On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients
Suficient conditions for regular solvability of the boundary value problem for an elliptic operator-differential equation of second order considered on the positive semi-axis are obtained. Note that the principal part of the equation contains a discontinuous coefficient, and the boundary condition i...
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nasplib_isofts_kiev_ua-123456789-1067462025-02-09T13:53:58Z On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients Mirzoev, S.S. Aliev, A.R. Rustamova, L.A. Suficient conditions for regular solvability of the boundary value problem for an elliptic operator-differential equation of second order considered on the positive semi-axis are obtained. Note that the principal part of the equation contains a discontinuous coefficient, and the boundary condition involves a linear operator. Найдены достаточные условия регулярной разрешимости краевой задачи для эллиптического операторно-дифференциального уравнения второго порядка, рассматриваемого на положительной полуоси. Отмечено, что главная часть уравнения содержит разрывной коэффициент, а в краевом условии участвует линейный оператор. 2013 Article On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients / S.S. Mirzoev, A.R. Aliev, L.A. Rustamova // Журнал математической физики, анализа, геометрии. — 2013. — Т. 9, № 2. — С. 207-226. — Бібліогр.: 22 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106746 en Журнал математической физики, анализа, геометрии application/pdf Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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Suficient conditions for regular solvability of the boundary value problem for an elliptic operator-differential equation of second order considered on the positive semi-axis are obtained. Note that the principal part of the equation contains a discontinuous coefficient, and the boundary condition involves a linear operator. |
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Mirzoev, S.S. Aliev, A.R. Rustamova, L.A. |
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Mirzoev, S.S. Aliev, A.R. Rustamova, L.A. On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients Журнал математической физики, анализа, геометрии |
| author_facet |
Mirzoev, S.S. Aliev, A.R. Rustamova, L.A. |
| author_sort |
Mirzoev, S.S. |
| title |
On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients |
| title_short |
On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients |
| title_full |
On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients |
| title_fullStr |
On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients |
| title_full_unstemmed |
On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients |
| title_sort |
on the boundary value problem with the operator in boundary conditions for the operator-differential equation of second order with discontinuous coeficients |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2013 |
| url |
https://nasplib.isofts.kiev.ua/handle/123456789/106746 |
| citation_txt |
On the Boundary Value Problem with the Operator in Boundary Conditions for the Operator-Differential Equation of Second Order with Discontinuous Coeficients / S.S. Mirzoev, A.R. Aliev, L.A. Rustamova // Журнал математической физики, анализа, геометрии. — 2013. — Т. 9, № 2. — С. 207-226. — Бібліогр.: 22 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT mirzoevss ontheboundaryvalueproblemwiththeoperatorinboundaryconditionsfortheoperatordifferentialequationofsecondorderwithdiscontinuouscoeficients AT alievar ontheboundaryvalueproblemwiththeoperatorinboundaryconditionsfortheoperatordifferentialequationofsecondorderwithdiscontinuouscoeficients AT rustamovala ontheboundaryvalueproblemwiththeoperatorinboundaryconditionsfortheoperatordifferentialequationofsecondorderwithdiscontinuouscoeficients |
| first_indexed |
2025-11-26T12:54:36Z |
| last_indexed |
2025-11-26T12:54:36Z |
| _version_ |
1849857600755073024 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2013, vol. 9, No. 2, pp. 207�226
On the Boundary Value Problem with the Operator
in Boundary Conditions for the Operator-Di�erential
Equation of Second Order with Discontinuous
Coe�cients
S.S. Mirzoev
Baku State University
23 Z. Khalilov Str., Baku AZ1148, Azerbaijan
E-mail: mirzoyevsabir@mail.ru
A.R. Aliev
Baku State University
23 Z. Khalilov Str., Baku AZ1148, Azerbaijan
Institute of Mathematics and Mechanics of NAS of Azerbaijan
9 F. Agayev Str., Baku AZ1141, Azerbaijan
E-mail: alievaraz@yahoo.com
L.A. Rustamova
Institute of Applied Mathematics, Baku State University
23 Z. Khalilov Str., Baku AZ1148, Azerbaijan
E-mail: lamia_rus@rambler.ru
Received January 25, 2012
Su�cient conditions for regular solvability of the boundary value problem
for an elliptic operator-di�erential equation of second order considered on
the positive semi-axis are obtained. Note that the principal part of the
equation contains a discontinuous coe�cient, and the boundary condition
involves a linear operator.
Key words: Hilbert space, operator-di�erential equation, discontinuous
coe�cient, regular solvability, Fourier transformation, intermediate deriva-
tives.
Mathematics Subject Classi�cation 2010: 34B40, 35J25, 47D03.
This work was supported by the Science Development Foundation under the President of
the Republic of Azerbaijan - Grant No. EIF-2011-1(3)-82/28/1.
c© S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova, 2013
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
Dedicated to the 80th birthday of Professor F.S. Rofe-Beketov
1. Introduction
Let A be a self-adjoint positive de�nite operator in a separable Hilbert space
H. It is known that the domain of the operator Aθ (θ ≥ 0) becomes a Hilbert
space Hθ with respect to the scalar product (x, y)θ = (Aθx, Aθy), x, y ∈ Dom(Aθ).
For θ = 0, we consider that H0 = H.
We denote by L2((a, b);H),−∞ ≤ a < b ≤ +∞, the Hilbert space of all vector
functions de�ned on (a, b), with the values in H, which have the �nite norm
‖f‖L2((a,b);H) =
b∫
a
‖f(t)‖2
H dt
1
2
(see [1]). Further, we denote by L(X, Y ) a set of the linear bounded operators
acting from the Hilbert space X to another Hilbert space Y . For Y = X, we
consider L(X,X) = L(X). We also denote the spectrum of the operator (·)
by σ(·).
We introduce the linear space
W 2
2 ((a, b);H) =
{
u(t) : A2u(t) ∈ L2((a, b);H), u′′(t) ∈ L2((a, b);H)
}
with the norm
‖u‖W 2
2 ((a,b);H) =
(∥∥A2u
∥∥2
L2((a,b);H)
+
∥∥u′′
∥∥2
L2((a,b);H)
) 1
2
.
The lineal becomes a Hilbert space [2]. Here and further the derivatives are
understood in the sense of distribution theory. For a = −∞, b = +∞, we will
assume that
L2((−∞, +∞);H) ≡ L2(R; H), W 2
2 ((−∞, +∞);H) ≡ W 2
2 (R; H), R = (−∞,+∞).
For a = 0, b = +∞, we will suppose that
L2((0,+∞);H) ≡ L2(R+; H), W 2
2 ((0, +∞);H) ≡ W 2
2 (R+; H), R+ = (0, +∞).
Besides the spaces introduced, we will use the following subspaces:
o
W 2
2 (R+; H) =
{
u(t) : u(t) ∈ W 2
2 (R+; H), u(0) = u′(0) = 0
}
,
o
W 2
2,T (R+; H) =
{
u(t) : u(t) ∈ W 2
2 (R+; H), u(0) = Tu′(0), T ∈ L(H 1
2
, H 3
2
)
}
.
208 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2
On the Boundary Value Problem with the Operator in Boundary Conditions
Now we pass to the statement of the boundary value problem studied. We
consider the operator di�erential equation of the form
−u′′(t) + ρ(t)A2u(t) + A1u
′(t) = f(t), t ∈ R+, (1)
satisfying the boundary conditions at zero
u(0) = Tu′(0), (2)
where f(t), u(t) are the functions with values in H, T ∈ L(H 1
2
, H 3
2
), A1 is a linear
unbounded operator, A is a self-adjoint positive de�nite operator in H, ρ(t) = α
if t ∈ (0, 1), and ρ(t) = β if t ∈ (1, +∞), and α, β are positive unequal numbers.
For de�niteness, we suppose that α ≤ β.
De�nition. Problem (1), (2) is called regularly solvable if for every function
f(t) ∈ L2(R+; H) there exists a function u(t) ∈ W 2
2 (R+; H) satisfying equation
(1) almost everywhere in R+, boundary condition (2) holds in the sense of con-
vergence of the space H 3
2
, i.e.,
lim
t→0
∥∥u(t)− Tu′(t)
∥∥
H 3
2
= 0,
and the estimate
‖u‖W 2
2 (R+;H) ≤ const ‖f‖L2(R+;H)
takes place.
A review article of A.A. Shkalikov [3] contains a detailed analysis of the results
of both the author himself and other authors obtained on the problems of sol-
vability and spectral problems, mainly the boundary value problems for opera-
tor di�erential equations when the coe�cients in the boundary conditions are
only complex numbers. Among the results, we especially mark out the papers of
M.G. Gasymov [4�6] and his followers. Note that these boundary value problems
do not lose their actuality (see, for example, the recent papers of S.S. Mirzoev and
M.Yu. Salimov [7], A.R. Aliev and S.S. Mirzoev [8], A.R. Aliev [9]). Nevertheless,
there are rather few works on solvability and the studying of the spectrum, the
completeness of root vectors and elementary solutions of boundary value problems
for operator-di�erential equations when the coe�cients of the boundary conditions
are operators. The �rst works on this subject were the papers of F.S. Rofe-Beketov
[10], V.A. Ilyin and A.F. Filippov [11], M.L.Gorbachuk [12], S.Y. Yakubov and
B.A. Aliev [13]. Later in this direction there appeared an interesting paper of
M.G. Gasymov and S.S. Mirzoev [14], in which both the problems of solvability
and some spectral aspects of the boundary value problems for elliptic type ope-
rator di�erential equations of the second order considered on the semiaxis were
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2 209
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
studied by using original methods. This work found its proper development in
the papers of S.S. Mirzoev and his followers (see [15�19]).
The present paper aims to obtain appropriate solvability results of the paper
by M.G. Gasymov and S.S. Mirzoev [14] for the case when the principal part of
the equation contains discontinuous (piecewise constant) coe�cient. Such prob-
lems are of interest not only because they contain appropriate boundary value
problems, in the boundary conditions in which the coe�cients are complex num-
bers, but also because they can be applied to a wider range of the problems for
partial di�erential equations and a number of problems in mechanics, in particu-
lar, non-standard problems in the theory of elasticity of multilayered bodies. For
simplicity, a point of discontinuity is taken. Here we note that a regular solva-
bility of the boundary value problems for operator di�erential equations of the
second order with discontinuous coe�cients, when the coe�cients in the boun-
dary condition are only complex numbers, is studied in paper [20] and developed
in [21].
2. Main Results
We begin with considering the problem
−u′′(t) + ρ(t)A2u(t) = f(t), t ∈ R+, (3)
u(0) = Tu′(0), (4)
where f(t) ∈ L2(R+; H), u(t) ∈ W 2
2 (R+; H).
As can be seen, equation (3) is obtained from (1) at A1 = 0.
The following theorem is true.
Theorem 1. Let the operator
Tα,β = E +
√
αTA +
√
β −√α√
α +
√
β
(√
αTA− E
)
e−2
√
αA
have a bounded inverse operator in H 3
2
, where E is an identity operator in H and
e−tA is a semigroup of bounded linear operators generated by the operator −A.
Then the operator P0, acting from the space
o
W 2
2,T (R+; H) to the space L2(R+;H),
P0u(t) ≡ −u′′(t) + ρ(t)A2u(t), u(t) ∈
o
W 2
2,T (R+; H),
induces an isomorphism between these spaces.
P r o o f. We will show that the equation P0u(t) = 0 has only the trivial
solution in the space
o
W 2
2,T (R+;H). Indeed, the general solution of the equation
210 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2
On the Boundary Value Problem with the Operator in Boundary Conditions
P0u(t) = 0 from the space W 2
2 (R+; H) has the form
u0(t) =
{
v1(t) = e−
√
αtAϕ1 + e−
√
α(1−t)Aϕ2 , t ∈ (0, 1),
v2(t) = e−
√
β(t−1)Aϕ3 , t ∈ (1,+∞),
where the vectors ϕ1, ϕ2, ϕ3 ∈ H 3
2
. From the condition u0(t) ∈
o
W 2
2,T (R+;H) we
have
v1(0) = Tv′1(0), v1(1) = v2(1), v′1(1) = v′2(1).
Thus for the vectors ϕ1, ϕ2, ϕ3 we get the system of equations
ϕ1 + e−
√
αAϕ2 = T
(
−√αAϕ1 +
√
αAe−
√
αAϕ2
)
,
e−
√
αA ϕ1 + ϕ2 = ϕ3,
−√αAe−
√
αAϕ1 +
√
αAϕ2 = −√βAϕ3.
From this system, in turn, it implies that
ϕ3 =
√
α
β
e−
√
αA ϕ1 −
√
α
β
ϕ2, ϕ2 =
√
α−√β√
α +
√
β
e−
√
αA ϕ1,
ϕ1 +
√
α−√β√
α +
√
β
e−2
√
αAϕ1 +
√
αTAϕ1 −
√
α−√β√
α +
√
β
√
αTAe−2
√
αAϕ1 = 0.
Consequently,
Tα,βϕ1 ≡
(
E +
√
αTA +
√
β −√α√
α +
√
β
(√
αTA− E
)
e−2
√
αA
)
ϕ1 = 0.
Since by the condition of the theorem the operator Tα,β has a bounded inverse
operator in H 3
2
, then from the last equation it follows that ϕ1 = 0. Consequently,
ϕ2 = ϕ3 = 0, i.e., u0(t) = 0. From the condition of the theorem it is clear that at
any f(t) ∈ L2(R+; H) the equation P0u(t) = f(t) has a solution from the space
o
W 2
2,T (R+; H) and this solution has the representation
u(t) =
{
u1(t) , t ∈ (0, 1),
u2(t) , t ∈ (1, +∞),
where
u1(t) =
1
2
√
α
1∫
0
e−
√
α|t−s|AA−1f(s)ds + T−1
α,β(
√
αTA−E)e−
√
α(t+1)A
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2 211
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
×
√
α−√β
2
√
α
(√
α +
√
β
)
1∫
0
e−
√
α(1−s)AA−1f(s)ds +
1
2
√
α
1∫
0
e−
√
α(s−1)AA−1f(s)ds
+
1√
α +
√
β
+∞∫
1
e−
√
β(s−1)AA−1f(s)ds
+ T−1
α,β(E +
√
αTA)e−
√
α(1−t)A
×
√
α−√β
2
√
α
(√
α +
√
β
)
1∫
0
e−
√
α(1−s)AA−1f(s)ds +
1√
α +
√
β
+∞∫
1
e−
√
β(s−1)AA−1f(s)ds
+T−1
α,β(
√
αTA− E)e−
√
α(2−t)A
√
α−√β
2
√
α
(√
α +
√
β
)
1∫
0
e−
√
αsAA−1f(s)ds,
u2(t) =
1
2
√
β
+∞∫
1
e−
√
β|t−s|AA−1f(s)ds + T−1
α,β(E +
√
αTA)
e−
√
β(t−1)A
√
α +
√
β
×
1∫
0
e−
√
α(1−s)AA−1f(s)ds +
√
β −√α
2
√
β
+∞∫
1
e−
√
β(s−1)AA−1f(s)ds
+T−1
α,β(
√
αTA−E)e−
√
β(t−1)Ae−2
√
αA
1√
α +
√
β
1∫
0
e−
√
α(s−1)AA−1f(s)ds
+
1
2
√
β
+∞∫
1
e−
√
β(s−1)AA−1f(s)ds
.
We note that the ful�llment of boundary condition (4) is veri�ed directly. In
addition, for any u(t) ∈
o
W 2
2,T (R+; H) we have
‖P0u‖2
L2(R+;H) =
∥∥u′′ + ρA2u
∥∥2
L2(R+;H)
≤ 2
(∥∥u′′
∥∥2
L2(R+;H)
+ max(α2;β2)
∥∥A2u
∥∥2
L2(R+;H)
)
≤ 2max(1;α2; β2) ‖u‖2
W 2
2 (R+;H) ,
i.e., the operator P0 :
o
W 2
2,T (R+; H) → L2(R+; H) is bounded. Then the assertion
of the theorem follows from the Banach theorem on the inverse operator. The
theorem is proved.
212 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2
On the Boundary Value Problem with the Operator in Boundary Conditions
From Theorem 1 it implies that the norms ‖P0u‖L2(R+;H) and ‖u‖W 2
2 (R+;H)
are equivalent in the space
o
W 2
2,T (R+;H).
The following theorem is true.
Theorem 2. Suppose that the operator Tα,β has a bounded inverse operator
in H 3
2
, and the operator B = A1A
−1 is bounded in H, moreover, ‖B‖ < N−1
T ,
where
NT = sup
06=u(t)∈
o
W 2
2,T (R+;H)
∥∥Au′
∥∥
L2(R+;H)
‖P0u‖−1
L2(R+;H) .
Then problem (1), (2) is regularly solvable.
P r o o f. Denoting by P1 the operator acting from the space
o
W 2
2,T (R+; H)
to the space L2(R+; H) in the following way:
P1u(t) ≡ A1u
′(t), u(t) ∈
o
W 2
2,T (R+;H),
we can rewrite problem (1), (2) in the form of the operator equation
P0u(t) + P1u(t) = f(t),
where f(t) ∈ L2(R+; H), u(t) ∈
o
W 2
2,T (R+; H). Since by Theorem 1 the operator
P0 has a bounded inverse P−1
0 acting from the space L2(R+; H) to the space
o
W 2
2,T (R+; H), then after substitution u(t) = P−1
0 v(t) we obtain the equation
(E + P1P
−1
0 )v(t) = f(t)
in the space L2(R+; H). And due to the fact that for any v(t) ∈ L2(R+;H)
∥∥P1P
−1
0 v
∥∥
L2(R+;H)
= ‖P1u‖L2(R+;H) ≤ ‖B‖∥∥Au′
∥∥
L2(R+;H)
≤ NT ‖B‖ ‖P0u‖L2(R+;H) = NT ‖B‖ ‖v‖L2(R+;H) ,
then for NT ‖B‖ < 1 the operator E +P1P
−1
0 is invertible in the space L2(R+; H)
and
u(t) = P−1
0 (E + P1P
−1
0 )−1f(t),
thus
‖u‖W 2
2 (R+;H) ≤ const ‖f‖L2(R+;H) .
The theorem is proved.
We note that the problem of estimating the number NT arises here. For this
purpose we will use the idea of the procedure o�ered in [22].
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2 213
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
First we prove the following statement.
Lemma 1. For any u(t) ∈
o
W 2
2,T (R+; H) there takes place the inequality
‖P0u‖2
L2(R+;H) ≥
α
β
(∥∥u′′
∥∥2
L2(R+;H)
+ αβ
∥∥A2u
∥∥2
L2(R+;H)
+2β
∥∥Au′
∥∥2
L2(R+;H)
+ 2βRe(ATx, x) 1
2
)
,
where x = u′(0) ∈ H 1
2
.
P r o o f. Multiplying both sides of equation (3) by ρ−
1
2 (t), we get
∥∥∥ρ−
1
2 f
∥∥∥
2
L2(R+;H)
=
∥∥∥ρ−
1
2 P0u
∥∥∥
2
L2(R+;H)
=
∥∥∥−ρ−
1
2 u′′ + ρ
1
2 A2u
∥∥∥
2
L2(R+;H)
=
∥∥∥ρ−
1
2 u′′
∥∥∥
2
L2(R+;H)
+
∥∥∥ρ
1
2 A2u
∥∥∥
2
L2(R+;H)
− 2Re
(
u′′, A2u
)
L2(R+;H)
.
Now, integrating by parts, we have
−2Re
(
u′′, A2u
)
L2(R+;H)
= 2Re
(
A
1
2 u′(0), A
3
2 u(0)
)
+ 2
∥∥Au′
∥∥2
L2(R+;H)
= 2Re (ATx, x) 1
2
+ 2
∥∥Au′
∥∥2
L2(R+;H)
.
Then ∥∥∥ρ−
1
2 P0u
∥∥∥
2
L2(R+;H)
=
∥∥∥ρ−
1
2 u′′
∥∥∥
2
L2(R+;H)
+
∥∥∥ρ
1
2 A2u
∥∥∥
2
L2(R+;H)
+2
∥∥Au′
∥∥2
L2(R+;H)
+ 2Re (ATx, x) 1
2
. (5)
On the other hand, for u(t) ∈
o
W 2
2,T (R+; H) we have
‖P0u‖2
L2(R+;H) ≥ α
∥∥∥ρ−
1
2 P0u
∥∥∥
2
L2(R+;H)
. (6)
Taking into account equality (5) in (6), we obtain
‖P0u‖2
L2(R+;H) ≥ α
(∥∥∥ρ−
1
2 u′′
∥∥∥
2
L2(R+;H)
+
∥∥∥ρ
1
2 A2u
∥∥∥
2
L2(R+;H)
+2
∥∥Au′
∥∥2
L2(R+;H)
+ 2Re(ATx, x) 1
2
)
≥ α
(
1
β
∥∥u′′
∥∥2
L2(R+;H)
+ α
∥∥A2u
∥∥2
L2(R+;H)
+ 2
∥∥Au′
∥∥2
L2(R+;H)
+2Re(ATx, x) 1
2
)
=
α
β
S(u),
214 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2
On the Boundary Value Problem with the Operator in Boundary Conditions
where
S(u) =
∥∥u′′
∥∥2
L2(R+;H)
+ αβ
∥∥A2u
∥∥2
L2(R+;H)
+2β
∥∥Au′
∥∥2
L2(R+;H)
+ 2βRe(ATx, x) 1
2
. (7)
Lemma 1 is proved.
For further operations we factorize the considered in the space H4 polynomial
operator pencil of the form
P (λ; γ; A) = λ4E + αβA4 − 2βλ2A2 + γλ2A2.
Lemma 2. For γ ∈ [0, 2β
1
2 (α
1
2 + β
1
2 )) the operator pencil P (λ; γ; A) is in-
vertible on an imaginary axis and is represented in the form
P (λ; γ;A) = F (λ; γ;A)F (−λ; γ;A),
where
F (λ; γ; A) = (λE − ω1(γ)A)(λE − ω2(γ)A)
≡ λ2E + a1(γ)λA + a2(γ)A2, (8)
and Reωk(γ) < 0, k = 1, 2, a1(γ) =
√
2β
1
2 (α
1
2 + β
1
2 )− γ, a2(γ) = α
1
2 β
1
2 .
P r o o f. Let γ ∈ [0, 2β
1
2 (α
1
2 + β
1
2 )). Then for σ ∈ σ(A) and λ = iξ, ξ ∈ R,
we have
P (iξ; γ;σ) = ξ4 + αβσ4 + 2βξ2σ2 − γξ2σ2
= σ4
(
ξ4
σ4
+ αβ + 2β
ξ2
σ2
− γ
ξ2
σ2
)
= σ4
(
ξ4
σ4
+ αβ + 2β
ξ2
σ2
)
×
(
1− γ
ξ2
σ2
ξ4
σ4 + αβ + 2β ξ2
σ2
)
≥ σ4
(
ξ4
σ4
+ αβ + 2β
ξ2
σ2
)
×
1− γ sup
ξ2
σ2≥0
ξ2
σ2
ξ4
σ4 + αβ + 2β ξ2
σ2
.
Since
sup
ξ2
σ2≥0
ξ2
σ2
ξ4
σ4 + αβ + 2β ξ2
σ2
=
1
2β
1
2 (α
1
2 + β
1
2 )
,
it follows that P (iξ; γ; σ) > 0 for γ ∈ [0, 2β
1
2 (α
1
2 + β
1
2 )). Consequently, the
polynomial P (iξ; γ;σ) has no roots on the imaginary axis for γ ∈ [0, 2β
1
2 (α
1
2 +
β
1
2 )). Therefore, this polynomial has two roots ω1(γ)σ and ω2(γ)σ from the left
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2 215
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
half-plane and two roots −ω1(γ)σ and −ω2(γ)σ from the right half-plane, i.e.,
Reωk(γ) < 0, k = 1, 2, and ω1(γ) = ω2(γ). Thus,
P (λ; γ; σ) = F (λ; γ; σ)F (−λ; γ;σ), (9)
where
F (λ; γ; σ) = (λ− ω1(γ)σ)(λ− ω2(γ)σ)
= λ2 + a1(γ)λσ + a2(γ)σ2.
Since Reωk(γ) < 0, k = 1, 2, then
a1(γ) = −(ω1(γ) + ω2(γ)) = −(ω1(γ) + ω1(γ)) = −2Reω1(γ) > 0.
And as a2
2(γ) = αβ and a2(γ) = ω1(γ)ω2(γ) = ω1(γ)ω1(γ) = |ω1(γ)|2 > 0, we ob-
tain that a2(γ) = α
1
2 β
1
2 . Further, from equality (9) it follows that a2
1(γ)−2a2(γ) =
−γ + 2β, i.e., a2
1(γ) = 2a2(γ) + 2β − γ = 2α
1
2 β
1
2 + 2β − γ > 0. Consequently,
a1(γ) =
√
2β
1
2 (α
1
2 + β
1
2 )− γ. Now, using the spectral decomposition of the ope-
rator A, from equality (9) we obtain the assertion of the lemma. Lemma 2 is
proved.
Now we prove the lemma which plays an important role in our investigation.
Lemma 3. For any u(t) ∈
o
W 2
2,T (R+; H) and γ ∈ [0, 2β
1
2 (α
1
2 + β
1
2 )) the
following identity holds:
S(u)− γ
∥∥Au′
∥∥2
L2(R+;H)
= 〈R(γ)x, x〉+
∥∥∥∥F
(
d
dt
; γ; A
)
u
∥∥∥∥
2
L2(R+;H)
, (10)
where F (λ; γ;A) is de�ned from equality (8), and
〈R(γ)x, x〉 = 2β
1
2 (α
1
2 + β
1
2 )Re(ATx, x) 1
2
+
√
2β
1
2 (α
1
2 + β
1
2 )− γ
×
(
‖x‖2
H 1
2
+ α
1
2 β
1
2 ‖ATx‖2
H 1
2
)
. (11)
P r o o f. First we denote by DT (R+; H2) the lineal of all in�nitely di�eren-
tiable functions with values in H2, having compact supports on R+ and satisfying
boundary condition (2) at zero. By density and trace theorems [2], this lineal is
dense everywhere in
o
W 2
2,T (R+; H). Consequently, it is enough to prove (10) for
the functions from DT (R+; H2). It is obvious that for u(t) ∈ DT (R+; H2) and
γ ∈ [0, 2β
1
2 (α
1
2 + β
1
2 )) the following equality holds:
∥∥∥∥F
(
d
dt
; γ; A
)
u
∥∥∥∥
2
L2(R+;H)
=
∥∥u′′ + a1(γ)Au′ + a2(γ)A2u
∥∥2
L2(R+;H)
=
∥∥u′′
∥∥2
L2(R+;H)
216 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2
On the Boundary Value Problem with the Operator in Boundary Conditions
+a2
1(γ)
∥∥Au′
∥∥2
L2(R+;H)
+ a2
2(γ)
∥∥A2u
∥∥2
L2(R+;H)
+ 2a1(γ)Re
(
u′′, Au′
)
L2(R+;H)
+2a2(γ)Re
(
u′′, A2u
)
L2(R+;H)
+ 2a1(γ)a2(γ)Re
(
Au′, A2u
)
L2(R+;H)
. (12)
On the other hand, applying integration by parts, we obtain the equalities
Re(u′′, A2u)L2(R+;H) = −∥∥Au′
∥∥2
L2(R+;H)
−Re(ATx, x) 1
2
,
2Re(u′′, Au′)L2(R+;H) = −‖x‖2
H 1
2
, 2Re(Au′, A2u)L2(R+;H) = −‖ATx‖2
H 1
2
.
Taking into account equalities from (12) and the values a1(γ) and a2(γ) from
Lemma 2, we have
∥∥∥∥F
(
d
dt
; γ; A
)
u
∥∥∥∥
2
L2(R+;H)
=
∥∥u′′
∥∥2
L2(R+;H)
+ αβ
∥∥A2u
∥∥2
L2(R+;H)
+(2β
1
2 (α
1
2 +β
1
2 )−γ)
∥∥Au′
∥∥2
L2(R+;H)
−2α
1
2 β
1
2
∥∥Au′
∥∥2
L2(R+;H)
−2α
1
2 β
1
2 Re(ATx, x) 1
2
−
√
2β
1
2 (α
1
2 + β
1
2 )− γ ‖x‖2
H 1
2
− α
1
2 β
1
2
√
2β
1
2 (α
1
2 + β
1
2 )− γ ‖ATx‖2
H 1
2
= S(u)− γ
∥∥Au′
∥∥2
L2(R+;H)
− 〈R(γ)x, x〉 , (13)
where S(u) and 〈R(γ)x, x〉 are de�ned from (7) and (11), respectively. The va-
lidity of (10) follows from (13). Lemma 3 is proved.
Obviously that S(u) is the norm in the space
o
W 2
2 (R+;H) which is equivalent
to the initial norm ‖u‖W 2
2 (R+;H).
The studies above allow us to assert that the following theorem is true.
Theorem 3. The number S0, de�ned as
S0 = sup
0 6=u∈
o
W 2
2 (R+;H)
∥∥Au′
∥∥
L2(R+;H)
S−
1
2 (u),
is �nite and S0 = 1√
2β
1
2 (α
1
2 +β
1
2 )
.
P r o o f. Clearly, for u(t) ∈
o
W 2
2 (R+; H), for any γ ∈ [0, 2β
1
2 (α
1
2 + β
1
2 )) from
equality (10) we obtain
S(u)− γ
∥∥Au′
∥∥2
L2(R+;H)
=
∥∥∥∥F
(
d
dt
; γ; A
)
u
∥∥∥∥
2
L2(R+;H)
. (14)
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2 217
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
Passing in (14) to the limit as γ → 2β
1
2 (α
1
2 + β
1
2 ), we have
S(u) ≥ 2β
1
2 (α
1
2 + β
1
2 )
∥∥Au′
∥∥2
L2(R+;H)
,
i.e., ∥∥Au′
∥∥
L2(R+;H)
≤ 1√
2β
1
2 (α
1
2 + β
1
2 )
S
1
2 (u).
Consequently, the number S0 ≤ 1√
2β
1
2 (α
1
2 +β
1
2 )
. We show that here the equality
holds, i.e., S0 = 1√
2β
1
2 (α
1
2 +β
1
2 )
. For this purpose, for any ε > 0 it is su�cient to
construct a vector function uε(t) ∈
o
W 2
2 (R+; H) such that
E(uε) ≡ S(uε)− (2β
1
2 (α
1
2 + β
1
2 ) + ε)
∥∥Au′
∥∥2
L2(R+;H)
< 0.
Let ψ ∈ H4, ‖ψ‖ = 1, and g(t) be a scalar function having a twice continuous
derivative in R, and g(t), g′′(t) ∈ L2(R). Then, using Plancherel's theorem on
the Fourier transformation, we obtain
E(g(t)ψ) ≡ S(g(t)ψ)− (2β
1
2 (α
1
2 + β
1
2 ) + ε)
∥∥Ag′(t)ψ
∥∥2
L2(R;H)
=
+∞∫
−∞
([ξ4E + αβA4 + 2βξ2A2 − (2β
1
2 (α
1
2 + β
1
2 ) + ε)ξ2A2]ψ,ψ) |ĝ(ξ)|2 dξ
=
+∞∫
−∞
qε(ξ, ψ) |ĝ(ξ)|2 dξ (‖ψ‖ = 1), (15)
where
qε(ξ, ψ) = ξ4 + αβ
∥∥A2ψ
∥∥2 − (2α
1
2 β
1
2 + ε)ξ2 ‖Aψ‖2 .
It is obvious that for �xed ψ, at the points ξ = ±
(
2α
1
2 β
1
2 +ε
2
) 1
2
‖Aψ‖, the
function qε(ξ, ψ) takes its minimum value which is equal to
hε(ψ) = αβ
∥∥A2ψ
∥∥2 −
(
α
1
2 β
1
2 +
ε
2
)2
‖Aψ‖4 .
Since ψ ∈ H4 is an arbitrary vector (‖ψ‖ = 1), it can be chosen. Namely, if the
operator A has an eigenvector, then we may chose this vector as ψ. Indeed, in this
case Aψ = µψ, and hε(ψ) = αβµ4 −
(
α
1
2 β
1
2 + ε
2
)2
µ4 < 0. If µ is a continuous
spectrum, then it is possible to �nd a vector ψ ∈ H4 such that Akψ = µkψ + o(δ)
218 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2
On the Boundary Value Problem with the Operator in Boundary Conditions
at δ → 0, k = 1, 2, 3, 4. Then hε(ψ) = αβµ4 −
(
α
1
2 β
1
2 + ε
2
)2
µ4 + o(δ) < 0 for
su�ciently small δ > 0. Thus, it is always possible to �nd a vector ψ ∈ H4
(‖ψ‖ = 1) such that min
ξ
qε(ξ, ψ) < 0. After choosing the vector ψ, by using the
continuity of the function qε(ξ, ψ) at ξ, there exists an interval (η1(ε), η2(ε)) in
which qε(ξ, ψ) < 0. Now we construct the function g(t). Let ĝ(ξ) be an arbitrary
twice continuously di�erentiable function in R whose support is contained in the
interval (η1(ε), η2(ε)). Then from equality (15) we get
E(g(t)ψ) =
η2(ε)∫
η1(ε)
qε(ξ, ψ) |ĝ(ξ)|2 dξ < 0,
and
g(t) =
1√
2π
η2(ε)∫
η1(ε)
ĝ(ξ)eiξtdξ.
The theorem is proved.
R e m a r k 1. We note that, in general, S(u) is not a positive number for
all u(t) ∈
o
W 2
2,T (R+; H). Indeed, let u(t) = e−λ0tϕ0, and ϕ0 be some eigenvector
of the operator A corresponding to λ0, ‖ϕ0‖ = 1. Then,
S(u) =
∥∥u′′
∥∥2
L2(R+;H)
+ αβ
∥∥A2u
∥∥2
L2(R+;H)
− 2βRe
(
u′′, A2u
)
L2(R+;H)
=
∥∥∥λ2
0e
−λ0tϕ0
∥∥∥
2
L2(R+;H)
+ αβ
∥∥∥λ2
0e
−λ0tϕ0
∥∥∥
2
L2(R+;H)
−2βRe
(
λ2
0e
−λ0tϕ0, λ
2
0e
−λ0tϕ0
)
L2(R+;H)
= (1 + αβ)λ4
0
∥∥∥e−λ0tϕ0
∥∥∥
2
L2(R+;H)
− 2βλ4
0
∥∥∥e−λ0tϕ0
∥∥∥
2
L2(R+;H)
= (1 + αβ − 2β)λ4
0
∥∥∥e−λ0tϕ0
∥∥∥
2
L2(R+;H)
= (1 + αβ − 2β)
λ3
0
2
.
Obviously, we can require T = −A−1, i.e., the condition Tu′(0) = u(0) to be
satis�ed. Then it is clear that for 2β − αβ > 1 the expression S(u) is negative,
and for 2β − αβ = 1 S(u) = 0, u(t) ∈
o
W 2
2,T (R+;H). Thus, for S(u) in the
space
o
W 2
2,T (R+; H) to be equivalent to the initial norm ‖u‖W 2
2 (R+;H), additional
conditions should be imposed on the operator T .
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2 219
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
Lemma 4. Let x ∈ H 1
2
. If min
‖x‖H 1
2
=1
〈R(0)x, x〉 > 0, then
S(u) ≥ const ‖u‖2
W 2
2 (R+;H)
for any u(t) ∈
o
W 2
2,T (R+; H).
P r o o f. From (10), for γ = 0 we get
S(u) ≥ 〈R(0)x, x〉 .
Since from the condition of the lemma min
‖x‖H 1
2
=1
〈R(0)x, x〉 = c > 0, it is obvious
that for ‖u‖W 2
2 (R+;H) = 1 the inequality
S(u) ≥ c
holds.
Since S(u) = ‖u‖2
W 2
2 (R+;H) S
(
u
‖u‖
W2
2 (R+;H)
)
, then for all u(t) ∈
o
W 2
2,T (R+; H)
S(u) ≥ c ‖u‖2
W 2
2 (R+;H)
is valid. The lemma is proved.
From Lemma 4 it follows that
ST = sup
06=u∈
o
W 2
2,T (R+;H)
∥∥Au′
∥∥
L2(R+;H)
S−
1
2 (u)
≤ 1
c
1
2
sup
06=u(t)∈
o
W 2
2,T (R+;H)
∥∥Au′
∥∥
L2(R+;H)
‖u‖−1
W 2
2 (R+;H)
= d,
and d < ∞ by the theorem on intermediate derivatives [2]. Consequently, ST < ∞.
Since
o
W 2
2,T (R+; H) ⊃
o
W 2
2 (R+; H), then ST ≥ S0 = 1√
2β
1
2 (α
1
2 +β
1
2 )
.
Theorem 4. Let the conditions of Lemma 4 be satis�ed and ReAT ≥ 0 in
H 1
2
. Then
ST =
1√
2β
1
2 (α
1
2 + β
1
2 )
.
220 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2
On the Boundary Value Problem with the Operator in Boundary Conditions
P r o o f. If the conditions of the theorem are satis�ed, then 〈R(γ)x, x〉 > 0
for any γ ∈ [0, 2β
1
2 (α
1
2 + β
1
2 )). Consequently, by Lemma 3 it follows that for any
u(t) ∈
o
W 2
2,T (R+;H) and γ ∈ [0, 2β
1
2 (α
1
2 + β
1
2 )) the inequality
S(u) ≥ γ
∥∥Au′
∥∥2
L2(R+;H)
is true. Passing in the last inequality to the limit as γ → 2β
1
2 (α
1
2 + β
1
2 ), we have
ST ≤ 1√
2β
1
2 (α
1
2 + β
1
2 )
.
Thereby,
ST =
1√
2β
1
2 (α
1
2 + β
1
2 )
.
The theorem is proved.
Basing on the obtained results, estimating the norm of the intermediate deriva-
tive operator A d
dt :
o
W 2
2,T (R+;H) → L2(R+; H) with respect to the norm
‖P0u‖L2(R+;H) and taking into account Lemma 1, we give the exact formulation
of the conditional Theorem 2 in the following form.
Theorem 5. Suppose that the conditions of Lemma 4 are satis�ed,
ReAT ≥ 0 in H 1
2
and the operator B = A1A
−1 is bounded in H, moreover,
‖B‖ <
√
2α
(
1 + α
1
2
β
1
2
)
. Then problem (1), (2) is regularly solvable.
R e m a r k 2. In Theorem 5, the condition ReAT ≥ 0 in H 1
2
provides
invertibility of the operator Tα,β in the space H 3
2
.
Now we will specify the value of ST under the condition min
‖x‖H 1
2
=1
Re(ATx, x) 1
2
<0.
The following theorem holds.
Theorem 6. Suppose that min
‖x‖H 1
2
=1
Re(ATx, x) 1
2
< 0 and the conditions of
Lemma 4 are satis�ed. Then
ST =
1√
2β
1
2 (α
1
2 + β
1
2 )
1− 2β
1
2 (α
1
2 + β
1
2 )
∣∣∣∣∣∣∣
min
‖x‖H 1
2
=1
Re(ATx, x) 1
2
1 + α
1
2 β
1
2 ‖ATx‖2
H 1
2
∣∣∣∣∣∣∣
2
− 1
2
.
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2 221
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
P r o o f. First we note that since the pencil F (λ; γ; A) for γ ∈ [0, 2β
1
2 (α
1
2 +
β
1
2 )) is represented in the form
F (λ; γ; A) = (λE − ω1(γ)A)(λE − ω2(γ)A),
where Reωk(γ) < 0, k = 1, 2, then the Cauchy problem
F
(
d
dt
; γ; A
)
u = 0, (16)
u(0) = Tx, u′(0) = x, x ∈ H 1
2
, (17)
has the unique solution u(t; γ;x) ∈ W 2
2 (R+; H) represented in the form
u(t; γ; x)
=
1
ω2(γ)− ω1(γ)
[
eω1(γ)tA(ω2(γ)Tx−A−1x) + eω2(γ)tA(A−1x− ω1(γ)Tx)
]
.
It follows easily that
‖u(t; γ;x)‖W 2
2 (R+;H) ≤ d1(γ) ‖x‖H 1
2
.
Further, using the uniqueness of the solution of Cauchy problem (16), (17) and
taking into account the Banach theorem on the inverse operator, we have
‖u(t; γ;x)‖W 2
2 (R+;H) ≥ d2(γ) ‖x‖H 1
2
, d2(γ) > 0. (18)
Now we note that from Theorem 4 it follows that
ST >
1√
2β
1
2 (α
1
2 + β
1
2 )
.
Consequently, S−2
T ∈ (0, 2β
1
2 (α
1
2 + β
1
2 )). Then, in equality (10), taking for u(t)
the solution u(t; γ; x) of Cauchy problem (16), (17), for ‖x‖H 1
2
= 1, we obtain
〈R(γ)x, x〉 = S(u(t; γ; x))− γ
∥∥Au′(t; γ; x)
∥∥2
L2(R+;H)
= S(u(t; γ; x))(1− γS2
T )>0
(19)
for γ ∈ [0, S−2
T ). Taking into account Lemma 4 and inequality (18), we obtain
〈R(γ)x, x〉 ≥ cd2
2(γ)(1− γS2
T ) > 0.
By that, min
‖x‖H 1
2
=1
〈R(γ)x, x〉 > 0 for γ ∈ [0, S−2
T ). The same argument can be used
for the case ω1(γ) = ω2(γ).
222 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2
On the Boundary Value Problem with the Operator in Boundary Conditions
Continuing, by de�nition ST , for all γ ∈ (S−2
T , 2β
1
2 (α
1
2 + β
1
2 )) there exists a
function v(t; γ) ∈
o
W 2
2,T (R+;H) such that
S(v(t; γ)) < γ
∥∥Av′(t; γ)
∥∥2
L2(R+;H)
.
Consequently, for γ ∈ (S−2
T , 2β
1
2 (α
1
2 + β
1
2 )) from equality (10) it follows that
〈R(γ)xγ , xγ〉+
∥∥∥∥F
(
d
dt
; γ;A
)
v(t; γ)
∥∥∥∥
2
L2(R+;H)
< 0, xγ = v′(0; γ),
i.e.,
min
‖x‖H 1
2
=1
〈R(γ)x, x〉 ≤ 〈R(γ)xγ , xγ〉 < 0.
Thus, since min
‖x‖H 1
2
=1
〈R(γ)x, x〉 is a continuous function and it changes its sign at
the point γ = S−2
T , then min
‖x‖H 1
2
=1
〈
R(S−2
T )x, x
〉
= 0.
To complete the proof, we consider the functional of the form
Q(γ;x) =
√
2β
1
2 (α
1
2 + β
1
2 )− γ+2β
1
2 (α
1
2 +β
1
2 )
Re(ATx, x) 1
2
1 + α
1
2 β
1
2 ‖ATx‖2
H 1
2
, ‖x‖H 1
2
= 1.
Since
Q(γ; x) =
〈R(γ)x, x〉
1 + α
1
2 β
1
2 ‖ATx‖2
H 1
2
≤ 〈R(γ)x, x〉 ,
then
min
‖x‖H 1
2
=1
Q(S−2
T ; x) ≤ min
‖x‖H 1
2
=1
〈
R(S−2
T )x, x
〉
= 0.
On the other hand,
〈
R(S−2
T )x, x
〉
= Q(S−2
T ;x)
(
1 + α
1
2 β
1
2 ‖ATx‖2
H 1
2
)
≤ Q(S−2
T ;x)
(
1 + α
1
2 β
1
2 ‖AT‖2
H 1
2
→H 1
2
)
and, consequently,
min
‖x‖H 1
2
=1
Q(S−2
T ; x) ≥
(
1 + α
1
2 β
1
2 ‖AT‖2
H 1
2
→H 1
2
)−1
min
‖x‖H 1
2
=1
〈
R(S−2
T )x, x
〉
= 0.
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 2 223
S.S. Mirzoev, A.R. Aliev, and L.A. Rustamova
Thus we have
min
‖x‖H 1
2
=1
Q(S−2
T ; x) = 0.
So,
√
2β
1
2 (α
1
2 + β
1
2 )− S−2
T = −2β
1
2 (α
1
2 + β
1
2 ) min
‖x‖H 1
2
=1
Re(ATx, x) 1
2
1 + α
1
2 β
1
2 ‖ATx‖2
H 1
2
.
Then, taking into account min
‖x‖H 1
2
=1
Re(ATx, x) 1
2
< 0, from the above we get
ST =
1√
2β
1
2 (α
1
2 + β
1
2 )
1− 2β
1
2 (α
1
2 + β
1
2 )
∣∣∣∣∣∣∣
min
‖x‖H 1
2
=1
Re(ATx, x) 1
2
1 + α
1
2 β
1
2 ‖ATx‖2
H 1
2
∣∣∣∣∣∣∣
2
1
2
.
The theorem is proved.
In the case when min
‖x‖H 1
2
=1
Re(ATx, x) 1
2
< 0, Theorem 2 is formulated as
follows.
Theorem 7. Suppose that the conditions of Theorem 6 are satis�ed, the
operator Tα,β is bounded invertible in the space H 3
2
and the operator B = A1A
−1
is bounded in H, moreover
‖B‖ <
√√√√2α
(
1 +
α
1
2
β
1
2
)
1− 2β
1
2 (α
1
2 + β
1
2 )
∣∣∣∣∣∣∣
min
‖x‖H 1
2
=1
Re(ATx, x) 1
2
1 + α
1
2 β
1
2 ‖ATx‖2
H 1
2
∣∣∣∣∣∣∣
2
1
2
.
Then problem (1), (2) is regularly solvable.
Despite the fact that equation (1) was presented in a more general form in the
paper of A.R.Aliev [18], in this paper only the case when ReAT ≥ 0 is studied.
Moreover, our results improve the results obtained in [18].
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