Gen. Math. Notes, Vol. 5, No. 1, July 2011, pp.27-33
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
On Generalized Solutions of Boundary Value
Problems for Some Class of Fourth Order
Operator-Differential Equations on the Segment
R.Z. Humbataliyev
Institute of Mathematics and Mechanics of NAS of Azerbaijan, Baku
E-mail: rovshangumbataliev@rambler.ru
(Received: 3-6-11/Accepted:17-6-11)
Abstract
In the paper we give definition of a generalized solution of boundary value
problems for some fourth order operator-differential equations, and prove the
existence of generalized solution of this problem by the coefficients of the equations on the segment.
Keywords: Hilbert spaces, existence of generalized solution, operatordifferential equation.
1
Introduction
Let H be a separable Hilbert space, A be a positive-definite self-adjoint operator in H with domain of definition D(A). By Hγ we denote a scale of Hilbert
space that is generated by the operator A, i.e. Hγ = D (Aγ ) , (γ ≥ 0) , (x, y)γ =
(Aγ x, Aγ y) , x, y ∈ D (Aγ ) . By L2,γ ((a, b); Hγ ) we denote a Hilbert space of
the vector-function f (t) defined in (a, b) almost everywhere, with values from
H, measurable square integrable in the Bochner sense

Zb
kf kL2 ((a,b);H) = 
a
1/2
kf k2γ dt
.
28
R.Z. Humbataliyev
Then we determine the Hilbert spaces
W22 ((a, b); H) = u|u00 ∈ L2 ((a, b); H), A2 u ∈ L2 ((a, b); H2 )
with the norm
2
kukW 2 ((a,b);H) = ku00 kL2 ((a,b);H) + A2 u
2
1/2
L2 ((a,b);H)
.
Here and in future the derivatives are understood in the sense of distribution
[1]. By D4 ((a, b); H) we denote a linear set of infinitely-differentiable functions
with values from H4 = D(A4 ) having compact supports in (a, b). Further we
◦ 2
determine the space W 2 ((a, b); H) ⊂ W22 ((a, b); H) in the following way.
◦ 2
W 2 ((a, b); H) = u|u ∈ W22 ((a, b); H), u(j) (a) = u(j) (b) = 0, j = 0, 1 .
In sequel, we’ll assume that (a, b) is as [0, 1]. Assume that
L2 ((a, b); H) ≡ L2 ([0, 1]; H),
◦ 2
W22 ((a, b); H) ≡ W22 ([0, 1]; H),
◦ 2
W 2 ((a, b); H) ≡ W 2 ([0, 1]; H), D4 ((a, b); H) ≡ D4 ([0, 1]; H).
In the separable Hilbert space H we consider a polynomial operator bundle
in the following form
2 2
2
P (λ) = −λ E + A
+
4
X
Aj λ(4−j) ,
(1)
j=0
and the related boundary value problem
P (d/dt) u(t) ≡
d2
− 2 + A2
dt
2
u(t) +
4
X
Aj u(4−j) (t) = f (t), t ∈ [0, 1], (2)
j=1
u(j) (0) = ϕj , u(j) (1) = ψj , j = 0, 1
(3)
where A is a self-adjoint positive-definite operator, Aj j = 1, 4 are linear,
generally speaking, unbounded operators in H, f ∈ L2 ([0, 1]; H), ϕj , ψj ∈
H, j = 0, 1.
In the paper we’ll give a generalized solution of boundary value problem
(2), (3) and prove for it theorems on the existence of a generalized solution in
terms of coefficients of the investigated fourth order differential equation on
the seqment.
29
On Generalized Solutions of Boundary Value...
2
Auxiliary Facts
Denote
P0
d
dt
P1
d
dt
2
d2
2
u(t) = − 2 + A
u(t) ,
dt
u(t) =
4
X
◦ 2
u(t) ∈ W 2 ([0, 1]; H),
◦ 2
Aj u(4−j) (t) ,
u(t) ∈ W 2 ([0, 1]; H),
j=0
and
◦ 2
u(t) ∈ W 2 ([0, 1]; H).
P u(t) = P0 u(t) + P1 u(t) ,
Now we formulate a lemma that shows
the conditions on operator coeffi d
cients (1) under which the expression P1
u(t) makes sense for the function
dt
◦ 2
from the class W 2 ([0, 1]; H).
Lemma 2.1. [2]. Let A be a self-adjoint positive-definite operator, B0 = A0 ,
−2
B1 = A1 A−1 , B2 = A−1 A2 A−1 , B3 = A−2 A3 A−1, B
A−2 A
be bounded
4 = 4A
d
u, ξ
deterope-rators in H. Then the bilinear functional P1
dt
L2 ([0,1];H)
◦ 2
mined on D4 ([0, 1]; H) ⊕ W 2 ([0, 1]; H) continues to the space W22 ([0, 1]; H) ⊕
◦ 2
W 2 ([0, 1]; H) continuously up to the bilinear functional P1 (u, ξ) acting in the
following way
P1 (u, ξ) =
1
X
(−1)j Aj u(2−j) , ξ 00
L2 ([0,1];H)
− (A2 u0, ξ 0 )L2 ([0,1];H) +
j=0
+
4
X
Aj u(4−j) , ξ
L2 ([0,1];H)
.
j=3
Lemma 2.2.
conditions of lemma 1 be fulfilled. Then a bilin the Let
all
d
u, ξ
determined on the space D4 ([0, 1]; H) ⊕
ear functional P
dt
L2 ([0,1];H)
◦ 2
W 2 ([0, 1]; H) of the bilinear functional
P (u, ξ) = (u, ξ)W22 ([0,1];H) + P1 (u, ξ) + 2 (Au0 , ξ 0 )L2 ([0,1];H)
where P1 (u, ξ) is determined as in lemma 1.
30
R.Z. Humbataliyev
Proof. Really
P
d
dt
u(t) = P0 u(t) + P1 u(t).
Integrating the following expression by parts we get
d
P0
u, ξ
=
dt
L2 ([0,1];H)
=
!
2
d2
− 2 + A2 u, ξ
dt
=
L2 ([0,1];H)
2
d4
2 d
4
= (u00 , ξ 00 )L2 ([0,1];H) +
− 2A 2 + A u, ξ
4
dt
dt
L2 ([0,1];H)
+2 (Au0 , Aξ 0 )L2 ([0,1];H) + A2 u, A2 ξ L2 ([0,1];H) .
Since Au0 (t) ∈ L2 ([0, 1]; H), Aξ 0 (t) ∈ L2 ([0, 1]; H), the right hand side of the
◦ 2
last equality continues by continuity from the space D4 ([0, 1]; H)⊕W 2 ([0, 1]; H)
◦ 2
to W22 ([0, 1]; H) ⊕ W 2 ([0, 1]; H). The lemma is proved.
Determination. The vector function u(t) ∈ W22 ([0, 1]; H) is said to be a
generalized solution of boundary value problem (2), (3) if it satisfies conditions
◦ 2
of (3) and for any function ξ(t) ∈ W 2 ([0, 1]; H) the equality
P (u, ξ) = (f, ξ)L2 ([0,1];H)
is fulfilled.
Now, let’s cite a lemma on the solvability of a boundary value problem for
an operator-differential equation representing the principal part of equation
(3).
Lemma 2.3. Let ϕ0 , ψ0 ∈ H3/2 = D A3/2 , ϕ1 , ψ1 ∈ H1/2 = D A1/2 .
Then the boundary value problems
P0
d
dt
u(t) =
d2
− 2 + A2
dt
2
u(t) = f (t), t ∈ [0, 1],
u(j) (0) = ϕj , u(j) (1) = ψj , j = 0, 1
(4)
(5)
has a unique solution.
3
Basic Results
Assuming the above mentioned facts we prove a theorem on the existence of
a generalized solution of boundary value problem (2), (3).
31
On Generalized Solutions of Boundary Value...
Theorem 3.1. Let A be a self-adjoint positive-definite operator, B0 = A0 ,
B1 = A1 A−1 , B2 = A−1 A2 A−1 , B3 = A−2 A3 A−1 , B4 = A−2 A4 A−2 be bounded
operators in H and it holds the inequality
δ=
4
X
γj kBj k < 1,
(6)
j=0
where γ0 = γ4 = 1, γ1 = γ3 = 1/2, γ2 = 1/4. Then boundary value problem
(2), (3) has a unique generalized solution.
◦ 2
Proof. Show that by fulfilling inequality (6) for any ξ ∈ W 2 ([0, 1]; H) it
holds the inequality
d
2
ξ, ξ
≥ d kξkW22 ([0,1];H) , (d > 0).
P
dt
L2 ([0,1];H)
Obviously
d
P
= kξk2W 2 ([0,1];H) +
ξ, ξ
2
dt
L2 ([0,1];H)
d
0 2
+2 kAξ kL2 ([0,1];H) + P1
ξ, ξ
.
dt
L2 ([0,1];H)
Therefore
Since
d
2
ξ, ξ
P
≥ kξkW22 ([0,1];H) +
dt
L2 ([0,1];H)
d
0 2
+2 kAξ kL2 ([0,1];H) − P1
.
ξ, ξ
dt
L2 ([0,1];H)
(7)
1
X
d
(2−j) 00
ξ, ξ
Aj ξ
, ξ L2 ([0,1];H) +
P1
≤
dt
L2 ([0,1];H)
j=0
4
X
+ (A2 ξ 0 , ξ)L2 ([0,1];H) + Aj ξ (4−j) , ξ L2 ([0,1];H) ,
j=3
then
0 0
0
0
(A2 ξ , ξ )L2 ([0,1];H) = (B2 Aξ , Aξ )L2 ([0,1];H) ≤
2
≤ kB2 k kAξ 0 kL2 ([0,1];H) kAξ 0 kL2 ([0,1];H) = kB2 k kAξ 0 kL2 ([0,1];H) .
On the other hand
2
kAξ 0 kL2 ([0,1];H) ≤
2
1 00 2
kξ kL2 ([0,1];H) + A2 ξ L2 ([0,1];H) .
2
(8)
32
R.Z. Humbataliyev
Then
2
2 kAξ 0 kL2 ([0,1];H) ≤
2
1 00 2
2
kξ kL2 ([0,1];H) + 2 kAξ 0 kL2 ([0,1];H) + A2 ξ L2 ([0,1];H)
2
or
2
kAξ 0 kL2 ([0,1];H) ≤
2
1 00 2
2
kξ kL2 ([0,1];H) + 2 kAξ 0 kL2 ([0,1];H) + A2 ξ L2 ([0,1];H) .
4
Allowing for the last inequality in (8) we get
2
k(A2 ξ 0 , ξ 0 )kL2 ([0,1];H) ≤ kB2 k
1 2
2
kξkW 2 ([0,1];H) + 2 kAξ 0 kL2 ([0,1];H) .
2
4
(9)
In the same way, for j = 0 we have
2
00 00
(A0 ξ , ξ )L2 ([0,1];H) ≤ kB0 k kξ 00 kL2 ([0,1];H) ≤
≤ kB0 k
2
kξ 00 kW 2 ([0,1];H)
2
+
2
2 kAξ 0 kL2 ([0,1];H)
,
(10)
for j = 1
(A1 ξ 0 , ξ 0 )L2 ([0,1];H) ≤ (B1 Aξ 0 , ξ 00 )L2 ([0,1];H) ≤
≤ kB1 k kAξ 0 kL2 ([0,1];H) kξ 0 kL2 ([0,1];H) ≤
1 00 2
≤ kB1 k
kξ kL2 ([0,1];H) + kAξ 0 kL2 ([0,1];H) ≤
2
1
2
≤ kB1 k kξk2W 2 ([0,1];H) + 2 kAξ 0 kL2 ([0,1];H) ,
2
2
(11)
for j = 3
(A3 ξ 0 , ξ 0 )L2 ([0,1];H) = B3 Aξ 0 , A2 ξ L2 ([0,1];H) ≤
≤ kB3 k kAξkL2 ([0,1];H) A2 ξ L2 ([0,1];H) ≤
2 2
1
0 2
≤ kB3 k
kAξ kL2 ([0,1];H) + A ξ L2 ([0,1];H) ≤
2
1
2
≤ kB3 k kξk2W 2 ([0,1];H) + 2 kAξ 0 kL2 ([0,1];H) ,
2
2
(12)
for j = 4
2 2
2
2
A ξ (A
ξ,
ξ)
=
B
A
ξ,
A
ξ
≤
kB
k
=
4
4
4
L2 ([0,1];H) L2 ([0,1];H)
L2 ([0,1];H)
2
= kB4 k kξk2W 2 ([0,1];H) + 2 A2 ξ 0 L2 ([0,1];H) .
2
(13)
33
On Generalized Solutions of Boundary Value...
Thereby, allowing for inequalities (9)-(13) in (7) we get
d
2
2
ξ, ξ
≥ (1 − δ) kξkW22 ([0,1];H) + 2 kAξ 0 kL2 ([0,1];H) .
P
dt
L2 ([0,1];H)
Now, we search for the generalized solution of boundary value problem (2),
(3) in the form of
u(t) = v0 (t) + v(t)
where v0 (t) is a generalized solution of boundary value problem (4), (5) and
◦ 2
v(t) ∈ W 2 ([0, 1]; H). Then, to determine v(t) we get
P (v0 + v, ξ) = (v0 + v, ξ)W 2 ([0,1];H) +P1 (v0 + v, ξ)+2 (Av00 + Av 0 , Aξ)L2 ([0,1];H) ≡
2
≡ (v0 , ξ)W22 ([0,1];H) + (v, ξ)W22 ([0,1];H) + P1 (v0 , ξ) + P1 (v, ξ) +
+2 (Av00 , Aξ 0 )L2 ([0,1];H) = (f, ξ)L2 ([0,1];H) .
Hence we get
(v0 , ξ)W22 ([0,1];H) + P1 (v, ξ) + 2 (Av 0 , Aξ)L2 ([0,1];H) = −P1 (v0 , ξ).
(14)
The right hand side of relation (14) determines a continuous functional in
W22 ([0, 1]; H)
◦ 2
⊕ W 2 ([0, 1]; H), the left-hand side uses equality (7) and satisfies
the conditions of Lax-Milgram theorem [3]. Therefore there exists a unique
◦ 2
vector - function v(t) ∈ W 2 ([0, 1]; H) satisfying equality (14), i.e. u(t) =
v(t) + v0 (t) is a generalized solution of boundary value problem (2), (3). The
theorem is proved.
References
[1] J.L. Lions and E. Majenes, Inhomogeneous Boundary Value Problems and
their Applications, Mir, (1971).
[2] R.Z. Humbataliyev, On generalized solution of fourth order operatordifferential equations, Izv. AN Azerb., Ser. Fiz.-Tech. and Math.,
XVIII(1998), 18-21.
[3] L. Bers, F. John and M. Schechter, Partial Equations. / M., Mir, (1966).