Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 31, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
REGULARIZED TRACE FORMULA FOR HIGHER ORDER
DIFFERENTIAL OPERATORS WITH UNBOUNDED
COEFFICIENTS
ERDOĞAN ŞEN, AZAD BAYRAMOV, KAMIL ORUÇOĞLU
Abstract. In this work we obtain the regularized trace formula for an evenorder differential operator with unbounded operator coefficient.
1. Introduction
The first work about the theory of regularized traces of differential operators
belongs to Gelfand and Levitan [1]. They considered the Sturm-Liouville operator
−y 00 + [q(x) − λ]y = 0,
with boundary conditions
y 0 (0) = y 0 (π) = 0,
Rπ
where q(x) ∈ C 1 [0, π]. Under the condition 0 q(x)dx = 0 they obtained the
formula
∞
X
1
(µn − λn ) = (q(0) + q(π)).
4
n=0
Gul [2] obtained the formula
nm
X
1
(λk − µk ) = [tr Q(π) − tr Q(0)]
lim
m→∞
4
k=1
for the regularized trace of the second order differential operator
l[y] = −y 00 (x) + Ay(x) + Q(x)y(x)
with unbounded operator coefficient and with the boundary conditions y(0) =
y 0 (π) = 0. Here λk and µk are the eigen-elements of the operators
l0 [y] = −y 00 (x) + Ay(x)
l[y] = −y 00 (x) + Ay(x) + Q(x)y(x)
with the same boundary conditions y(0) = y 0 (π) = 0 respectively.
Adıgüzelov and Sezer [3] obtained a regularized trace formula for a self-adjoint
differential operator of higher order with unbounded operator coefficient. Articles
2010 Mathematics Subject Classification. 47A10, 47A55.
Key words and phrases. Hilbert space; self-adjoint operator; spectrum; trace class operator;
regularized trace.
c
2016
Texas State University.
Submitted April 1, 2015. Published January 20, 2016.
1
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E. ŞEN, A. BAYRAMOV, K. ORUÇOĞLU
EJDE-2016/31
[4, 5, 6, 7, 8] are devoted to study of regularized trace formulas of differential operators with bounded operator coefficient. Bayramov et al [9] obtained the second regularized trace formula for the differential operator equation with the semi-periodic
boundary conditions. Makin [10] established a formula for the first regularized trace
of the Sturm-Liouville equation with a complex-valued potential and with irregular
boundary conditions.
Let H be a separable Hilbert space and let H1 = L2 (H; [0, π]) denote the set of
all strongly measurable functions f with values in H and such that
Z π
kf (t)k2H dt < ∞.
0
and the scalar function (f (t), g) is Lebesgue measurable for every g ∈ H in the
interval [0, π]. Here (·, ·) denotes the inner product in H and k · k denotes the norm
in H.
If the inner product of two arbitrary elements f and g of the space H1 is defined
by
Z π
(f, g)H1 =
(f (t), g(t))H dt, f, g ∈ H1
0
then H1 becomes a separable Hilbert space [11]. σ∞ (H) denotes the set of all
compact operators from H into H. If A ∈ σ∞ (H), then AA∗ is a nonnegative selfadjoint operator and (A∗ A)1/2 ∈ σ∞ (H). Let the non-zero eigen-elements of the
operator (A∗ A)1/2 be s1 ≥ s2 ≥ · · · ≥ sq (0 ≤ q ≤ ∞). Here, each eigen-element
is counted according to its own multiplicity. The numbers s1 , s2 , . . . , sq are called
s-numbers of the the operator A. σ1 (H) is the set of allPthe operators A ∈ σ∞ (H)
∞
such that the s-numbers of which satisfy the condition q=1 sq < ∞. An operator
is called a trace class operator if it belongs to σ1 (H).
Let us consider the operators l0 and l in H1 defined by
l0 [u] = (−1)m u(2m) (t) + Au(t),
m (2m)
l[u] = (−1) u
(t) + Au(t) + Q(t)u(t)
(1.1)
(1.2)
with the same boundary conditions y (2i−2) (0) = y (2i−1) (π) = 0 (i = 1, 2, . . . , m)
respectively. Here A : Ω(A) → H is a densely defined self-adjoint operator in H
with A = A∗ ≥ E where E : H → H is identity operator and A−1 ∈ σ∞ (H).
We also should note that our problem’s boundary conditions are different from the
considered problem’s boundary conditions in [3] which arise new difficulties.
Let η1 ≤ η2 ≤ · · · ≤ ηn ≤ . . . be the eigen-elements of the operator A and
ϕ1 , ϕ2 , . . . , ϕn , . . . be the orthonormal eigenvectors corresponding to these eigenelements. Here, each eigenvalue is counted according to its own multiplicity number. Let Ω(L00 ) denote the set of the functions u(t) of the space H1 satisfying the
following conditions:
(a) u(t) has continuous derivative of the 2m order with respect to the norm in
the space H in the interval [0, π];
(b) u(t) ∈ Ω(A) for every t ∈ [0, π] and Au(t) is continuous with respect to the
norm in the space H.
(c) y (2i−2) (0) = y (2i−1) (π) = 0 (i = 1, 2, . . . , m).
Here Ω(L00 ) = H1 . Let us consider the linear operator L00 u = l0 u from D(L0 ) to
H1 . L00 is a symmetric operator. The eigen-elements of L00 are ( 12 + k)2m + ηj
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TRACE FORMULA FOR AN EVEN-ORDER DIFFERENTIAL OPERATOR
3
(k = 0, 1, 2, . . . ; j = 1, 2, . . . ). and the orthonormal eigenvectors corresponding to
these eigen-elements are
r
2
1
sin ( + k)t ϕj (k = 0, 1, 2, . . . ; j = 1, 2, . . . ).
π
2
2. Some relations about the eigen-elements and resolvents
Let the eigenvalues of the operators L0 and L be µ1 ≤ µ2 ≤ · · · ≤ µn ≤ . . . and
λ1 ≤ λ2 ≤ · · · ≤ λn ≤ . . . respectively. Let N (µ) be the number of eigen-elements
of operator L0 which is not greater than a positive number µ. If ηj ∼ aj α as j → ∞
2m
(a > 0, α > 2m−1
) that is, if
ηj
=1
lim
j→∞ aj α
then using the same method in [12] it can be found that N (µ) ∼ dµ
Z π/2
1
2
2
d=
(sin τ ) α −1 (cos τ )1+ m dτ
1/a
αa
0
2m+α
2mα
, where
and hence
2mα
µn ∼ d0 n 2m+α
2mα
as j → ∞ (d0 = d 2m+α ).
(2.1)
Let Q(t) be an operator function satisfying the following conditions:
(1) Q(t) : H → H is a self-adjoint operator for every t ∈ [0, π];
(2) Q(t) is weakly measurable in the interval [0, π];
(3) The norm function kQ(t)k is bounded in the interval [0, π];
(4) Q(t) has weak derivative of the second order in the interval [0, π];
(5) The function (Q0 (t)f, g) is continuous for every f, g ∈ H;
(6) Q(i) (t) : H → H (i = 0, 1, 2) are self-adjoint trace class operators and the
functions kQ(i) (t)kσ1 (H) (i = 0, 1, 2) are bounded and measurable in the
interval [0, π].
Since Q is a self-adjoint operator from H1 to H1 for every y ∈ H1 we have
|(Qy, y)H1 | ≤ kQykH1 kykH1 ≤ kQkH1 kyk2H1
or
(−kQky, y)H1 ≤ (Qy, y)H1 ≤ (kQky, y)H1 .
This means that
−kQkH1 E ≤ Q ≤ kQkH1 E.
And so
L0 − kQkH1 E ≤ L = L0 + Q ≤ L0 + kQkH1 E.
In this situation, it is well-known that (Smirnov, [13])
µn − kQkH1 ≤ λn ≤ µn + kQkH1 .
According to this, we can write
1−
λn
kQkH1
kQkH1
≤
≤1+
.
µn
µn
µn
By applying limit to each side of this inequality and by considering the equality
µn
lim
2mα ,
n→∞ d n 2m+α
0
4
E. ŞEN, A. BAYRAMOV, K. ORUÇOĞLU
EJDE-2016/31
we get limn→∞ λn /µn = 1. Thus we have
λn
lim
n→∞
d0 n
= lim
2mα
2m+α
n→∞
λn
λn
µn
µn
= lim
=1
lim
2mα
2mα
n→∞ µn n→∞ d n 2m+α
µn d0 n 2m+α
0
or as n → ∞,
2mα
λn ∼ d0 n 2m+α .
(2.2)
By using the formula (2.1), it is easily seen that the sequence {µn } has a subsequence µn1 < µn2 < · · · < µnp < . . . such that
2mα 2mα 2mα
q = np + 1, np + 2, . . . ; d0 = d 2m+α . (2.3)
µq − µnp > d0 q 2m+α − np2m+α
Let Rλ0 = (L0 − λE)−1 , Rλ = (L − λE)−1 be the resolvents of the operators L0
2m
and L respectively. If α > 2m−1
then, by the formulas (2.1) and (2.2), Rλ0 and Rλ
are trace class operators for λ 6= λq , µq (q = 1, 2, . . . ). In this situation
tr(Rλ − Rλ0 ) = trRλ − trRλ0 =
∞
X
(
q=1
1
1
−
),
λq − λ µq − λ
(2.4)
see Cohberg and Krein [14].
Let |λ| = dp = 2−1 (µnp +1 + µnp ). It is easy to see that for large values of p
the inequalities µnp < dp < µnp +1 and λnp < dp < λnp +1 are satisfied. The series
P∞
P∞
λ
λ
q=1 λq −λ and
q=1 µq −λ are uniform convergent on the circle |λ| = dp . Hence
with the help of inequality (2.3), we obtain
Z
np
X
1
(λq − µq ) = −
λ tr(Rλ − Rλ0 )dλ,
(2.5)
2πi
|λ|=d
p
q=1
where i2 = −1.
Lemma 2.1. If ηj ∼ aj α as j → ∞ (a > 0, α >
(2δ+1)
2d−1
0 δnδ−1
p
2mα
2m+α
(δ =
2m
2m−1 )
then kRλ0 kσ1 (H1 ) <
− 1) on the circle |λ| = dp .
0
0
Proof. For λ ∈
/ {µq }∞
q=1 , since Rλ is a normal operator we have kRλ kσ1 (H1 ) =
P∞
1
q=1 |µq −λ| [14]. On the circle |λ| = dp we have
kRλ0 kσ1 (H1 )
≤
∞
X
q=1
np
∞
X
X
1
2
2
=
+
||λ| − µq | q=1 µnp + µnp +1 − 2µq q=n +1 2µq − µnp − µnp +1
p
np
≤
X
2
q=1
µnp +1 − µq
where Dp =
obtain
∞
X
+
q=np
P∞
k=np +1 (µk
(2.6)
np
X
2
2
=
+ 2Dp ,
2µ
−
µ
µ
− µq
q
n
n
+1
p
p
+1
q=1
− µnp )−1 (p = 1, 2, . . . ). By using the inequality (2.3) we
np
X
2
q=1
µnp +1 − µq
<
np
np
<
µnp +1 − µnp
d0 ((np + 1)1+δ − n1+δ
p )
n1−δ
np
p
<
<
,
d0 (np + 1)δ
d0
(2.7)
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TRACE FORMULA FOR AN EVEN-ORDER DIFFERENTIAL OPERATOR
∞
X
Dp =
(µk − µnp )−1 <
k=np +1
=
d−1
0 [
∞
X
1
d0
k=np +1
k 1+δ
5
1
− n1+δ
p
(2.8)
∞
X
1
+
].
1+δ − n1+δ
((np + 1)1+δ − n1+δ
k
)
p
p
k=np +2
1
It is easy to see that
∞
X
1
≤
1+δ
k
− n1+δ
p
k=np +2
Z
∞
np +1
t1+δ
Z
∞
np +1
t1+δ
dt
,
− n1+δ
p
dt
1
<
δ .
1+δ
− np
δ[((np + 1)1+δ − n1+δ
p )] 1+δ
Taking into account the inequality (2.8) we obtain
Dp < d−1
0
δ+1
δ[((np + 1)1+δ −
δ
n1+δ
p )] 1+δ
< d−1
0
δ+1
δ2
.
(2.9)
δnp1+δ
With the help of (2.6), (2.7) and (2.9), it follows that on the circle |λ| = dp ,
kRλ0 kσ1 (H1 ) < 2d−1
0
(2δ + 1)
.
δnδ−1
p
Lemma 2.2. If the operator function Q(t) satisfies the conditions (1)–(3), and
2m
ηj ∼ aj α as j → ∞ (a > 0, α > 2m−1
), then for |λ| = dp and large values of p,
−δ
kRλ kH1 < 4d−1
0 np .
Proof. Since the s-numbers of the trace class operator Rλ are { λ11−λ , λ21−λ , . . . ,
1
λq −λ , . . . }, it follows that
kRλ kH1 = max{
1
1
1
,
,...,
, . . . }.
λ1 − λ λ2 − λ
λq − λ
On the circle |λ| = dp ,
|λq | − |λ| = |λq | − 2−1 (µn + µn +1 ) = 2−1 µn + µn +1 − 2|λq |.
p
p
p
p
(2.10)
(2.11)
Using the inequality q ≤ np and for the large values of p, since |λq | ≤ λnp , we have
µnp + µnp +1 − 2|λq |
≥ µnp + µnp +1 − 2λnp = µnp +1 − µnp + 2(µnp − λnp )
(2.12)
≥ µnp +1 − µnp − 2|µnp − λnp |.
Considering |µq − λq | ≤ kQkH1 (q = 1, 2, . . . ) by (2.12) we obtain
µnp + µnp +1 − 2|λq | ≥ µnp +1 − µnp − 2kQkH1 (q ≤ np ).
(2.13)
With the help of inequality q ≥ np + 1 and for the large values of p, since |λq | =
λq ≥ λnp +1 then
2|λq | − µnp − µnp +1 ≥ 2λnp +1 − µnp − µnp +1
= 2(λnp +1 − µnp +1 ) + µnp +1 − µnp
≥ µnp +1 − µnp − 2|λnp +1 − µnp +1 |.
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E. ŞEN, A. BAYRAMOV, K. ORUÇOĞLU
EJDE-2016/31
Using the above inequality,
2|λq | − µnp − µnp +1 ≥ µnp +1 − µnp − 2kQkH1
(q ≥ np + 1).
(2.14)
Taking into account that limp→∞ (µnp +1 − µnp ) = ∞ and by (2.11), (2.13) and
(2.14), on the circle |λ| = dp we have
|λq | − |λ| > 4−1 (µnp +1 − µnp ).
(2.15)
By (2.3) and (2.15) we obtain
−1
|λq | − |λ| > 4−1 d0 ((np + 1)1+δ − n1+δ
d0 (np + 1)δ .
p )>4
From the above inequality and |λ| = dp for the sufficiently large values of p, we
have
|λq − λ| > 4−1 d0 nδp .
−δ
From (2.10) and the above inequality we have 4d−1
0 np .
3. Regularized trace formula
We know from operator theory that for the resolvents of the operators L0 and
L the following formula holds:
Rλ = Rλ0 − Rλ QRλ0
(λ ∈ ρ(L0 ) ∩ ρ(L)).
Using the above formula and (2.5), it can be easily shown that
np
s
X
X
(λq − µq ) =
Upj + Up(s) ,
q=1
(3.1)
j=1
where
Upj =
(−1)j
2πij
Up(s) =
Z
tr[(QRλ0 )j ]dλ (i2 = −1; j = 1, 2, . . . ),
(3.2)
λtr[Rλ (QRλ0 )s+1 ]dλ (i2 = −1).
(3.3)
|λ|=dp
s Z
(−1)
2πi
|λ|=dp
Theorem 3.1. If the operator function Q(t)
satisfies the the conditions (1)–(3)
√
2m(1+ 2)
α
and ηj ∼ aj as j → ∞ (a > 0, α > 2√2m−√2−1 ) then
lim Upj = 0
p→∞
(j = 2, 3, 4, . . . ).
Proof. According to (3.2) for Up2 we have the equality
Z
np
∞
i
dλ
1 X X h
(Qψj , ψk )H1 (Qψk , ψj )H1 . (3.4)
Up2 =
2πi j=1
|λ|=dp (λ − µj )(λ − µk )
k=np +1
Therefore,
|Up2 | ≤ kQk2H1 Dp .
(3.5)
By (2.9) and (3.5) we obtain
lim Up2 = 0
p→∞
(α >
2m
).
2m − 1
(3.6)
Let us show that
lim Up3 = 0.
p→∞
(3.7)
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TRACE FORMULA FOR AN EVEN-ORDER DIFFERENTIAL OPERATOR
7
By using (3.2) it follows that
Up3 =
np np
∞
X
X X
[G(j, k, s) + G(s, k, j) + G(j, s, k]
J=1 k=1 s=np +1
+
np
∞
X
X
(3.8)
∞
X
[G(j, k, s) + G(s, k, j) + G(k, j, s],
J=1 k=np +1 s=np +1
where
G(j, k, s) = g(j, k, s)(Qψj , ψk )H1 (Qψk , ψs )H1 (Qψs , ψj )H1 ,
Z
1
dλ
g(j, k, s) =
.
6πi |λ|=dp (λ − µj )(λ − µk )(λ − µs )
Taking into account g(j, k, s) = g(j, k, s) and Q = Q∗ we obtain
G(s, k, j) = G(j, k, s),
G(k, j, s) = G(j, k, s),
G(j, s, k) = G(j, k, s).
(3.9)
With the help of (3.8) and (3.9) we obtain
Up3 = E1 + E2
with
E1 =
np np
∞
X
X X
(G(j, k, s) + 2G(j, k, s)),
J=1 k=1 s=np +1
E2 =
np
∞
X
X
∞
X
(G(j, k, s) + 2G(j, k, s))
J=1 k=np +1 s=np +1
and
E1 = E11 + E11 ,
E2 = E21 + E21 ,
(3.10)
with
E11 =
np np
∞
X
X X
G(j, k, s),
J=1 k=1 s=np +1
E21 =
np
∞
X
X
∞
X
G(j, k, s).
J=1 k=np +1 s=np +1
It is not hard to see that the following inequalities hold:
2
1−2δ
1+δ
kQk3H1 np 1+δ ,
2
d0 δ
−2δ 2
1+δ 2
|E21 | ≤ (
) kQk3H1 np1+δ .
d0 δ
|E11 | ≤
It follows that
lim Up3 = 0.
p→∞
(3.11)
(3.12)
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E. ŞEN, A. BAYRAMOV, K. ORUÇOĞLU
EJDE-2016/31
Now, let us show that the equality limp→∞ Upj = 0 (j = 4, 5, . . . ) holds. According
to (3.2),
Z
1
|Upj | ≤
| tr(QRλ0 )j ||dλ|
2πj |λ|=dp
Z
≤
k(QRλ0 )j kσ1 (H1 ) |dλ|
|λ|=dp
Z
k(QRλ0 )kσ1 (H1 ) k(QRλ0 )j−1 kH1 |dλ|
≤
|λ|=dp
Z
(3.13)
kQkH1 kRλ0 kσ1 (H1 ) k(QRλ0 )j−1 kH1 |dλ|
≤
|λ|=dp
Z
≤ kQkH1
|λ|=dp
Z
≤ const.
|λ|=dp
kRλ0 kσ1 (H1 ) k(QRλ0 )kj−1
H1 |dλ|
kRλ0 kσ1 (H1 ) kRλ0 kj−1
H1 |dλ|.
Since Rλ = Rλ0 for Q(t) ≡ 0 according to Lemma 2.2, on the circle |λ| = dp ,
2mα
− 1).
2m + α
Using Lemma 2.1 and the inequalities (3.13) and (3.14) one obtains
Z
|Upj | < const.n1−δj
|dλ| < const.n1−δj
dp .
p
p
−δ
kRλ0 kH1 < 4d−1
0 np
(δ =
(3.14)
|λ|=dp
For the sufficiently large values of p, since dp = 2−1 (µnp + µnp +1 ) ≤ const.n1+δ
p ,
then we obtain
|Upj | < const.np2−δ(j−1) .
It is easy to see that if δ >
2
3
or α >
10m
6m−5 ,
lim Upj = 0
p→∞
However, if
then
(j = 4, 5, . . . ).
(3.15)
√
10m
2m(1 + 2)
√
√
,
>
6m
−5
2 2m − 2 − 1
considering (3.6) and (3.7) as α >
2, 3, . . . ).
√
2m(1+ 2)
√
√
2 2m− 2−1
one obtains limp→∞ Upj = 0 (j =
Since the eigen-elements of the operator L0 are
1 2m
k+
+ ηj (k = 0, 1, 2, . . . ; j = 1, 2, . . . ),
2
then for q = 1, 2, . . . ,
1 2m
µq = kq +
+ ηjq .
(3.16)
2
Theorem 3.2. If the operator function
Q(t) satisfies the conditions (4)–(6)and
√
2m(1+ 2)
√
ηj ∼ aj α as j → ∞ (a > 0, α > 2√
) then
2m− 2−1
lim
p→∞
np h
X
q=1
λq − µq − π −1
Z
0
π
i
(Q(t)ϕjq , ϕjq )dt = 4−1 [tr Q(π) − tr Q(0)]
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TRACE FORMULA FOR AN EVEN-ORDER DIFFERENTIAL OPERATOR
9
where {jq }∞
q=1 is a set of natural numbers satisfying (3.16).
Proof. From (3.2),
Up1
1
=−
2πi
Z
tr(QRλ0 )dλ.
(3.17)
|λ|=dp
Since QRλ0 is a trace class operator for each λ ∈ ρ(L0 ) and {Ψ1 (t), Ψ2 (t), . . . } is an
orthonormal basis of the space H1 , then
∞
X
tr(QRλ0 ) =
(QRλ0 Ψq , Ψq )H1 .
q=1
By putting
tr(QRλ0 )
into (3.17) and considering
Rλ0 Ψq = (L0 − λE)−1 Ψq = (µq − λ)−1 Ψq ,
one obtains
Up1
1
=−
2πi
Z
1
2πi
Z
=−
[
∞
X
(QRλ0 Ψq , Ψq )H1 ]dλ
|λ|=dp q=1
∞
X
(µq − λ)−1 (QΨq , Ψq )H1 ]dλ
[
(3.18)
|λ|=dp q=1
Z
∞
X
1
= [ (QΨq , Ψq )H1 ]
(λ − µq )−1 dλ.
2πi
|λ|=d
p
q=1
Since the orthonormal eigenvectors according to the eigen-elements (k + 12 )2m + ηj
q
1
2
(k = 0, 1, 2, . . . ; j = 1, 2, . . . ) of the operator L0 are
π sin((k + 2 )t)ϕj (k =
0, 1, 2, . . . ; j = 1, 2, . . . ), it follows that
r
2
1 Ψq (t) =
sin (k + )t ϕjq q = 1, 2, . . . .
(3.19)
π
2
Further,
(
Z
1, q ≤ np ,
1
−1
(λ − µq ) dλ =
(3.20)
2πi |λ|=dp
0, q > np
and by using (3.18)–(3.20) we find that
np
np Z π
X
X
Up1 =
(QΨq , Ψq )H1 =
(Q(t)Ψq (t), Ψq (t))
=
=
=
=
q=1
np Z π
X
q=1
r
Q(t)
q=1 0
np Z π
X
2
π
q=1 0
np Z π
1X
π q=1
1
π
2
1 sin (kq + )t ϕjq ,
π
2
r
2
1 sin (kq + )t ϕjq dt
π
2
1 sin2 (kq + )t (Q(t)ϕjq , ϕjq )dt
2
1 − cos (2kq + 1)t
Q(t)ϕjq , ϕjq dt
0
np
XZ
q=1
0
0
π
(Q(t)ϕjq , ϕjq )dt −
np Z
1X π
cos((2kq + 1)t)(Q(t)ϕjq , ϕjq )dt.
π q=1 0
10
E. ŞEN, A. BAYRAMOV, K. ORUÇOĞLU
EJDE-2016/31
By subtracting and adding the expression (Q(t)ϕjq , ϕjq ) cos(2kq t) into the second
integral on the right side of last equality one obtains
np Z
np Z
1X π
1X π
Up1 =
(Q(t)ϕjq , ϕjq )dt +
cos(2kq t)(Q(t)ϕjq , ϕjq )dt
π q=1 0
π q=1 0
−
np Z
1X π
[cos((2kq + 1)t) + cos(2kq t)](Q(t)ϕjq , ϕjq )dt
π q=1 0
We can write the expresssion
np Z
1X π
cos(rq t)(Q(t)ϕjq , ϕjq )dt,
−
π q=1 0
instead of first term in the right-hand side of the above equality. Thus, we have
np Z
np Z
1X π
1X π
Up1 =
(Q(t)ϕjq , ϕjq )dt +
cos(2kq t)(Q(t)ϕjq , ϕjq )dt
π q=1 0
π q=1 0
−
np Z
1X π
cos(rq t)(Q(t)ϕjq , ϕjq )dt.
π q=1 0
We can write this equation in the form
∞ Z
∞ ∞ Z
1 XX π
1X π
cos(rt)(Q(t)ϕj , ϕj )dt
(Q(t)ϕj , ϕj )dt −
lim Up1 =
p→∞
π j=1 0
π j=1 r=1 0
Z
∞ ∞
1 XXh π
+
cos(kt)(Q(t)ϕj , ϕj )dt
2π j=1
0
k=1
Z π
i
+ (−1)k
cos(kt)(Q(t)ϕj , ϕj )dt
0
and so we have
lim Up1 =
p→∞
∞ Z
1X π
(Q(t)ϕj , ϕj )dt
π j=1 0
Z
∞
∞
i
o
1 XnXh2 π
(Q(t)ϕj , ϕj ) cos rt dt cos r0
−
2 j=1 r=1 π 0
Z
∞
∞
i
1 XnXh2 π
+
(Q(t)ϕj , ϕj ) cos ktdt cos k0
4 j=1
π 0
k=1
∞ h Z π
i
o
X
2
+
(Q(t)ϕj , ϕj ) cos kt dt cos kπ .
π 0
(3.21)
k=1
The sum with respect to r in the first term on the right side of this expression in the
value at 0 of Fourier
series according to functions {cos rx}∞
r=0 in the interval [0, π] of
Rπ
the function 0 (Q(t)ϕj , ϕj )H having the derivative of second order. Analogically,
the sums in the second term with respect to k are the values at the points 0 and π
respectively of Fourier series with respect to the functions {cos kx}∞
k=0 in the same
interval of that function.
EJDE-2016/31
TRACE FORMULA FOR AN EVEN-ORDER DIFFERENTIAL OPERATOR 11
Also
p
p
X
X
|(Q(t)ϕj , ϕj )| ≤ kQ(t)kσ1 (H) ,
(Q(t)ϕj , ϕj ) ≤
(p = 1, 2, . . . ).
(3.22)
j=1
j=1
Since the norm function kQ(t)kσ1 (H) is bounded and measurable in the interval
[0, π], we have
Z π
kQ(t)kσ1 (H) dt < ∞.
(3.23)
0
By using (3.22), (3.23) and Lebesgue theorem we obtain
Z π hX
Z π
∞ Z π
∞
i
X
(Q(t)ϕj , ϕj )dt =
(Q(t)ϕj , ϕj ) dt =
tr Q(t)dt.
j=1
0
0
j=1
(3.24)
0
Furthermore, as in the proof of (3.15) by Lemma 2.1 and Lemma 2.2, we can show
that
3
lim Up(s) = 0 (s > ) .
(3.25)
p→∞
δ
Therefore by (3.1), (3.21), (3.24) and (3.25), we obtain
Z π
np h
i
X
lim
λq − µq − π −1
(Q(t)ϕjq , ϕjq )dt = 4−1 [tr Q(π) − tr Q(0)]. (3.26)
p→∞
q=1
0
The limit on the left hand side of the equality (44) is said to be regularized trace
of the operator L (see [1]).
References
[1] I. M. Gelfand, B. M. Levitan; On a simple identity for the characteristic values of a differential operator of the second order, Doklady Akademii Nauk SSSR, 88 (1953), 593–596.
[2] E. Gul; A regularized trace formula for a differential operator of second order with unbounded
operator coefficients given in a finite interval, International Journal of Pure and Applied
Mathematics, 32 (2) (2006), 223–242.
[3] E. Adıgüzelov, Y. Sezer; The regularized trace of a self adjoint differential operator of
higher order with unbounded operator coefficient, Applied Mathematics and Computation,
218 (2011), 2113-2121.
[4] E. E. Adıgüzelov, A. Bayramov, O. Baykal; On the spectrum and regularized trace of the
Sturm–Liouville problem with spectral parameter on the boundary condition and with the
operator coefficient, International Journal of Differential Equations and Applications, 2 (3)
(2001), 317–333.
[5] E. E. Adıgüzelov, H. Avcı, E. Gul; The trace formula for Sturm–Liouville operator with
operator coefficient, Journal of Mathematical Physics, 42 (6) (2001), 1611–1624.
[6] E. Adıgüzelov, Y. Sezer; On Spektrum of a self adjoint differential operator of higher order with unbounded operator coefficient and asymptotic behaviour of eigenvalues, Journal of
Engineering and Natural Sciences Sigma, 4 (2006), 32–49.
[7] E. E. Adıgüzelov, O. Baksi; On the regularized trace of the differantial operator equation given
in a finite interval, Journal of Engineering and Natural Sciences Sigma, 1 (2004), 47–55.
[8] I. Albayrak, K. Koklu, A. Bayramov; A Regularized trace formula for differential equations
with trace class operator coefficients, Rocky Mountain Journal of Mathematics, 40 (4) (2010),
1095–1110.
[9] A. Bayramov, S. K. Çalişkan, F. A. Akgün; The second regularized trace of differential
operator equation with the semi-periodic boundary conditions, Mathematical Methods in the
Applied Sciences, 35 (2012) 2185-2197.
[10] A. Makin; Regularized trace of the Sturm-Liouville operator with irregular boundary conditions, Electronic Journal of Differential Equations (EJDE), 2009 (27) (2009), 1-8.
12
E. ŞEN, A. BAYRAMOV, K. ORUÇOĞLU
EJDE-2016/31
[11] A. A. Kirillov; Elements of the Theory of Representations, Springer of Verlag, New York,
1976.
[12] V. I. Gorbachuk, M. L. Gorbachuk; Classes of boundary value problems for the SturmLiouville equation with an operator potential, Ukrainian Mathematical Journal, 24 (3) (1972),
291-305.
[13] V. I. Smirnov; A Course of Higher Mathematics, Volume 5, Pergamon Press, New York,
1964.
[14] C. Cohberg, M. G. Krein; Introduction to the Theory of Linear Non-self Adjoint Operators,
Translation of Mathematical Monographs, Vol. 18, American Mathematical Society, Providence, Rhode Island, 1969.
Erdoğan Şen
Department of Mathematics, Faculty of Arts and Science, NamıkKemal University,
59030, Tekirdağ, Turkey
E-mail address: erdogan.math@gmail.com
Azad Bayramov
Department of Mathematics Education, Faculty of Education, Recep Tayyip Erdogan
University, Rize, Turkey
E-mail address: azadbay@gmail.com
Kamil Oruçoğlu
Department of Mathematics Engineering, Istanbul Technical University, Maslak, 34469,
Istanbul, Turkey
E-mail address: koruc@itu.edu.tr