arXiv:1608.05576v1 [math.SP] 19 Aug 2016
ON THE NORMING CONSTANTS OF THE STURM-LIOUVILLE
PROBLEM
TIGRAN HARUTYUNYAN, AVETIK PAHLEVANYAN
Abstract. We derive new asymptotic formulae for the norming constants of
Sturm-Liouville problem with summable potentials, which generalize and make
more precise previously known formulae. Moreover, our formulae take into
account the smooth dependence of norming constants on boundary conditions.
We also find some new properties of the remainder terms of asymptotics.
1. Introduction and Statement of the Results
Let L (q, α, β) denote the Sturm–Liouville problem
− y ′′ + q (x) y = µy, x ∈ (0, π) , µ ∈ C,
y (0) cos α + y ′ (0) sin α = 0, α ∈ (0, π] ,
(1.1)
(1.2)
′
y (π) cos β + y (π) sin β = 0, β ∈ [0, π) ,
(1.3)
1
where q is a real-valued, summable function on [0, π] (we write q ∈ LR [0, π]). By
L (q, α, β) we also denote the self-adjoint operator, generated by the problem (1.1)–
(1.3) in Hilbert space L2 [0, π] (see [1, 2]). It is well-known that the spectra of
L (q, α, β) is discrete and consists of real, simple eigenvalues (see [1, 2, 3]), which
we denote by µn (q, α, β) , n = 0, 1, 2 . . . , emphasizing the dependence of µn on q,
α and β. For µn the following asymptotic formula have been proven in [4]:
Z
1 π
2
q (t) dt + rn (q, α, β) ,
(1.4)
µn (q, α, β) = [n + δn (α, β)] +
π 0
where δn is the solution of the equation (for n ≥ 2)
δn (α, β) =
1
cos α
−
arccos q
π
(n + δn (α, β))2 sin2 α + cos2 α
−
cos β
1
arccos q
, (1.5)
π
(n + δn (α, β))2 sin2 β + cos2 β
and rn (q, α, β) = o (1) , when n → ∞, uniformly in α, β ∈ [0, π] and q from any
bounded subset of L1R [0, π] (we will write q ∈ BL1R [0, π]). It follows from (1.5) (for
details see [4]), that
δn (α, β) =
cot β − cot α
+ O 1/n2 , α, β ∈ (0, π) ,
πn
(1.5a)
2010 Mathematics Subject Classification. 34B24, 34L20.
Key words and phrases. Sturm-Liouville problem, norming constants, asymptotics of the solutions, asymptotics of spectral data.
1
2
TIGRAN HARUTYUNYAN, AVETIK PAHLEVANYAN
δn (π, β) =
δn (α, 0) =
1
cot β
1
+ O 1/n2 = + O (1/n) , β ∈ (0, π) ,
+
1
2 π n+ 2
2
1
cot α
1
+ O 1/n2 = + O (1/n) , α ∈ (0, π) ,
−
1
2 π n+ 2
2
δn (π, 0) = 1.
(1.5b)
(1.5c)
(1.5d)
Let y = ϕ (x, µ, α, q) and y = ψ (x, µ, β, q) be the solutions of (1.1) with initial
values
ϕ (0, µ, α, q) = sin α, ϕ′ (0, µ, α, q) = − cos α,
ψ (π, µ, β, q) = sin β, ψ ′ (π, µ, β, q) = − cos β.
The eigenvalues µn of L (q, α, β) are the solutions of the equation
Φ (µ) = ϕ (π, µ, α, q) cos β + ϕ′ (π, µ, α, q) sin β =
= − [ψ (0, µ, β, q) sin α + ψ ′ (0, µ, β, q) cos α] = 0.
It is easy to see that for arbitrary n = 0, 1, 2, . . . , ϕn (x) := ϕ (x, µn (q, α, β) , α, q)
and ψn (x) := ψ (x, µn (q, α, β) , β, q) are eigenfunctions, corresponding to the eigenvalue µn (q, α, β) . The squares of the L2 -norm of these eigenfunctions:
Z π
Z π
2
2
an (q, α, β) :=
|ϕn (x)| dx, bn (q, α, β) :=
|ψn (x)| dx
0
0
are called the norming constants.
The main results of this paper are the following theorems:
Theorem 1.1. For norming constants an and bn the following asymptotic formulae
hold (when n → ∞):
"
#
2 æn (q, α, β)
π
1+
an (q, α, β) =
+ rn sin2 α+
2
π [n + δ (α, β)]
"
#
π
2 æn (q, α, β)
2
+
2 1 + π [n + δ (α, β)] + r̃n cos α, (1.6)
2 [n + δn (α, β)]
"
#
2 æn (q, α, β)
π
1+
+ rn sin2 β+
bn (q, α, β) =
2
π [n + δ (α, β)]
"
#
π
2 æn (q, α, β)
2
+
2 1 + π [n + δ (α, β)] + r̃n cos β,
2 [n + δn (α, β)]
where
rn
Z
1 π
æn = æn (q, α, β) = −
(π − t) q (t) sin 2 [n + δn (α, β)] tdt,
(1.7)
2 0
!
!
1
1
and r̃n = r̃n (q, α, β) = O
, when n → ∞,
= rn (q, α, β) = O
n2
n2
uniformly in α, β ∈ [0, π] and q ∈ BL1R [0, π] .
ON THE NORMING CONSTANTS OF THE STURM-LIOUVILLE PROBLEM
3
Theorem 1.2. For both α, β ∈ (0, π) and α = π, β = 0 cases the function k,
defined as the series
∞
X
æn
cos [n + δn (α, β)] x
k(x) =
n + δn (α, β)
n=2
is absolutely continuous function on arbitrary segment [a, b] ⊂ (0, 2π) , i.e. k ∈
AC (0, 2π) .
The dependence of norming constants on α and β (as far as we know) hasn’t
been investigated before. The dependence of spectral data (by spectral data here
we understand the set of eigenvalues and the set of norming constants) on α and
β has been usually studied (see [1, 2, 3, 5, 6, 7, 8, 9]) in the following sense: the
boundary conditions are separated into four cases:
1) sin α 6= 0, sin β 6= 0, i.e. α, β ∈ (0, π) ;
2) sin α = 0, sin β 6= 0, i.e. α = π, β ∈ (0, π) ;
3) sin α 6= 0, sin β = 0, i.e. α ∈ (0, π) , β = 0;
4) sin α = 0, sin β = 0, i.e. α = π, β = 0,
and results are formulated separately for each case. For eigenvalues, formula (1.4)
generalizes and unites four different formulae that were known before in four mentioned cases (see [4]).
So far, for norming constants the following is known.
In the case sin α 6= 0 it is known that for smooth q
!
an (q, α, β) π
1
.
(1.8)
= +O
2
n2
sin2 α
For absolutely continuous q (we will write q ∈ AC [0, π]) the proof of !(1.8) can
1
, and it
be found in [2]. Let us note, that if q ∈ AC [0, π] , then æn = O
n
is easy to see, that in this case (1.6) takes the form (1.8). In [10], under the
dF (x)
condition q (x) =
(almost everywhere), and sin α 6= 0, where F is a function
dx
of bounded variation (we will write F ∈ BV [0, π]), the author asserts that
π
+ αn ,
(1.9)
2
sin α
is characterized by the condition that the function
an (q, α, β)
2
=
where the sequence {αn }∞
n=0
∞
X
f (x) :=
αn cos nx has a bounded variation on [0, π] , i.e. f ∈ BV [0, π] . Our
n=0
result is similar to this, but there are some differences, in particular, we assert that
k ∈ AC (0, 2π) .
In [9], for q ∈ L2R [0, π] , it was proved that
an (q, α, β)
2
sin α
∞
X
=
π κn
+ ,
2
n
(1.10)
!
1
(see (1.7)).
n
n=0
It is also important to note that norming constants an (q, α, β) are analytic functions on α and β. It easily follows from formulae (3.1), (3.2) and (3.4) below
∞
where {κn }n=0 ∈ l2 (i.e.
2
|κn | < ∞), and κn = æn + O
4
TIGRAN HARUTYUNYAN, AVETIK PAHLEVANYAN
and from the result in [4], which states that λn (q, α, β) λ2n (q, α, β) = µn (q, α, β)
depend analytically on α and β.
In the case sin α = 0, sin β 6= 0 it is known that for smooth q (for q ∈ AC [0, π]
the proof of (1.11) can be found in [2])
"
!#
1
π
1+O
.
(1.11)
an (q, π, β) =
n2
2 (n + 1/2)2
!
1
1
(see (1.5b)), then it is easy to see that (1.11) follows
Since δn (π, β) = + O
2
n
from (1.6). Besides, we see that (1.6) smoothly turns into (1.11) when α → π (for
q ∈ AC [0, π]).
In the case sin α = 0, sin β = 0 the following result can be found in [2] for
q ∈ AC [0, π]:
"
!#
π
1
an (q, π, 0) = 2 1 + O
.
2n
n2
We think that it is more correct to write this result in the form (note that
δn (π, 0) = 1)
"
!#
1
π
(1.12)
an (q, π, 0) =
2 1+O
n2
2 (n + 1)
to keep the beginning of the enumeration of eigenvalues and norming constants
starting from 0, but not from 1, as in [2].
Our proofs of the theorems are based on the detailed study of the dependence of
eigenfunctions ϕn and ψn on parameters α and β. We will present it in the sections
3 and 4. But first we need to prove some properties of the solutions of the equation
(1.1).
2. Asymptotics of the solutions
Let q ∈ L1C [0, π] , i.e. q is a complex-valued, summable function on [0, π] , and
let us denote by yi (x, λ) , i = 1, 2, 3, 4, the solutions of the equation
− y ′′ + q (x) y = λ2 y,
satisfying the initial conditions
y1 (0, λ) = 1, y2 (0, λ) = 0, y3 (π, λ) = 1, y4 (π, λ) = 0,
y1′ (0, λ) = 0, y2′ (0, λ) = 1, y3′ (π, λ) = 0, y4′ (π, λ) = 1.
(2.1)
(2.2)
Let us recall that by a solution of (2.1) (which is the same as (1.1)) we understand the function y, such that y, y ′ ∈ AC [0, π] and which satisfies (2.1) almost
everywhere (see [1]).
The solutions y1 and y2 (as well as the second pair y3 and y4 ) form a fundamental
system of solutions of (1.1), i.e. any solution y of (1.1) can be represented in the
form:
y (x) = y (0) y1 (x, λ) + y ′ (0) y2 (x, λ) = y (π) y3 (x, λ) + y ′ (π) y4 (x, λ) .
(2.3)
The existence and uniqueness of the solutions yi , i = 1, 2, 3, 4 (under the condition q ∈ L1C [0, π]) were investigated in [1, 11, 12, 13, 14]. The following lemma
in some sense extends the results of the mentioned papers related to asymptotics
(when |λ| → ∞) of the solutions yi , i = 1, 2, 3, 4.
ON THE NORMING CONSTANTS OF THE STURM-LIOUVILLE PROBLEM
5
Lemma 2.1. Let q ∈ L1C [0, π] . Then for the solutions yi , i = 1, 2, 3, 4, the following
representations hold (when |λ| ≥ 1):
1
a (x, λ) ,
(2.4)
2λ
1
sin λx
(2.5)
− 2 b (x, λ) ,
y2 (x, λ) =
λ
2λ
1
y3 (x, λ) = cos λ (π − x) +
c (x, λ) ,
(2.6)
2λ
sin λ (π − x)
1
y4 (x, λ) =
(2.7)
− 2 d (x, λ) ,
λ
2λ
where a, b, c, d are twice differentiable with respect to x and entire functions with
respect to λ, and have the form
Zx
Zx
(2.8)
a (x, λ) = sin λx q (t) dt + q (t) sin λ (x − 2t) dt + R1 (x, λ, q) ,
y1 (x, λ) = cos λx +
b (x, λ) = cos λx
0
0
Zx
Zx
q (t) cos λ (x − 2t) dt + R2 (x, λ, q) ,
q (t) sin λ (2t − π − x) dt+ R3 (x, λ, q) , (2.10)
x
Zπ
Zπ
Zπ
q (t) cos λ (π + x − 2t) dt+R4 (x, λ, q) , (2.11)
0
c (x, λ) = sin λ (π − x)
q (t) dt −
Zπ
d (x, λ) = cos λ (π − x)
0
q (t) dt+
(2.9)
x
q (t) dt+
x
x
and Ri , i = 1, 2, 3, 4, satisfy the estimates (when |λ| ≥ 1)
R1 (x, λ, q) , R2 (x, λ, q) = O
R3 (x, λ, q) , R4 (x, λ, q) = O
uniformly with respect to q ∈ BL1C [0, π] .
e|Imλ|x
|λ|
!
e|Imλ|(π−x)
|λ|
,
!
(2.12)
,
(2.13)
Proof. In [14] the authors have proved that y2 (x, λ) can be obtained as a sum of
series
∞
X
y2 (x, λ, q) =
Sk (x, λ, q),
k=0
which converge to y2 (x, λ, q) uniformly on bounded subsets of the set [0, π] × C ×
sin λx
,
L1C [0, π] , and where S0 (x, λ, q) =
λ
Z x
sin λ (x − t)
q (t) Sk−1 (t, λ, q) dt, k = 1, 2, . . . .
Sk (x, λ, q) =
λ
0
For Sk we have the estimate (when |λ| ≥ 1):
|Sk (x, λ, q)| ≤
e|Imλ|x σ0k (x)
, k = 0, 1, 2, . . . ,
k+1
k!
|λ|
(2.14)
6
TIGRAN HARUTYUNYAN, AVETIK PAHLEVANYAN
where σ0 (x) ≡
x
Z
0
|q (t)| dt (see [14]). To prove (2.5), (2.9) and the estimate (2.12),
we write S1 in the form
Z x
sin λ (x − t) sin λt
S1 (x, λ, q) =
q (t) dt =
λ
λ
0
Z x
1
[cos λ (x − 2t) − cos λx] q (t) dt =
= 2
2λ 0
Z
Z x
cos λx x
1
=−
q
(t)
dt
+
cos λ (x − 2t)q (t) dt,
2λ2 0
2λ2 0
and note that
Sk′
(x, λ, q) =
Z
0
x
cos λ (x − t)q (t) Sk−1 (t, λ, q) dt, k = 1, 2, . . . .
This implies that Sk′ ∈ AC [0, π] . By writing y2 (x, λ, q) = S0 + S1 +
we obtain
∞
X
Sk (x, λ, q) ,
k=2
sin λx
1
− 2 b (x, λ) ,
λ
2λ
y2 (x, λ, q) =
∞
X
1
b
(x,
λ)
=
Sk (x, λ, q) and therefore b (x, λ) has the form (2.9), where
2λ2
k=1
∞
X
R2 (x, λ) = −2λ2
Sk (x, λ) . Now, from the estimate (2.14), we obtain that
where −
k=2
∞
X
k=2
|Sk (x, λ, q)| ≤
<
e
|Imλ|x
∞
∞
X
σ0k−2 (x)
e|Imλ|x σ0k (x) e|Imλ|x σ02 (x) X
=
<
3
k+1
k−2
k!
|λ|
|λ|
|λ|
k!
|λ|
k=2
k=2
σ02
3
∞
(x) X
k=2
|λ|
σ0k−2
k−2
(x)
(k − 2)!
=
e
|Imλ|x
=
σ02
3
|λ|
e
|Imλ|x
σ02
3
|λ|
(x)
σ0 (x)
e |λ|
∞
(x) X σ0n (x)
=
n
|λ| n!
n=0
=
σ02 (x)
|λ|3
e
σ0 (x)
|Imλ|x+ |λ|
.
This implies (2.12) for |λ| ≥ 1. Since y2′ ∈ AC [0, π] and S0 (x, λ) =
sin λx
, then
λ
1
b (x, λ) = y2 − S0 is also a twice differentiable function (more
2λ2
′
precisely b ∈ AC [0, π]). Assertions for y1 , y3 , y4 can be proven similarly.
we obtain that −
3. The proof of the Theorem 1.1
According to (2.3), the solution ϕ (x, µ, α, q) which we will denote by ϕ x, λ2 , α
for brevity, has the form
ϕ x, λ2 , α = y1 (x, λ) sin α − y2 (x, λ) cos α,
(3.1)
and according to (2.4) and (2.5) we arrive at:
#
#
"
"
1
sin λx
1
2
a (x, λ) sin α −
− 2 b (x, λ) cos α.
ϕ x, λ , α = cos λx +
2λ
λ
2λ
ON THE NORMING CONSTANTS OF THE STURM-LIOUVILLE PROBLEM
7
Taking the squares of both sides of the last equality, we obtain:
"
#
a2 (x, λ)
1
2
2
2
2
a (x, λ) cos λx +
sin2 α−
ϕ x, λ , α = cos λx sin α +
λ
4λ
"
#
2
b (x, λ) cos λx a (x, λ) sin λx a (x, λ) b (x, λ)
−
cos λx sin λx −
×
+
−
λ
2λ
2λ
4λ2
#
"
sin2 λx
b2 (x, λ) b (x, λ) sin λx
2
× sin α cos α +
cos2 α. (3.2)
cos α +
−
λ2
4λ4
λ3
Recalling the formulae cos2 λx =
1
1
(1 + cos 2λx) and sin2 λx = (1 − cos 2λx) ,
2
2
from (3.2), we obtain:
Z π
sin 2λπ
π
sin2 α+
ϕ2 x, λ2 , α dx = sin2 α +
2
4λ
0
!
Z π
Z
1
1 π 2
sin2 λπ
sin α cos α+
+
a (x, λ) cos λxdx +
a (x, λ) dx sin2 α −
λ
4λ 0
λ2
0
Z π
Z π
1
b (x, λ) cos λxdx −
a (x, λ) sin λxdx sin α cos α+
+ 2
λ
0
0
Z
sin 2λπ
π
sin α cos α π
cos2 α−
a (x, λ) b (x, λ) dx + 2 cos2 α −
+
2λ3
2λ
4λ3
0
!
Z π
Z
1 π 2
1
b (x, λ) sin λxdx −
b (x, λ) dx cos2 α. (3.3)
− 3
λ
4λ
0
0
We are going p
to receive the asymptotic formula (1.6) by the substitution λ =
λn (q, α, β) = µn (q, α, β) in (3.3). To this aim, we estimate each term of the
right-hand side of (3.3) for λ = λn . It can be easily deduced from (1.4) that for
√
λn = µn we have the following asymptotic formula:
[q]
+ ln ,
(3.4)
2(n + δn (α, β))
!
1
uniformly with respect to
= ln (q, α, β) = o
n
λn (q, α, β) = n + δn (α, β) +
1
where [q] :=
π
Z
π
q (t) dt, ln
0
α, β ∈ [0, π] and q ∈ BL1R [0, π] (see [15]).
It follows from (1.5a)–(1.5d) that sin 2πδn (α, β) = O
sin 2πλn (q, α, β) = O
!
1
and
n
!
1
, cos 2πλn (q, α, β) = 1 − O
n
1
n2
!
,
(3.5)
for all (α, β) ∈ (0, π] × [0, π) .
Thus, the second term
sin 2πλn
sin2 α = O
λn
1
n2
!
sin2 α.
(3.6)
8
TIGRAN HARUTYUNYAN, AVETIK PAHLEVANYAN
Important is the third term:
(2.12) we have
1
λn
Z
π
a (x, λn ) cos λn xdx. According to (2.8) and
0
a (x, λn ) = A (x, λn ) + O
where
A (x, λn ) =
Z
x
q (t) dt sin λn x +
0
Z
0
1
λn
!
,
(3.7)
x
q (t) sin λn (x − 2t) dt.
(3.8)
After multiplying both sides by cos λn x, integrating over [0, π] and changing the
order of integration we get
Z
Z π
sin2 λn π π
q (t) cos2 λn tdt−
A (x, λn ) cos λn xdx =
λn
0
0
Z
Z
sin 2λn π π
1 π
−
(π − t) q (t) sin 2λn tdt. (3.9)
q (t) sin 2λn tdt −
4λn
2 0
0
Taking into account the formulae (3.5) and denoting (see (1.7))
Z
1 π
æ̃n ≡ æ̃n (q, α, β) := −
(π − t) q (t) sin 2λn tdt,
2 0
we can rewrite (3.9) in the form
Z π
A(x, λn ) cos λn xdx = æ̃n + O
0
1
n
!!
!
1
.
n
(3.10)
1
n
!
t = sin 2 (n + δn ) t + O
holds uni!
1
formly with respect to t ∈ [0, π] , then æ̃n = æn + O
, and therefore the third
n
term of (3.3) has the form
!
Z
1
1 π
æn
.
+O
a(x, λn ) cos λn xdx =
λn 0
n + δn (α, β)
n2
Since sin 2λn t = sin 2 n + δn + O
Now, let us focus on the remained terms of the equality (3.3) for λ = λn . The
1
terms from the fourth to the eighth have the coefficient γ , where γ ≥ 2, and
λn
!
1
. Concerning the last four terms of (3.3),
therefore they have the order O
n2
Z
sin 2πλn
1 π 2
2
we observe that both
cos
α
and
b (x, λn ) dx cos2 α have the same
3
4
4λ
λ
0
n
n
!
Z
1 π
1
2
order O
cos
α.
An
important
term
is
b (x, λn ) sin λn xdx. According
λ4n
λ3n 0
to (2.9) and (2.12) we can write b (x, λn ) in the form
!
1
,
(3.11)
b (x, λn ) = B (x, λn ) + O
λn
ON THE NORMING CONSTANTS OF THE STURM-LIOUVILLE PROBLEM
where
B (x, λn ) =
Z
0
x
q (t) dt cos λn x −
A simple computation yields:
Z x
Z
B (x, λn ) sin λn x =
q (t) dt sin 2λn x −
0
Z
0
9
x
0
q (t) cos λn (x − 2t) dt.
(3.12)
x
q (t) sin 2λn tdt − A (x, λn ) cos λn x.
After integrating the latter equality from 0 to π, changing the order of integration
and taking into consideration (3.9) we get:
Z π
Z
cos2 λn π π
B (x, λn ) sin λn xdx = −
q (t) sin2 λn tdt+
λn
0
0
Z
Z
sin 2λn π π
1 π
+
(π − t) q (t) sin 2λn tdt =
q (t) sin 2λn tdt −
4λn
2 0
0
!
!
!
1
1
1
+ æ̃n = æn + O
,
+O
=O
λn
λ2n
n
and therefore the eleventh term of the equality (3.3) for λ = λn has the form
!!
Z
æn
1
1
1 π
.
+O
b (x, λn ) sin λn xdx = 2
λ3n 0
λn n + δn (α, β)
n2
!
1
1
1
Let us remark that from (3.4) we have
. Thus,
−
=O
λn n + δn
n3
"
!#
!
2 æn
1
π
1
2
1+
sin α + O
sin α cos α+
+O
an (q, α, β) =
2
π [n + δn (α, β)]
n2
n2
"
!#
π
2 æn
1
+
cos2 α. (3.13)
2 1 + π [n + δ (α, β)] + O
n2
n
2 [n + δn (α, β)]
!
!
1
1
If sin α 6= 0, then O
sin α cos α can be included into the term O
sin2 α,
n2
n2
and if sin α = 0, then these terms are absent. Finally, we can write (3.13) in the
form (1.6). For bn everything can be done similarly. Theorem 1.1 is proved.
4. The proof of the Theorem 1.2
In the sequel the following notations will be used:
Z
Z x
q̃ (t) dt =
q̃ (t) := (π − t) q (t) and σ (x) :=
0
0
x
(π − t) q (t) dt.
(4.1)
Now, we have
Z π
1
æn
=−
q̃ (t) sin 2 (n + δn ) tdt =
n + δn (α, β)
2 [n + δn (α, β)] 0
Z π
1
σ (π) sin 2πδn
=−
+
sin 2 (n + δn ) tdσ (t) = −
2 (n + δn ) 0
2 (n + δn )
Z π
σ (t) cos 2 (n + δn ) tdt.
+
0
(4.2)
10
TIGRAN HARUTYUNYAN, AVETIK PAHLEVANYAN
!
1
. If we denote by σ̃ (x) := σ
n
It was observed in (3.5) that sin 2πδn = O
!
sin 2πδn
1
and cn :=
, then we can rewrite k (x) in the form
=O
2 (n + δn )
n2
k (x) = k1 (x) + k2 (x) ,
x
2
!
(4.3)
where
k1 (x) = −σ (π)
k2 (x) =
∞ Z
X
n=2
∞
X
cn cos [n + δn (α, β)] x,
(4.4)
n=2
2π
σ̃ (t) cos [n + δn (α, β)] tdt cos [n + δn (α, β)] x.
(4.5)
0
!
1
, then the series in (4.4) converges absolutely and uniformly on
Since cn = O
n2
[0, 2π] , and k1 ∈ AC [0, 2π] .
Next, we consider two cases:
Case I: If α, β ∈ (0, π), then by (1.5a) we have
!
!
!
d
1
1
1
cot β − cot α
,
= +O
=O
+O
δn (α, β) =
πn
n2
n
n2
n
cot β − cot α
where d =
. Recalling the Maclaurin expansions of the functions sin x
π
and cos x around the point x = 0, we obtain
cos [n + δn (α, β)] x = cos nx − d · x
sin nx
+ en (x) ,
n
(4.6)
where en (x) , as all the other entries of (4.6), is a smooth function (en ∈ C ∞ ) and
!
1
en (x) = O
(4.7)
n2
uniformly on x ∈ [0, 2π] . Therefore k2 can be written in the form
k2 (x) = l1 (x) + l2 (x) + l3 (x) ,
where
l1 (x) = −d · x
Z
∞
X
1 2π
σ̃ (t) cos ntdt sin nx−
n 0
n=2
Z
Z
∞
∞
X
X
1 2π
1 2π
2
−d·
tσ̃ (t) sin ntdt cos nx + d · x
tσ̃ (t) sin ntdt sin nx+
n 0
n2 0
n=2
n=2
Z
∞
∞ Z 2π
X
X
1 2π
en (t) σ̃ (t) dt sin nx (4.8)
en (t) σ̃ (t) dt cos nx − d · x
+
n 0
n=2
n=2 0
ON THE NORMING CONSTANTS OF THE STURM-LIOUVILLE PROBLEM
l2 (x) =
∞
X
en (x)
2π
Z
σ̃ (t) cos ntdt−
0
n=2
∞
X
−d·
11
en (x)
n
n=2
Z
2π
tσ̃ (t) sin ntdt +
0
∞
X
en (x)
n=2
l3 (x) =
∞ Z
X
n=2
Z
2π
en (t) σ̃ (t) sin ntdt, (4.9)
0
2π
σ̃ (t) cos ntdt cos nx.
Since σ̃ ∈ AC[0, 2π], then Fourier coefficients are
! Z
Z 2π
2π
1
tσ̃ (t) sin ntdt = O
,
σ̃ (t) cos ntdt = O
n
0
0
Also we note that
Z
0
(4.10)
0
2π
en (t) σ̃ (t) dt = O
1
n2
!
!
1
.
n
.
(4.11)
(4.12)
Therefore the trigonometric series in (4.8) converges absolutely and uniformly on
[0, 2π] , and l1 ∈ AC (0, 2π) .
It follows from!(4.7), (4.11) and (4.12) that the terms of the series in (4.9) have
1
, and therefore l2 ∈ AC [0, 2π] .
the order O
n3
About l3 (x) we can say the following: Since σ̃ ∈ AC [0, 2π] , then the Fourier
series of σ̃
∞
a0 (σ̃) X
σ̃ (x) =
+
(an (σ̃) cos nx + bn (σ̃) sin x) ,
2
n=1
Z
Z
1 2π
1 2π
where an (σ̃) =
σ̃ (t) cos ntdt, bn (σ̃) =
σ̃ (t) sin ntdt, converges to σ̃ (x)
π 0
π 0
in every point of [0, 2π] and this series is a function from AC [0, 2π] . The same is
true for σ ⋆ (x) = σ̃ (2π − x) :
σ̃ (2π − x) =
∞
a0 (σ ⋆ ) X
(an (σ ⋆ ) cos nx + bn (σ ⋆ ) sin x) .
+
2
n=1
But it is easy to see, that an (σ ⋆ ) = an (σ̃) and bn (σ ⋆ ) = −bn (σ̃) . So
∞
1
a0 (σ̃) X
(σ̃ (x) + σ̃ (2π − x)) =
+
an (σ̃) cos nx,
2
2
n=1
i.e. this is ”the even part” of Fourier series of σ̃ (x) , and is absolutely continuous
on [0, 2π] . Thus, for the case α, β ∈ (0, π) Theorem 1.2 is proved.
Case II: If α = π, β = 0, then δn (π, 0) = 1, and the function k2 (·) takes
∞ Z 2π
X
the form k2 (x) =
σ̃ (t) cos ntdt cos nx and again it is ”the even part” of
n=3
0
Fourier series (without the zeroth, the first and the second terms) of an absolutely
continuous function. Theorem 1.2 is proved.
Acknowledgment. T.N. Harutyunyan was supported by State Committee of Science MES RA, in frame of the research project No. 15T–1A392.
12
TIGRAN HARUTYUNYAN, AVETIK PAHLEVANYAN
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Tigran Harutyunyan
Faculty of Mathematics and Mechanics, Yerevan State University, 1 Alex Manoogian,
0025, Yerevan, Armenia
E-mail address: hartigr@yahoo.co.uk
Avetik Pahlevanyan
Institute of Mathematics of National Academy of Sciences, 24/5 Baghramian ave., 0019,
Yerevan, Armenia
E-mail address: apahlevanyan@instmath.sci.am