Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 219, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
THIRD-ORDER OPERATOR-DIFFERENTIAL EQUATIONS
WITH DISCONTINUOUS COEFFICIENTS AND OPERATORS IN
THE BOUNDARY CONDITIONS
ARAZ R. ALIEV, NAZILA L. MURADOVA
Abstract. We study a third-order operator-differential equation on the semiaxis with a discontinuous coefficient and boundary conditions which include an
abstract linear operator. Sufficient conditions for the well-posed and unique
solvability are found by means of properties of the operator coefficients in a
Sobolev-type space.
1. Introduction
It is known that many problems of partial differential equations can be reduced to
problems for differential equations whose coefficients are unbounded operators in a
Hilbert space. Many articles are dedicated to the study of problems with operators
in the boundary conditions for operator-differential equations of second order (see,
for example, [1, 7, 10, 13, 16, 17, 18, 25] and the references therein); however, these
studies are far from the full completion. Note that only a few papers are dedicated
to the study of such boundary-value problems for operator-differential equations of
third order (see, for example, [5]).
This article is dedicated to the study of boundary-value problem for a class
of third-order operator-differential equations with a discontinuous coefficient; one
of the boundary conditions includes an abstract linear operator. Such equations
cover some non-classical problems of mathematical physics (see [8]), investigated in
inhomogeneous environments.
Let H be a separable Hilbert space with the scalar product (x, y), x, y ∈ H and
let A be a self-adjoint positive-definite operator in H (A = A∗ ≥ cE, c > 0, E is the
identity operator). By Hγ (γ ≥ 0) we denote the scale of Hilbert spaces generated
by the operator A; i.e., Hγ = D(Aγ ), (x, y)γ = (Aγ x, Aγ y), x, y ∈ D(Aγ ), for γ = 0
we consider that H0 = H, (x, y)0 = (x, y), x, y ∈ H.
2000 Mathematics Subject Classification. 47E05, 34B40, 34G10.
Key words and phrases. Operator-differential equation; discontinuous coefficient;
operator-valued boundary condition; self-adjoint operator; regular solvability;
Sobolev-type space; intermediate derivative operators.
c 2013 Texas State University - San Marcos.
Submitted June 7, 2013. Published October 4, 2013.
1
2
A. R. ALIEV, N. L. MURADOVA
EJDE-2013/219
We denote by L2 ([a, b]; H), −∞ ≤ a < b ≤ +∞, the Hilbert space of all vector
functions defined on [a, b] with values in H and endowed with the norm
Z b
1/2
kf kL2 ([a,b];H) =
kf (t)k2H dt
.
a
Following the book [14], we introduce the Hilbert space
W23 ([a, b]; H) = {u(t) : u000 (t) ∈ L2 ([a, b]; H), A3 u(t) ∈ L2 ([a, b]; H)}
endowed with the norm
kukW23 ([a,b];H) = ku000 k2L2 ([a,b];H) + kA3 uk2L2 ([a,b];H)
1/2
.
Hereafter, derivatives are understood in the sense of distributions in a Hilbert
space [14]. The spaces L2 ((−∞, +∞); H), W23 ((−∞, +∞); H), L2 ([0, +∞); H)
and W23 ([0, +∞); H) will be denoted by L2 (R; H), W23 (R; H), L2 (R+ ; H) and
W23 (R+ ; H), respectively.
Further, we denote by L(X, Y ) the space of all linear bounded operators acting
from a Hilbert space X to another Hilbert space Y , and we denote by σ(·) the
spectrum of the operator (·).
Consider the boundary value problem in the Hilbert space H
−u000 (t) + ρ(t)A3 u(t) +
3
X
Aj
j=1
u0 (0) = 0,
d3−j u(t)
= f (t), t ∈ R+ ,
dt3−j
u00 (0) = Ku(0),
(1.1)
(1.2)
∗
where A = A ≥ cE, c > 0, K ∈ L(H5/2 , H1/2 ), ρ(t) = α, if 0 ≤ t ≤ 1, ρ(t) = β, if
1 < t < +∞, here α, β are positive numbers, f (t) ∈ L2 (R+ ; H), u(t) ∈ W23 (R+ ; H).
Definition 1.1. If a vector function u(t) ∈ W23 (R+ ; H) satisfies (1.1) almost everywhere in R+ , then it is called a regular solution of equation (1.1).
Definition 1.2. If for any f (t) ∈ L2 (R+ ; H) there exists a regular solution of (1.1),
which satisfies the boundary conditions (1.2) in the sense that
lim ku0 (t)kH3/2 = 0,
t→0
lim ku00 (t) − Ku(t)kH1/2 = 0
t→0
and the following inequality holds
kukW23 (R+ ;H) ≤ const kf kL2 (R+ ;H) ,
then we say that the problem (1.1), (1.2) is regularly solvable.
Similar kind of problems on a semi-axis for elliptic operator-differential equations
of the second order is considered in papers [13, 16, 17]. We should especially note
the work [21] which considers the non-local boundary value problems for second
order elliptic operator-differential equations on the interval with the coefficients
belonging to a broader class of discontinuous functions, while the coefficients in
the boundary conditions are complex numbers. In [2, 8, 11, 12, 15, 19, 20, 22, 23]
along with other problems investigated the solvability of boundary-value problems
for elliptic operator-differential equations of the general form when the coefficients
in the boundary conditions are complex numbers and the equations do not contain
discontinuous coefficients. Such case also is considered in [3, 6] for the third and
fourth orders equations with multiple characteristics. Note that problem (1.1), (1.2)
EJDE-2013/219
THIRD-ORDER OPERATOR-DIFFERENTIAL EQUATIONS
3
is investigated in the case A3 = 0, K = 0 in [4] and when ρ(t) ≡ 1, t ∈ R+ , K = 0
in [15].
In this article, we obtain the conditions of regular solvability of the boundaryvalue problem (1.1), (1.2) by means of properties of operator coefficients.
2. Main results
Before proceeding to the consideration of the question posed, let us introduce
additional notation. Let
3
W2,K
(R+ ; H) = {u(t) : u(t) ∈ W23 (R+ ; H), u0 (0) = 0, u00 (0) = Ku(0)}
3
(R+ ; H) into
and denote by P0 , P1 and P the operators acting from the space W2,K
the space L2 (R+ ; H) by the following rules, respectively:
P0 u(t) = −u000 (t) + ρ(t)A3 u(t),
P1 u(t) = A1 u00 (t) + A2 u0 (t) + A3 u(t),
P u(t) = P0 u(t) + P1 u(t),
Put B = A1/2 KA−5/2 , κ(c1 , c2 , c3 ) = c1
p
3
3
u(t) ∈ W2,K
(R+ ; H).
√
√
3
β 2 + c2 3 αβ + c3 α2 and
√
1
3
ω2 A−2 K)(κ(1, 1, 1)ω2 E − κ(1, ω2 , ω1 )e α(ω2 −1)A )
Kα,β = (E + √
3
α2
√
1
3 α(ω −1)A
−2
1
ω
A
K
κ(1,
ω
,
ω
)e
+ E+ √
−
κ(1,
1,
1)ω
E
,
1
1
2
1
3
α2
where ω1 = − 12 +
√
3
2 i,
ω2 = − 12 −
√
3
2 i.
√
3
/ σ(B)
Lemma 2.1. Let A = A∗ ≥ cE, c > 0, K ∈ L(H5/2 , H1/2 ), − α2 ω2 ∈
and the operator Kα,β have a bounded inverse operator in H5/2 . Then the equation
3
P0 u = 0 has only the trivial solution in the space W2,K
(R+ ; H).
Proof. The general solution of the equation P0 u(t) = 0 in the space W23 (R+ ; H)
has the following form [24]:
(
u0 (t) =
√
3
√
3
√
3
u0,1 (t) = e αω1 tA ϕ0 + e αω2 tA ϕ1 + e− α(1−t)A ϕ2 , 0 ≤ t < 1,
√
√
3
3
u0,2 (t) = e βω1 (t−1)A ϕ3 + e βω2 (t−1)A ϕ4 ,
1 < t < +∞,
where the vectors ϕk ∈ H5/2 , k = 0, 1, 2, 3, 4, are determined from the boundary
conditions (1.2) and the condition u0 (t) ∈ W23 (R+ ; H). Therefore, to determine
the vectors ϕk , k = 0, 1, 2, 3, 4, we have the following relations:
u00,1 (0) = 0,
u000,1 (0) = Ku0,1 (0),
u00,1 (1)
=
u00,2 (1),
u000,1 (1)
u0,1 (1) = u0,2 (1),
= u000,2 (1).
4
A. R. ALIEV, N. L. MURADOVA
EJDE-2013/219
From these relations we obtain the following system of equations with respect to
ϕk , k = 0, 1, 2, 3, 4:
√
3
ω1 ϕ0 + ω2 ϕ1 + e− αA ϕ2 = 0 ,
√
√
1
3
3
ω12 ϕ0 + ω22 ϕ1 + e− αA ϕ2 = √
A−2 K(ϕ0 + ϕ1 + e− αA ϕ2 ),
3
2
α
√
√
3 αω A
3 αω A
(2.1)
1
2
ϕ0 + e
ϕ1 + ϕ2 = ϕ3 + ϕ4 ,
e
√
√
p
p
√
√
√
3
3
3
αω1 e αω1 A ϕ0 + 3 αω2 e αω2 A ϕ1 + 3 αϕ2 = 3 βω1 ϕ3 + 3 βω2 ϕ4 ,
p
p
√
√
√
√
√
3
3
3
3
3
α2 ω12 e αω1 A ϕ0 + α2 ω22 e αω2 A ϕ1 + α2 ϕ2 = 3 β 2 ω12 ϕ3 + 3 β 2 ω22 ϕ4 .
Taking into account ω1 ω2 = 1, ω1 + ω2 = −1, ω12 = ω2 , ω22 = ω1 , from the system
(2.1) after simple transformations with respect to ϕ0 we have
√
1
3
ω2 A−2 K)(κ(1, 1, 1)ω2 E − κ(1, ω2 , ω1 )e α(ω2 −1)A )ϕ0
(E + √
3
2
α
√
1
3
+ (E + √
ω1 A−2 K)(κ(1, ω1 , ω2 )e α(ω1 −1)A − κ(1, 1, 1)ω1 E)ϕ0 = 0.
3
2
α
Consequently,
h
√
1
3 α(ω −1)A −2
2
Kα,β ϕ0 ≡ E + √
ω
A
K
κ(1,
1,
1)ω
E
−
κ(1,
ω
,
ω
)e
2
2
2
1
3
2
α
√
1
3
+ E+ √
ω1 A−2 K κ(1, ω1 , ω2 )e α(ω1 −1)A − κ(1, 1, 1)ω1 E ]ϕ0 = 0.
3
α2
(2.2)
By the assumption of this lemma, Kα,β has a bounded inverse operator in the space
H5/2 , then from equation (2.2) follows that ϕ0 = 0. Considering ϕ0 = 0 in the first
and second equations of (2.1), we obtain
√
3
ϕ1 = −ω1 e− αA ϕ2 ,
(2.3)
√
1
3
(E + √
ω1 A−2 K)e− αA ϕ2 = 0.
(2.4)
3
α2
√
3
/ σ(B) from (2.4) it follows that ϕ2 = 0, and
In turn, by assumption − α2 ω2 ∈
therefore, from (2.3) ϕ1 = 0. Now, considering ϕ0 = ϕ1 = ϕ2 = 0 in the fourth and
fifth equations of (2.1), we obtain:
ω1 ϕ3 + ω2 ϕ4 = 0,
ω2 ϕ3 + ω1 ϕ4 = 0.
And from these equations we have that ϕ3 = ϕ4 = 0. Thus, u0 (t) = 0. The proof
is complete.
Let us consider the question of regular solvability of problem (1.1), (1.2) when
A1 = A2 = A3 = 0.
Lemma 2.2. In the conditions of Lemma 2.1, the problem
−u000 (t) + ρ(t)A3 u(t) = f (t),
0
u (0) = 0,
is regularly solvable.
00
t ∈ R+ ,
u (0) = Ku(0)
(2.5)
(2.6)
EJDE-2013/219
THIRD-ORDER OPERATOR-DIFFERENTIAL EQUATIONS
5
3
Proof. We show that the equation P0 u(t) = f (t) has a solution u(t) ∈ W2,K
(R+ ; H)
for any f (t) ∈ L2 (R+ ; H). First, we continue the vector-function f (t) in such a way
that f (t) = 0 for t < 0. We denote the new function by g(t). Let ĝ(ξ) be the Fourier
transform of the vector-function g(t); i.e.,
Z +∞
1
g(t)e−iξt dt,
ĝ(ξ) = √
2π −∞
where the integral on the right side is understood in the sense of convergence on
the average in H. Then, using the direct and inverse Fourier transforms, it is clear
that the vector-functions
Z +∞
Z +∞
1
3
3 −1
f (s)e−iξs ds eitξ dξ, t ∈ R,
υ1 (t) =
(iξ E + αA )
2π −∞
0
Z +∞
Z +∞
1
υ2 (t) =
(iξ 3 E + βA3 )−1
f (s)e−iξs ds eitξ dξ, t ∈ R,
2π −∞
0
satisfy the equations
d3 υ(t)
+ αA3 υ(t) = g(t),
dt3
d3 υ(t)
−
+ βA3 υ(t) = g(t),
dt3
respectively, almost everywhere in R. We prove that υ1 (t) and υ2 (t) belong to
W23 (R; H). By Plancherel’s theorem
−
kυ1 (t)k2W 3 (R;H) = kυ1000 (t)k2L2 (R;H) + kA3 υ1 (t)k2L2 (R;H)
2
= k − iξ 3 υ̂1 (ξ)k2L2 (R;H) + kA3 υ̂1 (ξ)k2L2 (R;H) ,
where υ̂1 (ξ) is the Fourier transform of the function υ1 (t). Since
υ̂1 (ξ) = (iξ 3 E + αA3 )−1 ĝ(ξ),
we have
k − iξ 3 υ̂1 (ξ)kL2 (R;H) = k − iξ 3 (iξ 3 E + αA3 )−1 ĝ(ξ)kL2 (R;H)
≤ sup k − iξ 3 (iξ 3 E + αA3 )−1 kH→H kĝ(ξ)kL2 (R;H)
ξ∈R
3
3 −1
3
= sup k − iξ (iξ E + αA )
ξ∈R
(2.7)
kH→H kg(t)kL2 (R;H) .
It follows from the spectral theory of self-adjoint operators that
k − iξ 3 (iξ 3 E + αA3 )−1 k = sup
− iξ 3 (iξ 3 + ασ 3 )−1
σ∈σ(A)
|ξ|3
≤ 1.
6
2 6 1/2
σ∈σ(A) (ξ + α σ )
(2.8)
= sup
Therefore, from (2.7) it follows that −iξ 3 υ̂1 (ξ) ∈ L2 (R; H). Since
kA3 υ̂1 (ξ)kL2 (R;H) = kA3 (iξ 3 E + αA3 )−1 ĝ(ξ)kL2 (R;H)
≤ sup kA3 (iξ 3 E + αA3 )−1 kH→H kĝ(ξ)kL2 (R;H)
ξ∈R
3
3
3 −1
= sup kA (iξ E + αA )
ξ∈R
kH→H kg(t)kL2 (R;H) ,
(2.9)
6
A. R. ALIEV, N. L. MURADOVA
EJDE-2013/219
again from the spectral theory of self-adjoint operators, we have
σ3
1
≤ .
6 + α2 σ 6 )1/2
α
(ξ
σ∈σ(A)
kA3 (iξ 3 E + αA3 )−1 k = sup |σ 3 (iξ 3 + ασ 3 )−1 | = sup
σ∈σ(A)
Thus, from (2.9) it follows that A3 υ̂1 (ξ) ∈ L2 (R; H). Hence, υ1 (t) ∈ W23 (R; H).
Thus υ2 (t) ∈ W23 (R; H).
Let us denote the restriction of the vector-function υ1 (t) on [0, 1) by uα (t) and
the restriction of the vector-function υ2 (t) on (1, +∞) by uβ (t). It is obvious, that
uα (t) ∈ W23 ([0, 1); H), uβ (t) ∈ W23 ((1, +∞); H). Then, from the theorem on traces
s
s
ds u (0) ds u (1)
[14, ch. 1] it follows that d udtαs(0) , d udtαs(1) , dtβs , dtβs ∈ H5/2−s , s = 0, 1, 2.
Now, we denote
(
√
√
√
3
3
3
u1 (t) = uα (t) + e αω1 tA ψ0 + e αω2 tA ψ1 + e− α(1−t)A ψ2 , 0 ≤ t < 1,
√
√
u(t) =
3
3
u2 (t) = uβ (t) + e βω1 (t−1)A ψ3 + e βω2 (t−1)A ψ4 ,
1 < t < +∞,
3
where ψk ∈ H5/2 , k = 0, 1, 2, 3, 4. The function u(t) belongs to W2,K
(R+ ; H), so
the vectors ψk , k = 0, 1, 2, 3, 4, can be determined from the following relations:
u01 (0) = 0,
u001 (0) = Ku1 (0),
u1 (1) = u2 (1),
u01 (1) = u02 (1),
u001 (1) = u002 (1).
From here with respect to ψk , k = 0, 1, 2, 3, 4, we have the system of equations
√
√
√
√
3
u0α (0) + 3 αω1 Aψ0 + 3 αω2 Aψ1 + 3 αAe− αA ψ2 = 0 ,
√
√
√
3
3
3
u00α (0) + α2 A2 ω12 ψ0 + ω22 ψ1 + e− αA ψ2 = Kuα (0) + K ψ0 + ψ1 + e− αA ψ2 ,
√
3
√
3
uα (1) + e αω1 A ψ0 + e αω2 A ψ1 + ψ2 = uβ (1) + ψ3 + ψ4 ,
√
√
√
√
√
3
3
u0α (1) + 3 αω1 Ae αω1 A ψ0 + 3 αω2 Ae αω2 A ψ1 + 3 αAψ2
p
p
= u0β (1) + 3 βω1 Aψ3 + 3 βω2 Aψ4 ,
√
√
√
√
√
3
3
3
3
3
u00α (1) + α2 ω12 A2 e αω1 A ψ0 + α2 ω22 A2 e αω2 A ψ1 + α2 A2 ψ2
p
p
= u00β (1) + 3 β 2 ω12 A2 ψ3 + 3 β 2 ω22 A2 ψ4 .
From this system we obtain:
√
√
√
√
3
3
αω1 ψ0 + 3 αω2 ψ1 + 3 αe− αA ψ2 = −A−1 u0α (0),
√
√
√
3
3
3
α2 (ω12 ψ0 + ω22 ψ1 + e− αA ψ2 ) − A−2 K(ψ0 + ψ1 + e− αA ψ2 )
= A−2 (Kuα (0) − u00α (0)) ,
√
3
√
3
αω1 A
√
3
ψ0 + e αω2 A ψ1 + ψ2 − ψ3 − ψ4 = uβ (1) − uα (1),
√
√
p
p
√
√
√
3
3
3
αω1 e αω1 A ψ0 + 3 αω2 e αω2 A ψ1 + 3 αψ2 − 3 βω1 ψ3 − 3 βω2 ψ4
e
(2.10)
= A−1 (u0β (1) − u0α (1)) ,
p
p
√
√
√
√
3
3
3
3
α2 ω12 e αω1 A ψ0 + α2 ω22 e αω2 A ψ1 + α2 ψ2 − 3 β 2 ω12 ψ3 − 3 β 2 ω22 ψ4
= A−2 (u00β (1) − u00α (1)) .
Since uα (t) ∈ W23 ([0, 1); H) and uβ (t) ∈ W23 ((1, +∞); H), by the theorem on traces
[14, ch. 1] A−1 u0α (0), A−2 (Kuα (0) − u00α (0)), uβ (1) − uα (1), A−1 (u0β (1) − u0α (1))
and A−2 (u00β (1) − u00α (1)) belong to H5/2 . Then by these values acting also as in the
system (2.1), in this case, taking into account
that the operator Kα,β has a bounded
√
3
inverse operator in the space H5/2 and − α2 ω2 ∈
/ σ(B), obviously, from (2.10) it is
EJDE-2013/219
THIRD-ORDER OPERATOR-DIFFERENTIAL EQUATIONS
7
possible to find the vectors ψk , k = 0, 1, 2, 3, 4, where all ψk ∈ H5/2 , k = 0, 1, 2, 3, 4.
Therefore, u(t) ∈ W23 (R+ ; H) satisfies equation (2.5) almost everywhere in R+ and
conditions (2.6).
By lemma 2.1, the problem
−u000 (t) + ρ(t)A3 u(t) = 0,
u0 (0) = 0,
u00 (0) = Ku(0)
3
has only the trivial solution in the space W2,K
(R+ ; H).
3
Now we show that the operator P0 : W2,K (R+ ; H) → L2 (R+ ; H) is bounded.
3
Indeed, for u(t) ∈ W2,K
(R+ ; H) we have
kP0 uk2L2 (R+ ;H)
= ku000 k2L2 (R+ ;H) + kρ(t)A3 uk2L2 (R+ ;H) − 2 Re(u000 , ρ(t)A3 u)L2 (R+ ;H)
≤ ku000 k2L2 (R+ ;H) + kρ(t)A3 uk2L2 (R+ ;H) + 2ku000 kL2 (R+ ;H) kρ(t)A3 ukL2 (R+ ;H)
≤ 2 ku000 k2L2 (R+ ;H) + kρ(t)A3 uk2L2 (R+ ;H)
≤ 2 max(1; α2 ; β 2 )kuk2W 3 (R+ ;H) .
2
Thus, according to the Banach theorem on the inverse operator, there exists P0−1 :
3
(R+ ; H) and it is bounded. Hence, it follows that
L2 (R+ ; H) → W2,K
kukW23 (R+ ;H) ≤ constkf kL2 (R+ H) .
The proof is complete.
On the basis of Lemmas 2.1 and 2.2 we obtain the following conclusion.
Theorem 2.3. Let the conditions of Lemma 2.1 be satisfied. Then the operator P0
3
(R+ ; H) and L2 (R+ ; H).
is an isomorphism between the spaces W2,K
Let us prove the following coercive inequality which will be used further.
3
(R+ ; H), the following
Lemma 2.4. Let Re(B) ≥ 0. Then for any u(t) ∈ W2,K
inequality holds
kρ−1/2 (t)u000 k2L2 (R+ ;H) + kρ1/2 (t)A3 uk2L2 (R+ ;H) ≤
1
kP0 uk2L2 (R+ ;H) . (2.11)
min(α; β)
Proof. Consider the following equalities:
(P0 u, A3 u)L2 (R+ ;H) = (−u000 + ρ(t)A3 u, A3 u)L2 (R+ ;H)
= (−u000 , A3 u)L2 (R+ ;H) + (ρ(t)A3 u, A3 u)L2 (R+ ;H)
000
3
= (−u , A u)L2 (R+ ;H) + kρ
1/2
(t)A
3
(2.12)
uk2L2 (R+ ;H) ,
(P0 u, −ρ−1 (t)u000 )L2 (R+ ;H) = (−u000 + ρ(t)A3 u, −ρ−1 (t)u000 )L2 (R+ ;H)
= (−u000 , −ρ−1 (t)u000 )L2 (R+ ;H) − (A3 u, u000 )L2 (R+ ;H)
= kρ−1/2 (t)u000 k2L2 (R+ ;H) − (A3 u, u000 )L2 (R+ ;H) .
(2.13)
3
Note that by integrating by parts for u(t) ∈ W2,K
(R+ ; H), we have
− Re(u000 , A3 u)L2 (R+ ;H) = Re(BA5/2 u(0), A5/2 u(0)).
(2.14)
8
A. R. ALIEV, N. L. MURADOVA
EJDE-2013/219
By (2.12) and (2.13) and taking into account (2.14), we obtain
(P0 u, A3 u − ρ−1 (t)u000 )L2 (R+ ;H)
= kρ1/2 (t)A3 uk2L2 (R+ ;H) + kρ−1/2 (t)u000 k2L2 (R+ ;H) + 2 Re(BA5/2 u(0), A5/2 u(0)).
(2.15)
Applying the Cauchy-Schwarz inequality to the left side and then the Young’s
inequality and taking into account (2.14), we obtain
(P0 u, A3 u − ρ−1 (t)u000 )L2 (R+ ;H)
≤ kρ−1/2 (t)P0 ukL2 (R+ ;H) kρ1/2 (t)A3 u − ρ−1/2 (t)u000 kL2 (R+ ;H)
1
1
≤
kP0 uk2L2 (R+ ;H) + kρ1/2 (t)A3 u − ρ−1/2 (t)u000 k2L2 (R+ ;H)
2 min(α; β)
2
1
1
kP0 uk2L2 (R+ ;H) + kρ1/2 (t)A3 uk2L2 (R+ ;H)
=
2 min(α; β)
2
1
+ kρ−1/2 (t)u000 k2L2 (R+ ;H) + Re BA5/2 u(0), A5/2 u(0) .
2
(2.16)
Taking into account (2.16) into (2.15), we have
kρ−1/2 (t)u000 k2L2 (R+ ;H) + kρ1/2 (t)A3 uk2L2 (R+ ;H) + 2 Re(BA5/2 u(0), A5/2 u(0))
1
≤
kP0 uk2L2 (R+ ;H) .
min(α; β)
(2.17)
Since Re B ≥ 0, then from inequality (2.17), we obtain the validity of inequality
(2.11). The proof is complete.
Theorem 2.3 implies that the norm kP0 ukL2 (R+ ;H) is equivalent to the norm
3
(R+ ; H). Therefore, the norms of the intermedikukW23 (R+ ;H) in the space W2,K
3−j
d
3
ate derivative operators Aj dt
3−j : W2,K (R+ ; H) → L2 (R+ ; H), j = 1, 2, 3, can
be estimated with respect to kP0 ukL2 (R+ ;H) (by the continuity of these operators
[14]). Methods for solution of equations with scalar boundary conditions are often
inapplicable to the problems with boundary conditions which include abstract operators. For example, when K = 0, operator pencil factorization method for the
estimation of the norms of intermediate derivative operators has been developed in
[4] (this method was first mentioned in [15] when considering operator-differential
equations with constant coefficients). The estimates for the norms of intermediate
derivative operators are playing an important role in obtaining solvability conditions. But, the method of [4] is not applicable to the boundary value problems
for odd order operator-differential equations with the boundary conditions which
include abstract operators. In this work, to estimate the norms of intermediate
derivative operators we use the classical inequalities of mathematical analysis and
the coercive inequality (2.11).
3
Theorem 2.5. Let Re B ≥ 0. Then for any u(t) ∈ W2,K
(R+ ; H) the following
inequalities hold:
kAj
d3−j u
kL (R ;H) ≤ aj kP0 ukL2 (R+ ;H) ,
dt3−j 2 +
j = 1, 2, 3,
(2.18)
EJDE-2013/219
THIRD-ORDER OPERATOR-DIFFERENTIAL EQUATIONS
9
where
a1 =
21/3 max1/3 (α; β)
31/2
2/3
min
(α; β)
,
a2 =
21/3 max1/6 (α; β)
31/2
5/6
min
(α; β)
,
a3 =
1
.
min(α; β)
3
W2,K
(R+ ; H).
Proof. Let u(t) ∈
Integrating by parts and applying the CauchySchwarz inequality, and then the Young’s inequality, we obtain
Z +∞
(Au00 , Au00 )H dt
kAu00 k2L2 (R+ ;H) =
0
Z +∞
= (Au0 , Au00 )H |+∞
−
(Au0 , Au000 )H dt
0
0
Z +∞
(2.19)
2 0
000
(A u , u )H dt ≤ kA2 u0 kL2 (R+ ;H) ku000 kL2 (R+ ;H)
=−
0
≤ max ρ1/2 (t)kA2 u0 kL2 (R+ ;H) kρ−1/2 (t)u000 kL2 (R+ ;H)
t
1
ε
≤ max(α; β)kA2 u0 k2L2 (R+ ;H) + kρ−1/2 (t)u000 k2L2 (R+ ;H) ,
2
2ε
with ε > 0. Proceeding in a similar manner, we have
Z +∞
kA2 u0 k2L2 (R+ ;H) =
(A2 u0 , A2 u0 )H dt
0
Z +∞
= (A2 u, A2 u0 )H |+∞
−
(A2 u, A2 u00 )H dt
0
0
Z +∞
3
00
=−
(A u, Au )H dt
(2.20)
0
≤ kA3 ukL2 (R+ ;H) kAu00 kL2 (R+ ;H)
≤ max ρ−1/2 (t)kAu00 kL2 (R+ ;H) kρ1/2 (t)A3 ukL2 (R+ ;H)
t
1
η
1
kAu00 k2L2 (R+ ;H) + kρ1/2 (t)A3 uk2L2 (R+ ;H) ,
≤
2 min(α; β)
2η
with η > 0. Taking into account inequality (2.20) in (2.19):
kAu00 k2L2 (R+ ;H) ≤
ε
η
1
max(α; β)(
kAu00 k2L2 (R+ ;H)
2
2 min(α; β)
(2.21)
1
1
+ kρ1/2 (t)A3 uk2L2 (R+ ;H) ) + kρ−1/2 (t)u000 k2L2 (R+ ;H) .
2η
2ε
From this inequality we obtain
εη max(α; β) ε max(α; β) 1/2
1−
kAu00 k2L2 (R+ ;H) ≤
kρ (t)A3 uk2L2 (R+ ;H)
4 min(α; β)
4η
1
+ kρ−1/2 (t)u000 k2L2 (R+ ;H) .
2ε
Choosing η =
ε2 max(α;β)
,
2
(2.22)
from inequality (2.22) we have
kAu00 k2L2 (R+ ;H)
≤
1/2
4 min(α; β)
kρ (t)A3 uk2L2 (R+ ;H) + kρ−1/2 (t)u000 k2L2 (R+ ;H) .
8ε min(α; β) − ε4 max2 (α; β)
10
A. R. ALIEV, N. L. MURADOVA
Then, by minimizing ε, we find ε =
EJDE-2013/219
p
3
2 min(α; β)/ max2 (α; β). Therefore,
kAu00 k2L2 (R+ ;H)
22/3 max2/3 (α; β) 1/2
kρ (t)A3 uk2L2 (R+ ;H) + kρ−1/2 (t)u000 k2L2 (R+ ;H) .
1/3
3 min (α; β)
≤
(2.23)
Now, taking into account inequality (2.11), from inequality (2.23) we obtain
kAu00 k2L2 (R+ ;H) ≤
22/3 max2/3 (α; β)
3 min4/3 (α; β)
kP0 uk2L2 (R+ ;H) .
As a result,
kAu00 kL2 (R+ ;H) ≤
21/3 max1/3 (α; β)
31/2 min2/3 (α; β)
kP0 ukL2 (R+ ;H) .
To estimate the norm kA2 u0 kL2 (R+ ;H) , we take into account (2.19) in (2.20):
εη max(α; β) 2 0 2
kA u kL2 (R+ ;H)
4 min(α; β)
η
1
≤
kρ−1/2 (t)u000 k2L2 (R+ ;H) + kρ1/2 (t)A3 uk2L2 (R+ ;H) .
4ε min(α; β)
2η
1−
(2.24)
Choosing ε = η 2 /(2 min(α; β)), from inequality (2.24) we have
kA2 u0 k2L2 (R+ ;H)
−1/2
4 min2 (α; β)
kρ
(t)u000 k2L2 (R+ ;H) + kρ1/2 (t)A3 uk2L2 (R+ ;H) .
2
4
8η min (α; β) − η max(α; β)
q
In this case, minimizing η, we find η = 3 2 min2 (α; β)/ max(α; β). Therefore,
≤
kA2 u0 k2L2 (R+ ;H)
≤
22/3 max1/3 (α; β)
2/3
3 min
(α; β)
[kρ−1/2 (t)u000 k2L2 (R+ ;H) + kρ1/2 (t)A3 uk2L2 (R+ ;H) ].
(2.25)
From this inequality, taking into account inequality (2.11), we obtain
kA2 u0 k2L2 (R+ ;H) ≤
22/3 max1/3 (α; β)
3 min5/3 (α; β)
kP0 uk2L2 (R+ ;H) .
Thus,
kA2 u0 kL2 (R+ ;H) ≤
21/3 max1/6 (α; β)
kP0 ukL2 (R+ ;H) .
31/2 min5/6 (α; β)
Now we estimate the norm kA3 ukL2 (R+ ;H) . From inequality (2.11) we have
1
kP0 uk2L2 (R+ ;H) ≥ kρ1/2 (t)A3 uk2L2 (R+ ;H) ≥ min(α; β)kA3 uk2L2 (R+ ;H) .
min(α; β)
Hence, we obtain
1
kA3 uk2L2 (R+ ;H) ≤
kP0 uk2L2 (R+ ;H)
min2 (α; β)
or
1
kA3 ukL2 (R+ ;H) ≤
kP0 ukL2 (R+ ;H) .
min(α; β)
The proof is complete.
EJDE-2013/219
THIRD-ORDER OPERATOR-DIFFERENTIAL EQUATIONS
11
3
Now, we prove the boundedness of the operator P1 : W2,K
(R+ ; H) → L2 (R+ ; H).
Lemma 2.6. Let Aj A−j ∈ L(H, H), j = 1, 2, 3. Then P1 is a bounded operator
3
from the space W2,K
(R+ ; H) into the space L2 (R+ ; H).
3
Proof. For any u(t) ∈ W2,K
(R+ ; H) we have
kP1 ukL2 (R+ ;H) = kA1 u00 + A2 u0 + A3 ukL2 (R+ ;H)
≤ kA1 A−1 kH→H kAu00 kL2 (R+ ;H) + kA2 A−2 kH→H kA2 u0 kL2 (R+ ;H)
+ kA3 A−3 kH→H kA3 ukL2 (R+ ;H) .
Applying the theorem on intermediate derivatives [14, ch. 1], we obtain from the
last inequality that
kP1 ukL2 (R+ ;H) ≤ constkukW23 (R+ ;H) .
The proof is complete.
Let us consider the question of regular solvability of problem (1.1), (1.2).
√
3
/ σ(B),
Theorem 2.7. Let A = A∗ ≥ cE, c > 0, K ∈ L(H5/2 , H1/2 ), − α2 ω2 ∈
the operator Kα,β has a bounded inverse in the space H5/2 , Re B ≥ 0 and Aj A−j ∈
L(H, H), j = 1, 2, 3, moreover, the following inequality holds
a1 kA1 A−1 kH→H + a2 kA2 A−2 kH→H + a3 kA3 A−3 kH→H < 1,
where the numbers aj , j = 1, 2, 3, are defined in Theorem 2.5. Then the boundary
value problem (1.1), (1.2) is regularly solvable.
Proof. Boundary value problem (1.1), (1.2) can be represented in the operator form
P0 u(t) + P1 u(t) = f (t),
3
where f (t) ∈ L2 (R+ ; H), u(t) ∈ W2,K
(R+ ; H).
√
3
∗
Under conditions A = A ≥ cE, c > 0, K ∈ L(H5/2 , H1/2 ), − α2 ω2 ∈
/ σ(B),
the operator Kα,β has a bounded inverse in the space H5/2 , by Theorem 2.3 the
operator P0 has a bounded inverse P0−1 acting from the space L2 (R+ ; H) into the
3
(R+ ; H). If we put v(t) = P0 u(t) we obtain the following equation in
space W2,K
L2 (R+ ; H):
(E + P1 P0−1 )v(t) = f (t).
We show that under the conditions of the theorem, the norm of the operator P1 P0−1
is less than unity. Taking into account inequalities (2.18), we have
kP1 P0−1 vkL2 (R+ ;H)
= kP1 ukL2 (R+ ;H)
≤ kA1 u00 kL2 (R+ ;H) + kA2 u0 kL2 (R+ ;H) + kA3 ukL2 (R+ ;H)
≤ kA1 A−1 kH→H kAu00 kL2 (R+ ;H) + kA2 A−2 kH→H kA2 u0 kL2 (R+ ;H)
+ kA3 A−3 kH→H kA3 ukL2 (R+ ;H)
≤ a1 kA1 A−1 kH→H kP0 ukL2 (R+ ;H) + a2 kA2 A−2 kH→H kP0 ukL2 (R+ ;H)
+ a3 kA3 A−3 kH→H kP0 ukL2 (R+ ;H)
=
3
X
j=1
aj kAj A−j kH→H kvkL2 (R+ ;H) .
12
A. R. ALIEV, N. L. MURADOVA
EJDE-2013/219
Thus,
kP1 P0−1 kL2 (R+ ;H)→L2 (R+ ;H)
≤
3
X
aj kAj A−j kH→H < 1.
j=1
Therefore, the operator E + P1 P0−1 is invertible in the space L2 (R+ ; H) and u(t)
is defined by the formula
u(t) = P0−1 (E + P1 P0−1 )−1 f (t),
moreover
kukW23 (R+ ;H)
≤ kP0−1 kL2 (R+ ;H)→W23 (R+ ;H) k(E + P1 P0−1 )−1 kL2 (R+ ;H)→L2 (R+ ;H) kf kL2 (R+ ;H)
≤ constkf kL2 (R+ ;H) .
The proof is complete.
Corollary 2.8. In the conditions of Theorem 2.7, the operator P is an isomorphism
3
(R+ ; H) and L2 (R+ ; H).
between the spaces W2,K
In conclusion, we remark that our solvability results imply the results of [4] when
K = 0 and A3 = 0, and the results of [15] when K = 0 and α = β = 1.
Acknowledgements. We are very grateful to the referees for their careful reading
of the original manuscript, for their helpful comments which led to the improvement
of this article.
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Araz R. Aliev
Baku State University, Institute of Mathematics and Mechanics of NAS of Azerbaijan,
Baku, Azerbaijan
E-mail address: alievaraz@yahoo.com
Nazila L. Muradova
Nakhchivan State University, Nakhchivan, Azerbaijan
E-mail address: nazilamuradova@gmail.com