B0 (λ) ∈ σ1 (H ), Bi (λ) ∈ σ1 (H ), i ≥ 0 and (21) we have tr [(Bn (λ))0 ] =

n −1 X

tr [B0 (λ)Bn−1−i (λ)Bi (λ)]

i =0

or tr [(Bn (λ))0 ] = n[B0 (λ)Bn−1 (λ)]. Theorem 4 is proved.



Theorem 5. Let operator function B(λ) ∈ σ1 (H ) be analytic at the point λ = λ0 with respect to the norm in σ1 (H ). Then scalar function trB(λ) also is analytic at the point λ = λ0 and tr B0 (λ0 ) = [tr B(λ)]0λ=λ0 . Proof. According to the condition of Theorem 5



B(λ0 + 1λ) − B(λ0 )

0

lim − B (λ0 )

1λ→0 1λ

σ1 (H )

= 0.

(22)

On the other hand





tr B(λ0 + 1λ) − B(λ0 ) − B0 (λ0 ) ≤ B(λ0 + 1λ) − B(λ0 ) − B0 (λ0 )

1λ 1λ

σ1 (H )

or



tr B(λ0 + 1λ) − tr B(λ0 ) B(λ0 + 1λ) − B(λ0 )

0 0

− tr B (λ0 ) ≤ − B (λ0 ) .

1λ 1λ σ1 (H )

(23)

408

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

From (22) and (23) we have

tr B(λ0 + 1λ) − tr B(λ0 )

lim

1λ

1λ→0

Theorem 5 is proved.

− tr B0 (λ0 ) = 0.



Now using the Theorems 3–5 we prove Theorem 2. From (15) and (18) we have tr {[(QR0λ )j ]0 } = jtr [(QR0λ )j−1 Q (R0λ )2 ]

= jtr [(QR0λ )j R0λ ] = jtr [R0λ (QR0λ )j ].

(24)

From (12) and (24) we have

(−1)j+1 2π ij

Mpj =

Z |λ|=bp

 λtr [(QR0λ )j ]0 dλ

Z  (−1)j+1 = tr [λ(QR0λ )j ]0 − (QR0λ )j dλ 2π ij |λ|=bp Z j Z  (−1) (−1)j+1 0 j = tr [(QRλ ) ]dλ + tr [λ(QR0λ )j ]0 dλ. 2π ij |λ|=bp 2π ij |λ|=bp

(25)

Using Theorem 5 we have

Z

tr [λ(QR0λ )j ]0 dλ =



|λ|=bp



Z

0

dλ.

0

dλ +

tr [λ(QR0λ )j ]

 |λ|=bp

(26)

Right side of this equality may be written as

Z

 |λ|=bp

tr [λ(QR0λ )j ]

0

dλ =

Z

 |λ|=bp Im λ≥0

tr [λ(QR0λ )j ]

Z

 |λ|=bp Im λ≤0

tr [λ(QR0λ )j ]

0

dλ.

(27)

Let ε0 be an arbitrary number such that bp + ε0 < µp+1 . The function tr [λ(QR0λ )j ] is analytic in the domains G1 and G2 where G1 = {λ : bp − ε0 < |λ| < bp + ε0 , Im λ > −ε},

G2 = {λ : bp − ε0 < |λ| < bp + ε0 , Im λ < −ε0 },

{λ : |λ| = bp , Im λ ≥ 0} ⊂ G1 {λ : |λ| = bp , Im λ ≤ 0} ⊂ G2 and using Leibnitz’s formula, from (27) we have

Z

 |λ|=bp

tr [λ(QR0λ )j ]

0

dλ = tr [−bp (QR0−bp )j ] − tr [bp (QR0bp )j ] + tr [bp (QR0bp )j ] − tr [−bp (QR0−bp )j ] = 0.

(28)

From (25)–(28) we have Mpj =

(−1)j 2π ij

Z |λ|=bp

tr [(QR0λ )j ]dλ.

Theorem 2 is proved. From relations (2) and (14) we have Mp1 = −

1

Z

2π i |λ|=bp

tr (QR0λ )dλ = −

1

∞ X ∞ X

Z

2π i |λ|=bp m=1 n=1

0 0 (QR0λ ψmn , ψmn )H1 dλ.

(29)

0 0 0 We can show that βm (λ) = (m = 1, 2, . . .) and m=1 βm (λ) are absolute and uniform n=1 (QRλ ψmn , ψmn )H1 convergent series with respect to λ on the circle |λ| = bp . Therefore from (29), we have

P∞

Mp1 =

∞ X ∞ X

0 0 (Q ψmn , ψmn )H1

m=1 n=1

=

2p X ∞ X

1

P∞

Z

dλ

2π i |λ|=bp λ − µm

0 0 (Q ψmn , ψmn )H1

m=1 n=1

=

Z π p X ∞ X 2 (Q (x)ϕn , ϕn )H [cos2 (2m − 1)x + sin2 (2m − 1)x]dx. π 0 m=1 n=1

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

409

By condition (4) and Lebesque’s theorem we get π

∞ Z X n =1

(Q (x)ϕn , ϕn )H dx =

π

Z 0

0

π

" # Z ∞ X (Q (x)ϕn , ϕn )H dx =

tr Q (x)dx.

0

n =1

Then Mp1 =

2p

π

Z

π

tr Q (x)dx.

(30)

0

Now we shall show that lim Mpj = 0,

j≥2

(31)

lim MpN = 0,

N ≥ 4.

(32)

For j = 2 from (14) we write Mp2

=

Z

1

tr (QRλ ) dλ =

1

0 2

4π i |λ|=bp

"

Z

4π i |λ|=bp

∞ X ∞ X

# (QRλ ) ψ , ψ 0 2

0 mn

m=1 n=1

0 mn H1




dλ.

(33)

Moreover we know that 0 QR0λ ψmn =

0 Q ψmn

µm − λ

and 0 0 (QR0λ )2 ψmn = (µm − λ)−1 QR0λ Q ψmn ( ) ∞ X ∞ X −1 0 0 0 0 = (µm − λ) QRλ (Q ψmn , ψrq )H1 ψrq r =1 q =1

∞ X ∞ X 0 = (µm − λ)−1 (µr − λ)−1 (Q ψmn , ψrq0 )Q ψrq0 .

(34)

r =1 q =1

Substituting (34) in (33) we obtain Mp2

=

1

"

Z

4π i |λ|=bp

0 0 ∞ X ∞ X ∞ X ∞ X (Q ψmn , ψrq0 )H1 (Q ψrq0 , ψmn )H1

(λ − µm )(λ − µr )

m=1 n=1 r =1 q=1

# dλ.

(35)

It is easy to verify that the formula dλ =0 (λ − µm )(λ − µr )

Z |λ|=bp

(36)

is true when µm , µr < bp and it is also true forµm , µr > bp . Then from (35) and (36) we have Mp2

=

=

2p ∞ ∞ ∞ 1 XX X X

2π i m=1 n=1 r =2p+1 q=1

|(Q ψ , ψ ) | 0 mn

0 2 rq H1

Z

(λ − µm )−1 (λ − µr )−1 dλ |λ|=bp

2p X ∞ X ∞ X ∞ X 0 (µm − µr )−1 |(Q ψmn , ψrq0 )H1 |2 . m=1 n=1 r =2p+1 q=1

Then we get

|Mp2 | ≤

∞ X ∞ ∞ X ∞ X X 0 (µr − µ2p )−1 |(Q ψrq0 , ψmn )H1 |2 r =2p+1 q=1

=

∞ X

(µr − µ2p )−1

r =2p+1

m=1 n=1

∞ X

kQ ψrq0 k2H1 .

(37)

q =1

By using (2) and condition (3) we estimate

P∞

q=1

kQ ψrq0 k2H1 .

410

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

If r is odd then ∞ X

kQ ψ k = 0 rq

π

∞ Z X

q =1

π

∞ Z X q =1

kQ ( x )

0

q =1

≤

r

2

π

cos(2r − 1)xϕq k2H dx π

∞ Z X

kQ (x)ϕq k2H dx =

0

0

q =1

kQ (x)ϕq k2H1 dx < C .

(38)

Here C is a positive constant. This estimation is also true for even r. From (37) and (38) we get

|Mp2 | ≤ C

∞ X

∞ X

(µr − µ2p )−1 ≤ 2C

r =2p+1

(2r − 1)2 − (2p − 1)2


−1

.

r =p+1

Also one can show that the equality ∞ X

(2r − 1)2 − (2p − 1)2

1


−1

< 2p− 2

(39)

r =p+1

is true. Hence 1

|Mp2 | ≤ 4Cp− 2 . This gives lim Mp2 = 0.

p→∞

In a similar way we can prove lim Mp3 = 0.

p→∞

Now, let us prove formula (31) for j ≥ 4. For this we estimate the value of expression kQR0λ kσ1 (H1 ) on the circle |λ| = bp . As shown in [1]

kQR0λ kσ1 (H1 ) ≤

∞ ∞ X X

0 kQR0λ ψmn kH1 .

m=1 n=1

By (3) and condition (3) we get

kQR0λ kσ1 (H1 ) < C1

∞ X

|(2m − 1)2 − λ|−1

m=1

= C1

p X

|(2m − 1) − λ|

−1

2

+

m=1

≤ C1

p X

= C1

|(2m − 1) − λ|

(|λ| − (2m − 1) )

2 −1

+

!

∞ X

((2m − 1) − |λ|)

< C1

m=p+1

((2p − 1) + 4p − (2m − 1) )

2 −1

2

+

p

−1

+

Here C1 > 0 is a constant.

∞  X 1 m=p+1

∞  X 1 m=p+1

< C1 1 +

∞ X

! ((2m − 1) − (2p − 1) − 4p) 2

−1

2

m=p+1

m=1

< C1 1 +

−1

2

m=1 p X

−1

2

m=p+1

m=1 p X

!

∞ X

∞ X

2

2

1

((2m − 1) − (2p − 1) ) + ((2m − 1) − (2p − 1) ) − 4p 2

2

2

2

−1 !

2

1

((2m − 1) − (2p − 1) ) + (2(p + 1) − 1) − (2p − 1) ) − 4p 2

2

2

2

(2m − 1)2 − (2p − 1)2 m=p+1

! .

2

2

−1 !

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

411

By (39) and the last inequality we have

kQR0λ kσ1 (H1 ) < C2 ,

|λ| = bp C2 > 0.

(40)

Now we estimate kR0λ kH1 on the circle |λ| = bp . For m ≤ p

|(2m − 1)2 − λ| ≥ |λ| − (2m − 1)2 = (2p − 1)2 + 4p − (2m − 1)2 > p and for m ≥ p + 1

|(2m − 1)2 − λ| ≥ (2m − 1)2 − |λ| ≥ (2(p + 1) − 1)2 − (2p − 1)2 − 4p > p. Hence

|(2m − 1)2 − λ|−1 < p−1 ;

|λ| = bp .

Then, by (6) we have

kR0λ kH1 < p−1 .

(41)

Similarly (by Theorem 1 and condition (2)) we can easily show that for sufficiently large p

kRλ kH1 < C3 p−1 ,

C3 > 0

(42)

on the circle |λ| = bp . From (14), (40) and (41) and condition (2) we get

|

Mpj

Z



Z

1 1 tr (QR0λ )j dλ ≤ k(QR0λ )j kσ1 (H1 ) |dλ| |= 2π j |λ|=bp 2π j |λ|=bp

≤

Z

1

2π j |λ|=bp

kQR0λ kσ1 (H1 ) k(QR0λ )j−1 kH1 |dλ|

Z ≤ C2

kQ k |λ|=bp

j −1 H1

j −1 R0λ H1

k k

|dλ| ≤ C2

Z

41−j p1−j |dλ| |λ|=bp

< C4 p3−j C4 > 0. This gives lim Mpj = 0,

j ≥ 4.

p→∞

And so the formulas (31) are proved. Now let us prove the formula (32). From (13) and (40)–(42) we get

Z



Z

1 |MpN | = λtr [Rλ (QR0λ )N +1 ]dλ ≤ |λ| tr [Rλ (QR0λ )N +1 ] |dλ| 2π |λ|=bp |λ|=bp Z ≤ bp |λ|=bp

Z ≤ bp |λ|=bp

≤ C 3 bp p

−1

kRλ (QR0λ )N +1 kσ1 (H1 ) |dλ| kRλ kH1 k(QR0λ )N +1 kσ1 (H1 ) |dλ|

Z |λ|=bp

kQR0λ kNH1 kQR0λ kσ1 (H1 ) |dλ|

≤ C3 bp p−1 4N p−N C2 2π bp ≤ C5 p3−N , So we obtain lim MpN = 0,

p→∞

N ≥ 4.

And the formula (32) is proved.

C5 > 0.

412

A. Bayramov et al. / Mathematical and Computer Modelling 49 (2009) 403–412

By using the relations (11) and (30)–(32) we get

# " Z p ∞ X X 2p π 2 tr Q (x)dx = 0 lim (λmn − 2(2m − 1) ) − p→∞ π 0 m=1 n=1 or

# " Z π ∞ ∞ X X 2 2 tr Q (x)dx = 0. (λmn − 2(2m − 1) ) − π 0 m=1 n=1

(43)

The series on the left hand side of this equality is called regularized trace of the operator L. We have proved the main result of this article which is given by the following theorem. Theorem 6. If the operator function Q (x) satisfies the conditions (1)–(4) then formula (43) holds for regularized trace of L. References [1] I.C. Gohberg, M.G. Krein, Introduction to the Theory of Linear Non-self Adjoint Operators, in: Translation of Mathematical Monographs, vol. 18, AMS, RI, 1969. [2] I.M. Gelfand, B.M. Levitan, On a formula for eigenvalues of a differential operator of second order, Dokl. Akad. Nauk SSSR 88 (4) (1953) 593–596. [3] L.A. Dikiy, A new method of calculation of eigenvalues of Sturm–Liouville operator, Dokl. Akad. Nauk. 116 (1) (1957). [4] L.A. Dikiy, On the traces of Sturm–Liouville differential operators, Uspechi Mat. Nauk. 13 (3) (1958) 111–143. (81). [5] W. Magnus, W. Winkler, Hill’s Equation, Interscience Wiley, New York, 1966. [6] V.A. Marchenko, Operatory Shturma–Liuvillya i ikh Prilozheniya, in: Sturm–Liouville Operators and Their Applications, Kiev, 1977. [7] V.A. Sadovnichii, Teoriya Operatorov, Izd. Mosk. Universiteta, 1979. [8] B.M. Levitan, I.S. Sargsyan, Sturm–Liouville and Dirac Operators, Kluwer, Dordrecht, 1991. [9] R.Z. Khalilova, On arranging Sturm–Liouville operator equation’s trace, Funks, Analiz, Teoriya Funksiy i ik pril Mahachkala 3 (part I) (1976) 154–161. [10] F.G. Maksudov, M. Bayramoglu, E.E. Adiguzelov, On regularized trace of Sturm–Liouville operator on a finite interval with the unbounded operator coefficient, Dokl. Akad. Nauk SSSR 30 (1) (1984) 169–173. [11] M. Bayramoglu, E.E. Adiguzelov, On a regularized trace formula for the Sturm–Liouville operator with a bounded operator coefficient and with a singularity, Differential Equations 32 (12) (1997) 1581–1585 (1996). [12] I. Albayrak, M. Bayramoglu, E.E. Adiguzelov, The second regularized trace formula for the Sturm–Liouville problem with spectral parameter in a boundary condition, Methods Funct. Anal. Topology 6 (3) (2000) 1–8. [13] V.A. Sadovnichii, V.E. Podol’skii, Traces of operators with the relative compact perturbation, Mat. Sb. 193 (2) (2002) 129–152. [14] T. Kato, Perturbation Theory for Linear Operators, Springer, 1980.