Volume 13, 2009
21
ON SOLVABILITY OF FOURTH ORDER OPERATOR-DIFFERENTIAL
EQUATIONS OF ELLIPTIC TYPE ON WEIGHT SPACES
Ismailova M.F.
Baku State University, 23, Z.I.Khalilov str., AZ1148, Baku, Azerbaijan
Tel: (99412) 439 05 49 (off.), E-mail: mila 72@mail.ru
Abstract. Solvability of the operator-differential equations of the fourth order of elliptic type with partial derivatives in the weighted spaces are considered
in the paper. Solvability conditions in some weighted spaces, expressed by the
coefficients of given operator-differential equations are obtained. Connection
between lower bound of the spectrum of operator participating in the main
part of the equation and exponent of the weight function is also determined.
In this paper are found conditions oh the lower boundary of the spectrum of
the main operator for operationally-differential equation depending of two and
of the order of the weighted spaces which provide existence and uniqueness of
the regular solution of a class of the operationally-differential equation of the
fourth order in weighted spaces on all axis.
Key words: Hilbert space; operator-differential equation; self-adjoin operator; bounded operators.
MSC 2000 : 773H05; 73K15; 46C05; 46E22; 39A70; 49G99.
LetH – be a separable Hilbert space, A – a positive definite self-adjoin
operator in H. Let γ = (γ1 , γ2 ) ∈ R2 , R = (−∞, ∞). By L2,γ (R2 ; H) we
denote a Hilbert space of vector-function f (x, y) with values in H for which

 12
Z Z
kf kL2,γ = 
kf (x, y)k2 e−2γ1 x−2γ2 y dxdy  < ∞.
R2
On D (R2 ; H 4 )- set of infinitely differentiable in H vector-function u(x, y),
determines in R2 , with values H4 = D (A4 ) with compact supports, we determine the norm
 12

kukW 4
2,γ


X

=

 k, j = 0
k+j ≤4
°
°
° 4−(k+j) ∂ k+j u °
°A
°
°
∂xk ∂y j °



 .

L2,γ 
4
(R2 ; H) we denote a completion of D (R2 ; H 4 ) by the norm kukW 4
By W2,γ
2,γ
Notice that for γ = 0 = (0, 0) the space L2,0 (R2 ; H) = L2 (R2 ; H) and
W2 , 04 (R2 ; H) = W24 (R2 ; H). In the spaceHconsider the following operatordifferential equation
22
Bulletin of TICMI
4
X
∂4u ∂ 4u
+
+ A4 u +
∂x4 ∂y 4
Ak,j
k, j = 0
k+j ≤4
∂ k+j u
= f (x, y) , (x, y) ∈ R2 ,
∂xk ∂y j
(1)
where f (x, y), u(x, y) are vector functions with values in H, and the operators
A and Ak,j (k, j = 0, 4, k + j ≤ 4) satisfy the following conditions:
1) A is a positive-definite operator with lower boundary of the spectrum
µ0 , i.e., A ≥ µ0 E
2) the operators Bk,j = Ak,j A(k+j)−4 (k, j = 0, 4, k + j ≤ 4) are bounded
operators in H.
Definition. If for f (x, y) ∈ L2,γ (R2 ; H) there exists a vector-function
4
(R2 ; H), that satisfies equation (1) almost everywhere in R2 ,
u (x, y) ∈ W2,γ
we’ll call it a regular solution of equation (1). If for any f (x, y) ∈ L2,γ (R2 ; H)
there exists a regular solution of equation (1) that satisfies the estimation
kukW 4 ≤ const kf kL2,γ ,
2,γ
then equation (1) is said to be regularly solvable.
In the present paper, we’ll find conditions on coefficients of equation (1) and
the vector γ = (γ1 , γ2 ), that provide regular solvability of the given equation.
Note that for γ = 0 solvability of equation (1) was investigated in the
paper [1] for Ak,j = 0 (k, j = 0, 4, k + j ≤ 4) in the paper [2]. For one variable,
when A is an elliptic operator with discrete spectrum, and Aij ≡ aij are scalar
numbers, equation (1) was investigated in the paper [3], when A is a self-adjoint
operator and coefficients are unbounded operators see [4].
The following theorem was proved in [2]
Theorem 1.[2]. Let A-be a positive-definite
self-adjoin operator with lower
p
4
1
4
boundary of the spectrum µ0 and |γ| = γ1 + γ24 < √
4 µ0 , then the equation
8
µ
P0 u ≡ P0
∂ ∂
,
∂x ∂y
¶
u≡
∂ 4u ∂4u
+
+ A4 u = f (x, y) , (x, y) ∈ R2
∂x4 ∂y 4
(2)
is regularly solvable.
Further, denote
µ
P1 u = P1
∂ ∂
,
∂x ∂y
¶
u=
¡ 2 ¢
4
u (x, y) ∈ W2,γ
R ;H
and
4
X
k, j = 0
k+j ≤4
Akj
∂ k+j u
,
∂xk ∂y j
(3)
Volume 13, 2009
23
P u = P0 u + P1 u.
(4)
At first we prove the following theorem.
Theorem 2. Let the operator A and vector γ satisfy the conditions of
4
theorem 1. Then for any u (x, y) ∈ W2,γ
(R2 ; H) the following estimations
hold:
°
°
° 4−(k+j) ∂ k+j u °
°A
°
≤ Ck,j (γ; µ0 ) kP0 ukL2,γ (R2 ;H)
°
∂xk ∂y j °L2,γ (R2 ;H)
(5)
(k, j = 0, 4, k + j ≤ 4)
where
C0,0 (γ; µ0 ) =
µ40
,
µ40 − 8 (γ14 + γ24 )
(6)
4γ12
16γ14
C4,0 (γ; µ0 ) = 1 + p 4
+
,
µ0 − 8 (γ14 + γ24 ) µ40 − 8 (γ14 + γ24 )
(7)
4γ22
16γ24
C0,4 (γ; µ0 ) = 1 + p 4
,
+
µ0 − 8 (γ14 + γ24 ) µ40 − 8 (γ14 + γ24 )
(8)
for j = 0 , k = 1, 2, 3 and k = 0 , j = 1, 2, 3
Ck,0
µ ¶k µ
¶ 4−k
k
4−k 4 ³
4γ12
24γ14 + 8γ24 ´
=
1+ p 4
+
, (9)
4
4
µ0 − 8 (γ14 + γ24 ) µ40 − 8 (γ14 + γ24 )
C0,j
Ã
!
µ ¶µ
¶ 4−j
2
4
4
4
j
4−j
4γ2
24γ2 + 8γ1
=
1+ p 4
+ 4
, (10)
4
4
4
4
µ0 − 8 (γ1 + γ2 ) µ0 − 8 (γ14 + γ24 )
for k 6= 0 , j 6= 0, k + j = 4 (k = 1, j = 3, k = 2, j = 2, k = 3, j = 1)
µ ¶ k4 µ ¶ 4j ³
4(γ12 + γ22 )
k
j
16(γ14 + γ24 ) ´
1+ p 4
Ck,j (γ; µ0 ) =
+
. (11)
4
4
µ0 − 8 (γ14 + γ24 ) µ40 − 8 (γ14 + γ24 )
And for 2 ≤ k + j ≤ 3, k 6= 0 , j 6= 0, k = 1, j = 1; k = 2, j = 1; k = 1, j = 2)
µ
Ck,j (γ; µ0 ) =
4 − (k + j)
4
¶ 4−(k+j)
µ ¶ k4 µ ¶ 4j
4
k
j
.
4
4
(12)
Proof: Obviously, it suffices to prove these inequalities for a vector-function
from D (R2 , H4 ). Let u (x, y) ∈ D (R2 , H4 ). After substitution v (x, y) =
u (x, y) e−γ1 x−γ2 y we get following equivalents of the inequality
24
Bulletin of TICMI
°
°
´j
°
° 4−(k+j) ¡ ∂
¢k ³ ∂
°A
+ γ1
+ γ2 v (x, y)°
≤
∂x
∂y
°
°
2 ;H)
L
(R
2
° ¡
´ °
¢ ³∂
°
°
∂
≤ Ck,j (γ; µ0 ) °P0 ∂x
+ γ1 , ∂y
+ γ2 v °
,
(13)
L2 (R2 ;H)
where the numbers Ck,j (γ; µ0 ) are determined from equalities (6)-(12).
After Fourier transformation we get the following equivalent inequalities:
°
°
°
° 4−(k+j)
k
j
≤
(iξ + γ1 ) (iη + γ2 ) v̂ (ξ, η)°
°A
L2 (R2 ;H)
(14)
≤ Ck,j (γ; µ0 ) kP0 (iξ + v1 ) , (iξ + v2 ) v̂ (ξ, η)kL2 (R2 ;H) .
Let
P0 ((iξ + γ1 ) , (iη + γ2 )) v̂ (ξ, η) = ĝ (ξ, η) , (ξ, η) ∈ R2 .
Since the operator pencil
P0 ((iξ + γ1 ) , (iη + γ2 )) = (iξ + γ1 )4 E + (iη + γ2 )4 E + A4 , (ξ, η) ∈ R2 ,
p
1
is invertible in H for |γ| = 4 γ14 + γ24 < √
4 µ0 (see [2]), then
8
v̂ (ξ, η) = P0−1 ((iξ + γ1 ) , (iη + γ2 )) ĝ (ξ, η) , (ξ, η) ∈ R2 .
Then inequality (14) takes the following form
°
°
° 4−(k+j)
°
(iξ + γ1 )k (iη + γ2 )j P0−1 ((iξ + γ1 ) , (iξ + γ2 )) ĝ (ξ, η)°
°A
≤ Ck,j (γ; µ0 ) kĝ (ξ, η)kL2 (R2 ;H) .
L2 (R2 ;H)
(15)
Now, consider the case k = 0, j = 0. Then
° 4 −1
°
°A P0 ((iξ + γ1 ) , (iη + γ2 )) ĝ (ξ, η)°
L2
° 4 −1
°(R2 ;H)
°
≤ sup A P0 ((iξ + γ1 ) , (iη + γ2 ))° kĝ (ξ, η)k
(ξ,η)∈R2
L2 (R2 ;H) .
(16)
Since for any (ξ, η) ∈ R2
° 4 −1
°
°A P0 ((iξ + γ1 ) , (iη + γ2 ))°
° ¡
¢−1 °
°
°
= °A4 (iξ + γ1 )4 + (iη + γ2 )4 + A4 °
° ¡
¢ °
° 4
4
4
4 −1 °
≤ sup °µ (iξ + γ1 ) + (iη + γ2 ) + µ
°
µ∈σ(A) °
¢ °
¡
≤ sup °µ4 [Re (iξ + γ1 )4 + (iη + γ2 )4 + µ4 ]−1 °
µ∈σ(A) °
°
°
−1 °
≤ sup °µ4 (ξ 4 + γ14 + η 4 + γ24 − 6ξ 2 γ12 − 6η 2 γ22 ) °
µ∈σ(A)
°
°
°
−1 °
2
2
≤ sup °µ4 (ξ 2 − 3γ12 ) + (η 2 − 3γ22 ) + (µ4 − 8 (γ14 + γ24 )) °
µ≥µ0 °
°
° 4 4
µ4
4 −1 °
4
≤ sup °µ (µ − 8 (γ1 + γ2 )) ° ≤ µ4 −8 γ04 +γ 4 = C0,0 (γ; µ0 ) ,
( 1 2)
0
µ≥µ0
(17)
Volume 13, 2009
25
Then taking into account inequality (12) in inequality (16) we prove the validity
of inequality (15) for k = 0, j = 0.
Now, let’s consider the case j = 0, k = 1, 2, 3. Then
°
°
° 4−k
°
k −1
(iξ + γ1 ) P0 ((iξ + γ1 ) , (iη + γ2 )) ĝ (ξ, η)°
°A
L2 (R2 ;H)
°
°
°
°
k 4 1
sup °(iξ + γ1 ) A P0 ((iξ + γ1 ) , (iη + γ2 ))° · kĝ (ξ, η)kL2 (R2 ;H) .
(18)
(ξ,η)∈R2
On the other hand, for (ξ, η) ∈ R2
°
°
°
°
k 4 −1
°(iξ + γ1 ) A P0 ((iξ + γ1 ) , (iη + γ2 ))°
k
µ4−k (ξ 2 + γ12 ) 2
≤ sup
2
2
(ξ 2 − 3γ12 ) + (η 2 − 3γ22 ) + µ4 − 8 (γ14 + γ24 )
k
µ4−k (ξ 2 + γ12 ) 2
≤ sup
.
2 2
4
4
µ∈σ(A) (ξ 2 − 3γ1 ) + µ4 − 8 (γ1 + γ2 )
µ∈σ(A)
(19)
Let δ > 0. Then having applied Young’s inequality, we have:
µ
4−k
¡
2
ξ +
γ12
¢ k2
¡
= δµ
4
¢ 4−k
4
µ
¡
2
ξ +
1 ¡
δ
Now choose δ > 0.So, that
Then
4−k
µ
γ12
¢ k2
4−k
k
4−k
δ
4
=
2
ξ +
k
4
4−k
≤
4
1
4−k
δ 4
µ
¢2
γ12
¶ k4
≤δ
, i.e., δ =
k
4−k
¢
4−k 4 k 1 ¡ 2
µ + 4−k ξ + γ12 .
4
4δ k
¡
k
4−k
¢ k4
.
¶ k4 ³
¡
¢2 ´
µ4 + ξ 2 + γ12
µ ¶ k4 µ
¶ 4−k
¢´
k
4−k 4 ³ 4 ¡ 2
2 2
=
µ + ξ + γ1
.
4
4
Considering this inequality in (19) we get
°
°
°
°
k 4 −1
(iξ
+
γ
)
A
P
((iξ
+
γ
)
,
(iη
+
γ
))
°
1
1
2 °
0
µ ¶ k4 µ
¶ 4−k
2
k
µ4 + (ξ 2 + γ12 )
4−k 4
≤
sup
2 2
4
4
4
4
µ∈σ(A) (ξ 2 − 3γ1 ) + µ4 − 8 (γ1 + γ2 )
µ ¶ k4 µ
¶ 4−k
k
4−k 4
≤
4
4
2
× sup
µ∈σ(A)
µ4 − 8 (γ14 + γ24 ) + (ξ 2 − 3γ12 ) + 8γ12 (ξ 2 − 3γ12 ) + 24γ14 + 8γ24
2
(ξ 2 − 3γ12 ) + µ4 − 8 (γ14 + γ24 )
26
Bulletin of TICMI
µ ¶ k4 µ
¶ 4−k
k
4−k 4
≤
4
4
µ
¶
ξ 2 − 3γ12
24γ14 + 8γ24
2
× sup 1 + 8γ1 2
+
(ξ − 3γ12 ) + µ4 − 8 (γ14 + γ24 ) µ4 − 8 (γ14 + γ24 )
µ∈σ(A)
!
µ ¶ k4 µ
¶ 4−k Ã
4γ12
k
4−k 4
24γ14 + 8γ24
≤
1+ p 4
+
4
4
µ0 − 8 (γ14 + γ24 ) µ40 − 8 (γ14 + γ24 )
= Ck,0 (γ; µ0 ) .
Thus, inequality (15) is true also for j = 0, k = 1, 2, 3. Inequality (15) is
similarly proved for j = 0, k = 1, 2, 3.
Now, consider the case k 6= 0, j 6= 0, k + j = 4 (k = 1, j = 3; k = 2, j = 2;
k = 3, j = 2). In this case
°
°
°
°
k
j
−1
(iξ
+
γ
)
(iη
+
γ
)
P
((iξ
+
γ
)
,
(iη
+
γ
))
ĝ
(ξ,
η)
°
°
1
2
1
2
0
2
L2 (R ;H)
°
°
°
°
≤ sup °(iξ + γ1 )k (iη + γ2 )j P0−1 ((iξ + γ1 ) , (iη + γ2 ))°
(20)
(ξ,η)∈R2
× kĝ (ξ, η)kL2 (R2 ;H) .
For (ξ, η) ∈ R2 we have:
°
°
°
°
k
j
°(iξ + γ1 ) (iη + γ2 ) P0−1 ((iξ + γ1 ) , (iξ + γ2 )) ĝ (ξ, η)°
(21)
j
k
(ξ 2 + γ12 ) 2 (η 2 + γ22 ) 2
≤ sup
µ∈σ(A)
2
L2 (R2 ;H)
2
(ξ 2 − 3γ12 ) + (η 2 − 3γ22 ) + µ4 − 8 (γ14 + γ24 )
.
Since for any δ > 0
¡
ξ 2 + γ12
¢ k2 ¡
η 2 + γ22
¢ 2j
=
¡
ξ 2 + γ12
¢ k2 ¡
η 2 + γ22
¢ 4−k
2
¶ 4−k
µ
³ ¡
4
¢
¢ ´k
1 ¡ 2
2 2
2
2 2 4
η + γ2
= δ ξ + γ1
k
δ 4−k
¢2 4 − k 1 ¡ 2
¢
k ¡ 2
δ ξ + γ12 +
ξ + γ12 .
≤
k
4
4 δ 4−k
Now, assuming δ =
¡ 4−k ¢ 4−k
4
k
we get that
¡ 2
¢k ¡
¢j
k
ξ + γ12 2 η 2 + γ22 2 ≤
4
µ
4−k
k
¶ 4−k
4 ³¡
ξ 2 + γ12
¢2
¡
¢2 ´
+ η 2 + γ22
Volume 13, 2009
27
µ ¶ k4 µ ¶ 4j ³
¡ 2
¢2
¡
¢
k
j
=
ξ − 3γ12 + 8γ12 ξ 2 − 3γ12
4
4
´
¡
¢2
¡
¢
+16γ14 + η 2 − 3γ22 + 8γ22 η 2 − 3γ22 + 16γ24 .
Thus (21) implies
°
°
j
°
°
k
−1
°(iξ + γ1 ) (iη + γ2 ) P0 ((iξ + γ1 ) , (iη + γ2 ))° ≤
2
¡ ¢k ¡ ¢j
(ξ2 +γ 2 ) +(η2 +γ 2 )
≤ k4 4 4j 4 sup 2 2 2 21 2 2 42 4 4 =
µ∈σ(A) (ξ −3γ1 ) +(η −3γ2 ) +µ −8(γ1 +γ2 )
¡ k ¢ k4 ¡ j ¢ 4j (ξ2 −3γ12 )2 +(η2 −3γ22 )2 +8γ12 (ξ2 −3γ12 )+8γ22 (η2 −3γ22 )+16(γ14 +γ24 )
= 4
=
2
2
4
(ξ2 −3γ12 ) +(η2 −3γ22 ) +µ40 −8(γ14 +γ24 )
µ
¡ ¢k ¡ ¢j
ξ 2 −3γ 2
= k4 4 4j 4 1 + 8γ12 ξ2 −3γ 2 +µ4 −81 γ 4 +γ 4 +
(
( 1 2¶
)
1)
0
4
4
16(γ +γ )
η 2 −3γ 2
+ 8γ22 2 2 2 4 2 4 4 + µ4 −8 1γ 4 +γ2 4
≤
( 1 2)
(η −3γ2 )µ+µ0 −8(γ1 +γ2 )
0
¶
¡ k ¢ k4 ¡ j ¢ 4j
4(γ12 +γ22 )
16(γ14 +γ24 )
= Ck,j (γ; µ0 ) .
1+ q 4
+ µ4 −8 γ 4 +γ 4
≤ 4
4
( 1 2)
µ0 −8(γ14 +γ24 )
0
(22)
Now, consider the case 2 ≤ k + j ≤ 3, k 6= 0, j 6= 0. In this case
°
°
° 4−(k+j)
°
k
j
−1
A
(iξ
+
γ
)
(iη
+
γ
)
P
((iξ
+
γ
)
,
(iη
+
γ
))
ĝ
(ξ,
η)
°
°
1
2
1
2
0
L2 (R2 ;H)
°
°
°
°
≤ sup °A4−(k+j) (iξ + γ1 )k (iη + γ2 )j P01 ((iξ + γ1 ) , (iη + γ2 ))°
(23)
(ξ,η)∈R2
× kĝ (ξ, η)kL2 (R2 ;H) .
Since for (ξ, η) ∈ R2
°
°
° 4−(k+j)
°
k
j
−1
(iξ + γ1 ) (iη + γ2 ) P0 ((iξ + γ1 ) , (iη + γ2 ))°
°A
L2 (R2 ;H)
°
°
° 4−(k+j)
°
k
j
1
≤ sup °µ
(iξ + γ1 ) (iη + γ2 ) P0 ((iξ + γ1 ) , (iη + γ2 ))° (24)
µ∈σ(A)
k
≤ sup
µ∈σ(A)
j
µ4−(k+j) (ξ 2 + γ12 ) 2 (η 2 + γ22 ) 2
2
2
(ξ 2 − 3γ12 ) + (η 2 − 3γ22 ) + µ4 − 8 (γ14 + γ24 )
4−(k+j)
4
Let δ1 , δ2 , δ3 > 0. δ1
k
.
j
· δ24 · δ34 = 1. Then we have
¢ k2 ¡ 2
¡
¢j
µ4−(k+j) ξ 2 + γ12
η + γ22 2
³ ¡
¢ ´k ³ ¡ 2
¢ ´j
¡ 4 ¢ 4−(k+j)
2
2 2 4
2 2 4
4
= δ1 µ
δ 2 ξ + γ1
δ 3 η + γ2
≤ δ1
¢2
¢2
k¡
j¡
4 − (k + j) 4
µ + δ2 ξ 2 + γ12 + δ3 η 2 + γ22 .
4
4
4
28
Bulletin of TICMI
Additionally we require δ1 4−(k+j)
= δ2 k4 = δ3 4j . Then we get
4
µ
δ1 =
k
4 − (k + j)
¶ k+j
µ ¶ 4j
4
j
,
k
i.e.,
¡
¢k ¡
¢j ¡
¢j
µ4−(k+j) ξ 2 + 3γ12 2 η 2 + γ22 2 η 2 + γ22 2
µ ¶ 4j ³
µ
¶ k+j
4
¡
¢2 ¡
¢2 ´
4 − (k + j)
j
k
≤
µ4 + ξ 2 + γ12 + η 2 + γ22
4
4 − (k + j)
4
Consequently,
j
k
sup
µ4−(k+j) (ξ 2 + γ12 ) 2 (η 2 + γ22 ) 2
2
2
(ξ 2 − 3γ12 ) + (η 2 − 3γ12 ) + µ4 − 8 (γ14 + γ24 )
µ
¶ 4−(k+j)
µ ¶ k4 µ ¶ 4j
4
4 − (k + j)
k
j
≤
4
4
4
µ∈σ(A)
× sup
µ∈σ(A)
+
n µ4 + (ξ 2 − 3γ 2 )2 + (η 2 − 3γ 2 )2 + 8γ 2 (ξ 2 − 3γ 2 )2
1
(ξ 2
−
2
3γ12 )
+
2
(η 2
−
2
3γ22 )
+
1
µ4
8γ22 (η 2 − 3γ22 ) + 24 (γ14 + γ24 )
2
−
8 (γ14
+
1
4
γ2 )
o
2
(ξ 2 − 3γ12 ) + (η 2 − 3γ22 ) + µ4 − 8 (γ14 + γ24 )
µ
¶ 4−(k+j)
µ ¶ k4 µ ¶ 4j
4
4 − (k + j)
k
j
≤
4
4
4
µ
¶
4 (γ12 + γ22 )
24 (γ14 + γ24 )
× 1+ 4
+
= Ck,j (γ; µ0 ) .
µ0 − 8 (γ14 + γ24 ) µ40 − (γ14 + γ24 )
The theorem is proved.
Now, prove a theorem on solvability of equation (1).
Theorem 3. Let the conditions of theorem 1) and condition 2), moreover
α (γ; µ0 ) =
4
X
Ck.j (γ; µ0 ) kBk,j k < 1,
k+j =0
k+j ≤4
where the numbers Ck,j (γ; µ0 ) are determined from theorem 2 by equalities
(6)-(12). Operator Bk,j determined from condition 2). Then equation (1) is
regularly solvable.
Proof. Write equation (1) in the form P u = P0 u+P1 u = f wheref (x, y) ∈
4
(R¢2 ; H). After substitution of P0 u = v ∈ L2,γ (R2 ; H)
L2,γ (R2 ; H), u (x, y) ¡∈ W2,γ
we get the equation +P1 P0−1 v = f in L2,γ (R2 ; H). Since
Volume 13, 2009
°
°
°P1 P0−1 u°
≤
L2,γ
4
P
k+j =0
k+j ≤4
= kP1 ukL2,γ ≤
P4
29
°
°
k+j
°
°°
°
°Ak,j A(k+j)−4 ° °A4−(k+j) ∂ k4 uj °
∂x
∂y
°
°
k+j =0
L2,γ
k+j ≤4
Ck,j (γ; µ0 ) kBk,j k kP0 ukL2,γ = α (γ; µ0 ) kvkL2,γ .
As α (γ; µ0 ) < 1, the operator E+P1 P0−1 is invertible in the space L2,γ (R2 ; H),
¡
¢−1
¡
¢−1
then v = E + P1 P0−1
f bat u = P0−1 E + P1 P0−1
f
Hence it follows that
kukW 4 ≤ const kf kL2,γ .
2,γ
The theorem is proved.
References
[1] S.S. Mirzoyev, M.F. Ismailova. On solvability of fourth order partial operatordifferential equations in Hilbert. Vestnic BSU No. 4 (2006), 5-11 (in Russian)
[2] M.F. Ismailova. On solvability of fourth order operator-differential equations on
weight spaces. Vestnic BSU, No.3 (2008), 42-52 (in Russian)
[3] Yu.A. Doubinskiy. Mixed problems for some partial operator-differential equations.
Trudy Moskowskogo Matematicheskogo Obshestva, 20 (1969), 205-240 (in Russian)
[4] E.N. Mamedov. On a fourth order operator-differential equations in Hilbert space.
Proc. Inst. Math. Mech. Natl. Acad. Sci. Azerb., 24 (2006), 147-152.
Received 01.09.2009; revised 21.12.2009; accepted 28.12.2009