Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 2, Article 43, 2003
COERCIVENESS INEQUALITY FOR NONLOCAL BOUNDARY VALUE
PROBLEMS FOR SECOND ORDER ABSTRACT ELLIPTIC DIFFERENTIAL
EQUATIONS
AISSA AIBECHE
D EPARTMENT OF M ATHEMATICS ,
FACULTY OF S CIENCES
U NIVERSITY F ERHAT A BBAS ,
ROUTE DE S CIPION , 19000,
S ETIF, A LGERIA .
aibeche@wissal.dz
Received 09 December, 2002; accepted 17 March, 2003
Communicated by Z. Nashed
A BSTRACT. In this paper we give conditions which assure the coercive solvability of an abstract
differential equation of elliptic type with an operator in the boundary conditions, and the completeness of generalized eigenfunctions. We apply the abstract result to show that a non regular
boundary value problem for a second order partial differential equation of an elliptic type in a
cylindrical domain is coercive solvable.
Key words and phrases: Ellipticity, Abstract Differential Equation, Coerciveness, Non Regular Elliptic Problems.
2000 Mathematics Subject Classification. 35J05, 35P20, 34L10, 47E05.
1. I NTRODUCTION
Many works are devoted to the study of hyperbolic or parabolic abstract equations [16, 18, 9].
In [16, 20] regular boundary value problems for elliptic abstract equations are considered. A
few works are concerned with non regular problems.
In this paper we establish conditions guaranteeing that non local boundary value problem for
elliptic abstract differential equation of the second order in an interval is coercive solvable in
the Hilbert space L2 (0, 1; H). A coercive estimates, when the problem is regular, was proved
in [1, 3]. The considered problem is not regular, since the boundary conditions are non local,
similar problems have been considered in [4, 5, 7, 22]. Moreover, we prove the completeness
of root functions. The completeness of root functions for regular boundary value problems are
proved in [1, 6, 10, 14] and in the book [22].
The obtained results are then applied to study a non local boundary value problem for the
Laplace equation in a cylinder.
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
036-03
2
A ISSA A IBECHE
2. N OTATIONS AND D EFINITIONS
Let H be a Hilbert space, A a linear closed operator in H and D(A) its domain. We denote
by L(H) the space of bounded linear operators acting on H, with the usual operator norm, and
by Lp (0, 1; H) the space of strongly measurable functions x → u(x) : [0, 1] → H, whose
pth −power are summable, with the norm
Z 1
p
p
kuk0,p = kukLp (0,1;H) =
ku(x)kpH dx < ∞, p ∈ [1, ∞] .
0
Now, introduce the Lp (0, 1; H) vector-valued Sobolev spaces Wp2 (0, 1; H1 , H), where H1 , H
are Hilbert spaces such that H1 ⊂ H with continuous embedding
Wp2 (0, 1; H1 , H) = {u : u00 ∈ Lp (0, 1; H) and u ∈ Lp (0, 1; H1 )}
and
kukWp2 (0,1;H1 ,H) = kukLp (0,1;H1 ) + ku00 kLp (0,1;H) < ∞.
Let E0 , E1 be two Banach spaces, which are continuously injected in the Banach space E,
the pair {E0 , E1 } is said to be an interpolation couple. Consider the Banach space
E0 + E1 = {u : u ∈ E, ∃uj ∈ Ej , j = 0, 1, with u = u0 + u1 } ,
kukE0 +E1 =
inf
ku0 kE0 + ku1 kE1 ,
u=u0 +u1 ;uj ∈Ej
and the functional
K(t, u) =
inf
u=u0 +u1 ;uj ∈Ej
ku0 kE0 + t ku1 kE1 .
The interpolation space for the couple {E0 , E1 } is defined, by the K-method, as follows
)
(
p1
Z ∞
t−1−θp K p (t, u)dt
(E0 , E1 )θ,p = u : u ∈ E0 + E1 , kukθ,p =
<∞ ,
0
0 < θ < 1, 1 ≤ p ≤ ∞.
(
)
u : u ∈ E0 + E1 , kukθ,∞ = sup t−θ K(t, u)dt < ∞ ,
(E0 , E1 )θ,∞ =
t∈(0,∞)
0 < θ < 1. Let A be a closed operator in H. H(A) is the domain of A provided with the
Hilbertian graph norm
kuk2H(A) = kAuk2 + kuk2 , u ∈ D(A).
If −A is the infinitesimal generator of the semigroup exp(−xA) which is analytic for x > 0,
decreasing at infinity and strongly continuous for x ≥ 0, then the following holds [19, p. 96]:
n
o
(H, H(An ))θ,p = u : u ∈ H, kukpθ,p < ∞ ,
where
kukpθ,p
Z
=
∞
t−n(1−θ)p−1 kAn exp(−tA)ukp dt + kukp ,
0
0 < θ < 1, n ∈ N, 1 ≤ p < ∞ and kukθ,p its norm.
Let H and H1 be Hilbert spaces such that the imbedding H1 ⊂ H is continuous and H1 = H.
Then (H, H1 )θ,2 , is a Hilbert space, we denote it by (H, H1 )θ . It is known that (H, H1 )θ =
H(S θ ), where S is a self-adjoint positive-definite operator in H [17].
+∞
1 R
Let F f (σ) = (2π)− 2
exp(iσx)f (x)dx be the Fourier transform of the function f.
−∞
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
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C OERCIVENESS I NEQUALITY
3
Lemma 2.1. [22, p. 300] Let A be a self adjoint and positive definite operator in H. Then
h
i
h
i
1 1
(1) ∃ω > 0, Aα exp −x (A + λI) 2 ≤ C exp −ωx |λ| 2 for all α ∈ R, x ≥ x0 > 0,
|arg λ| ≤ ϕ < π, where C does not depend onx and λ.
i h
R1
1
α− 14 2
2α− 12
2
α
2
(2) 0 (A + λI) exp −x (A + λI) u ≤ C A
u + |λ|
kuk
for all α ≥
1
1
, |arg λ| ≤ ϕ < π, and u ∈ D Aα− 4 , where C does not depend on u and λ.
4
(3) Aα (A + λI)−β ≤ C (1 + |λ|)α−β , for all 0 ≤ α ≤ β, |arg λ| ≤ ϕ < π, where C
does not depend on λ.
3. S OLVABILITY OF THE P RINCIPAL P ROBLEM
3.1. Homogeneous Problem. Consider in the Hilbert space H the boundary value problem for
the second order abstract differential equation
(3.1)
L(D)u = −u00 (x) + Au(x) + A(x)u(x) = f (x)
L1 u = δu(0)
(3.2)
x ∈ (0, 1),
= f1 ,
L2 u = u0 (1) + Bu(0) = f2 ,
A, A(x), B are linear operators and δ is a complex number.
Looking to the principal part of the problem (3.1), (3.2) with a parameter
(3.3)
L(D)u = −u00 (x) + (A + λI)u(x) = 0
L1 u = δu(0)
(3.4)
x ∈ (0, 1),
= f1 ,
L2 u = u0 (1) + Bu(0) = f2 .
Theorem 3.1. Assume that the following conditions are satisfied
(1) A is a self-adjoint and positive-definite operator in H.
(2) δ 6= 0.
(3) B is continuous from H(A1/2 ) in H(A) and from H in H(A1/2 ).
Then the problem (3.3), (3.4) for f1 ∈ (H, H(A)) 1 , f2 ∈ (H, H(A)) 3 and for λ such that
4
4
|arg λ| ≤ φ < π, |λ| −→ ∞, has a unique solution in the space W22 (0, 1; H(A), H), and for
the solution of the problem (3.3), (3.4) the following coercive estimate holds
(3.5)
ku00 kL2 (0,1;H) + kAukL2 (0,1;H) + |λ| kukL2 (0,1;H)
1 3 3
1
≤ C A 4 f1 + |λ| 4 kf1 kH + A 4 f2 + |λ| 4 kf2 kH ,
H
H
where C does not depend on λ.
Proof. The solution u, belonging to W22 (0, 1; H(A), H), of the equation (3.3) is in the form
(3.6)
1
2
1
2
u(x) = e−xAλ g1 + e−(1−x)Aλ g2
with Aλ = A + λI and g1 , g2 ∈ (H, H(A)) 3 .
4
Indeed, let u ∈ W22 (0, 1; H(A), H) be a solution of (3.3). Then we have
1
1
2
D − Aλ D + Aλ2 u(x) = 0.
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
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4
A ISSA A IBECHE
Note by
1
2
v(x) = D + Aλ u(x).
1
From [22, p. 168] v ∈ W21 (0, 1; H(A 2 ), H) and
1
2
(3.7)
D − Aλ v(x) = 0.
So
1
2
v(x) = e−(1−x)Aλ v(1),
(3.8)
where, according to [19, p. 44],
1
1
v(1) ∈ H(A 2 ), H 1 = H, H(A 2 ) 1 .
2
2
From (3.7) , (3.8) we have
1
−xAλ2
Z
x
1
2
1
2
e−(x−y)Aλ e−(1−y)Aλ v(1)dy
0
1
1
1
1
1
− 12
−xAλ2
−(1−x)Aλ2
−xAλ2 −Aλ2
=e
u(0) + Aλ
−e
e
v(1),
e
2
u(x) = e
u(0) +
where u(0) ∈ (H(A), H) 1 [19, p. 44]. Now,
2
1
A : (H, H(A)) 3 → (H, H(A)) 1 = H, H(A 2 ) 1
1
2
4
4
2
is an isomorphism. Consequently the last inequality is in the form (3.6).
Let us show the reverse, i.e. the function u in the form (3.6) with g1 and g2 in (H, H(A)) 3 ,
4
belongs to W22 (0, 1; H(A), H). From interpolation spaces properties see [15], [19, p. 96] and
the expression (3.6) of the function u we have
kukW 2 (0,1;H(A),H)
2
+1
≤ AA−1
λ

2 ! 12
 Z 1
1
Aλ e−xAλ2 g1 dx
+
×

Z
0
0
(3.9)
1

2 ! 12 
1
Aλ e−(1−x)Aλ2 g2 dx

≤ C kg1 k(H,H(Aλ )) 3 + kg2 k(H,H(Aλ )) 3
4
4
≤ C(λ) kg1 k(H,H(A)) 3 + kg2 k(H,H(A)) 3 .
4
4
The function u satisfies the boundary conditions (3.4) if

1

 δg1 + δe−Aλ2 g2


1
1
2
= f1 ,
1
2
1
−Aλ2 e−Aλ g1 + Bg1 + Aλ2 g2 + Be−Aλ g2 = f2 ,
which we can write in matrix form as:

 
δI 0
0

+
(3.10)


1
1
1
2
B Aλ2
−Aλ2 e−Aλ
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
1
−Aλ2
δe
Be
1
−Aλ2



g1
!
f1
=
g2
!
.
f2
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C OERCIVENESS I NEQUALITY
5
The first matrix of operators is invertible, its inverse is
 1

I
0
δ

.
(3.11)
1
− 12
1 −2
− δ Aλ B Aλ
Multiplying the two members of (3.10) by the matrix inverse (3.11), we get the following
system:

1

 g1 + e−Aλ2 g2 = 1δ f1


1
2
−1
−1
e−Aλ g1 + g2 = − 1δ Aλ 2 Bf1 + Aλ 2 f2
we can solve it by Cramer’s method, because the coefficients of the linear system are bounded
1
2
linear operators.The determinant
is given by I+e−2Aλ which is invertible as a little perturbation
−2A 21 of unity, in fact e λ ≤ q < 1.
Hence the solution is written as

 g1 = 1δ f1 + R11 (λ)f1 + R12 (λ)f2 ,
(3.12)
−1
−1

g2 = − 1δ (I + T (λ)) Aλ 2 Bf1 + (I + T (λ)) Aλ 2 f2 + R21 (λ)f1 ,
where Rij (λ) are given by

1
1
−1
2
2

 R11 (λ) = − 1δ (I + T (λ)) e−2Aλ + 1δ (I + T (λ)) Aλ 2 e−Aλ B,




1
− 21 −A 2
λ ,
R
(λ)
=
−
(I
+
T
(λ))
A
e
12
λ




1


2
R21 (λ) = 1δ (I + T (λ)) e−Aλ ,
1
2
and satisfy kRij (λ)k → 0 when |λ| → ∞. (I + T (λ)) is the inverse of I + e−2Aλ obtained
from the corresponding Neumann series.
Finally the solution u is given by
1
1
−xAλ2
u(x) = e
f1 + R11 (λ)f1 + R12 (λ)f2
δ
1
1
− 12
− 12
−(1−x)Aλ2
+e
− (I + T (λ)) Aλ Bf1 + (I + T (λ)) Aλ f2 + R21 (λ)f1 .
δ
From the assumptions of Theorem 3.1 and the properties of interpolation spaces, the following
applications are continuous,
−1
(I + T (λ)) Aλ 2 B : (H, H(A)) 3 7−→ (H, H(A)) 3 ,
4
4
− 12
(I + T (λ)) Aλ : (H, H(A)) 1 7−→ (H, H(A)) 3 .
4
4
Then we have the estimates
− 12
(I + T (λ)) Aλ Bf1 (H,H(A)) 3
≤ C kf1 k(H,H(A)) 3
4
4
and
−1
(I + T (λ)) Aλ 2 f2 (H,H(A)) 3
4
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
≤ C kf2 k(H,H(A)) 1 .
4
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6
A ISSA A IBECHE
Set u(x) = u1 (x) + u2 (x) + u3 (x), where
1
1
2
u1 (x) = e−xAλ f1 ,
δ
1 −(1−x)A 12 −1
λ
(I + T (λ)) Aλ 2 Bf1 ,
u2 (x) = − e
δ
1 2
−1
u2 (x) = e−(1−x)Aλ (I + T (λ)) Aλ 2 f2 .
Then
ku1 kW 2 (0,1;H(A),H)
2
1
−xA 12 1
1 2
−xA
Aλ e λ f1 Ae λ f1 =
+
.
|δ| |δ| L2 (0,1;H)
L2 (0,1;H)
However, from Lemma 2.1, we have
1
Aλ e−xAλ2 f1 L2 (0,1;H)
3 3
≤ C A 4 f1 + |λ| 4 kf1 kH .
H
Similarly we obtain bounds for u2 and u3 .
3.2. Non Homogeneous Problem. Consider, now, the principal problem for the non homogeneous equation with a parameter
(3.13)
L0 (λ, D)u = −u00 (x) + Aλ u(x) = f (x)
L10 u = δu(0)
(3.14)
x ∈ (0, 1),
= f1 ,
L20 u = u0 (1) + Bu(0) = f2 .
We have the result.
Theorem 3.2. Suppose the following conditions satisfied
(1) A is a self-adjoint and positive-definite operator in H.
(2) B is continuous from H(A1/2 ) in H(A) and from H in H(A1/2 ).
(3) δ 6= 0.
Then the problem (3.13), (3.14), for f, f1 and f2 in L2 (0, 1; H), (H, H(A)) 3 and (H, H(A)) 1
4
4
respectively, and for λ such that |arg λ| ≤ φ < π, |λ| −→ ∞, has a unique solution belonging
to the space Wp2 (0, 1; H(A), H), for p ∈ (1, ∞), and the following coercive estimate holds
(3.15)
ku00 kL2 (0,1;H) + kAukL2 (0,1;H) + |λ| kukL2 (0,1;H)
3 1 3
1
4 4
≤ C kf kL2 (0,1;H) + A f1 + |λ| kf1 kH + A 4 f2 + |λ| 4 kf2 kH ,
H
H
where C does not depend on λ.
Proof. In Theorem 3.1, we proved the uniqueness. Let us now show that the solution of the
problem (3.13), (3.14) belonging to Wp2 (0, 1; H(A), H) can be written in the form u(x) =
u1 (x) + u2 (x), u1 (x) is the restriction to [0, 1] of u
e1 (x), where u
e1 (x) is the solution of the
equation
(3.16)
L0 (λ, D)e
u1 (x) = fe(x),
x ∈ R,
with fe(x) = f (x) if x ∈ [0, 1] and fe(x) = 0 otherwise. u2 (x) is the solution of the problem
(3.17)
L0 (λ, D)u2 = 0, L10 u2 = f1 − L10 u1 , L20 u2 = f2 − L20 u1 .
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
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C OERCIVENESS I NEQUALITY
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The solution of the equation (3.16) is given by the formula
Z
1
b
(3.18)
u
b1 (x) = √
eiµx L0 (λ, iµ)−1 fe(µ)dµ
2π R
b
where fe is the Fourier transform of the function fe(x), L0 (λ, s) is the characteristic pencil of the
equation (3.16) i.e. L0 (λ, s) = −s2 I + A + λI.
From (3.18) and Plancherel equality it follows that:
|λ| ku1 kL2 (0,1;H) + ku001 kL2 (0,1;H) + kAu1 kL2 (0,1;H)
(3.19)
≤ |λ| kb
u1 kL2 (R;H(A)) + kb
u001 kL2 (R;H) + kAb
u1 kL2 (R;H)
−1
−1
−1 e
2
−1 e
≤ |λ| F L0 (λ, iµ) F f (µ)
+ F (iµ) L0 (λ, iµ) F f (µ)
L2 (R;H)
L2 (R;H)
−1
+ F AL0 (λ, iµ)−1 F fe(µ)
,
L2 (R;H)
where F is the Fourier transform.
From condition (1) of Theorem 3.2, for |arg λ| ≤ ϕ < π and |λ| sufficiently large, we have
L0 (λ, iµ)−1 = (A + λI + µ2 I)−1 ≤ C(1 + λ + µ2 )−1 ≤ C |µ|−2
(3.20)
AL0 (λ, iµ)−1 = A(A + λI + µ2 I)−1 ≤ C
(3.21)
(3.22)
|λ| L0 (λ, iµ)−1 = |λ| (A + λI + µ2 I)−1 ≤ C |λ| (1 + λ + µ2 )−1 ≤ C.
Then it follows that
|λ| ku1 kL2 (0,1;H) + ku001 kL2 (0,1;H) + kAu1 kL2 (0,1;H) ≤ C kf kL2 (0,1;H) .
Since u1 ∈ W22 (0, 1; H(A), H) and from [19, p. 44] we have
u01 (0) ∈ (H(A), H) 3 = (H, H(A)) 1 ,
4
4
u1 (0) ∈ (H, H(A)) 3 .
4
Therefore L10 u1 ∈ (H, H(A)) 3 and L20 u1 ∈ (H, H(A)) 1 .
4
4
From Theorem 3.1, the problem (3.17) , when |arg λ| ≤ φ < π, |λ| → ∞, has a solution
u2 (x) which is in W22 (0, 1; H(A), H). Now, we have to find bounds for the following terms
3
3
4
4
A
L
u
=
A
u
(0)
,
10 1 1
H
H
3
4
3
4
|λ| kL10 u1 kH = |λ| ku1 (0)kH ,
1
1
1
4
4 0
4
A
L
u
≤
A
u
(1)
+
A
Bu
(0)
20 1 1
1
H
and
H
1
1
H
1
|λ| 4 kL20 u1 kH ≤ |λ| 4 ku01 (1)kH + |λ| 4 kBu1 (0)kH .
For example, we have
3
4
|λ| ku1 (0)kH ≤ C ku1 kW 2 (0,1;H(A),H) + |λ| ku1 kL2 (0,1;H)
2
≤ C kf kL2 (0,1;H) .
Similarly, we get the other bounds and by the same way the coerciveness estimate.
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
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8
A ISSA A IBECHE
4. S OLVABILITY OF THE G ENERAL P ROBLEM
Consider, now, the general problem with a parameter
(4.1)
L0 (λ, D)u = λu(x) − u00 (x) + Au(x) + A(x)u(x) = f (x)
L10 u = δu(0)
(4.2)
x ∈ (0, 1),
= f1 ,
L20 u = u0 (1) + Bu(0) = f2 .
We have the result.
Theorem 4.1. Suppose the following conditions satisfied
(1)
(2)
(3)
(4)
(5)
A is a self-adjoint and positive-definite operator in H.
The imbedding H(A) ⊂ H is compact.
B is continuous from H(A1/2 ) in H(A) and from H in H(A1/2 ).
δ 6= 0.
kA(.)ukL2 (0,1;H) ≤ kAukL2 (0,1;H) + C() kukL2 (0,1;H) .
Then the problem (4.1), (4.2), for f, f1 and f2 in L2 (0, 1; H), (H, H(A)) 3 and (H, H(A)) 1
4
4
respectively, and for λ such that |arg λ| ≤ φ < π, |λ| −→ ∞, has a unique solution belonging
to the space Wp2 (0, 1; H(A), H), for p ∈ (1, ∞), and the following coercive estimate holds
(4.3)
ku00 kL2 (0,1;H) + kAukL2 (0,1;H) + |λ| kukL2 (0,1;H)
1 3 3
1
4 4 4
4
≤ C kf kL2 (0,1;H) + A f1 + |λ| kf1 kH + A f2 + |λ| kf2 kH .
H
H
where C does not depend on u, f, f1, f2, and λ.
Proof. Let u be a solution of (4.1), (4.2) belonging to W22 (0, 1; H(A), H). Then u is a solution
of the problem

x ∈ (0, 1),

 L0 (λ, D)u = f (x) − A(x)u(x)
L
u
=
δu(0)
=
f1
10
(P0 )


L20 u = u0 (1) + Bu(0) = f2 .
From Theorem 3.2, we get the estimate
ku00 kL2 (0,1;H) + kAukL2 (0,1;H) + |λ| kukL2 (0,1;H)
1 3 3
1
≤ C kf − A(.)ukL2 (0,1;H) + A 4 f1 + |λ| 4 kf1 kH + A 4 f2 + |λ| 4 kf2 kH .
H
H
Using condition (5) of Theorem 3.2, we get
ku00 kL2 (0,1;H) + (1 − C) kAukL2 (0,1;H) + (|λ| − C.C()) kukL2 (0,1;H)
3 1 3
1
4 4
≤ C kf kL2 (0,1;H) + A f1 + |λ| kf1 kH + A 4 f2 + |λ| 4 kf2 kH .
H
H
Choosing such that C · < 1, the coerciveness estimates follows easily.
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
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C OERCIVENESS I NEQUALITY
9
5. C OMPLETENESS OF ROOT F UNCTIONS
Let us define the operator L by
Lu ≡ −u00 + Au,
D (L) = W22 (0, 1; H(A), H, Lk u = 0, k = 0, 1) .
Lemma 5.1. Suppose that sj (I, H(A), H) ' Cj −q then
− 11
sj I, W22 (0, 1; H(A), H) , L2 (0, 1; H) ' Cj 2 +q .
I (resp. I) is the imbedding of H(A) in H (resp. of W22 (0, 1; H(A), H) in L2 (0, 1; H)) and
sj (I, H(A), H) are the s-numbers of the operator I from H(A) to H.
Proof. Let S1 be the operator defined in L2 (0, 1) such that S1 = S1∗ ≥ γ 2 I, D(S1 ) = H(S1 ) =
W22 (0, 1). From [17], we know that if H1 ⊂ H and H1 is dense in H then there exists an
operator S1 such that S1 = S1∗ and D(S1 ) = H. Otherwise, let S2 be the operator defined by
S2 = S2∗ ≥ γ 2 I, D(S2 ) = H(A). If we define the operator S on L2 (0, 1) ⊗ H = L2 (0, 1; H)
by S = S1 ⊗ I2 + I1 ⊗ S2 , where I1 (resp. I2 ) is the identity operator in L2 (0, 1) (resp. H). We
have
sj S1−1 , L2 (0, 1), L2 (0, 1) ' sj (I, H(S1 ), L2 (0, 1)) ' Cj −2 ,
sj S2−1 , H, H ' sj (I, H(A), H) ' Cj −q .
Hence, from [11], we obtain sj (S −1 , L2 (0, 1; H), L2 (0, 1; H)) ' Cj
−
1
1 +q
2
.
Theorem 5.2. Let the conditions of Theorem 3.2 hold along with A−1 ∈ σq (H) , q > 0. Then,
the system of root functions of the operator L is complete in L2 (0, 1; H).
Proof. From Theorem 4.1, we have kR(λ, L)k ≤ C |λ|−1 for |arg λ| ≤ ϕ < π and |λ| sufficiently large. Using Lemma 5.1, we have R(λ, L) ∈ σp (L2 (0, 1; H)) for p > 12 + 1q , so, for the
operator L, all the conditions of [8, Theorem 126, 2.3, p. 50], are fulfilled. This achieves the
proof of the theorem.
Theorem 5.3. Suppose that the conditions of Theorem 5.2 are satisfied as well as the condition
D(A(x)) ⊂ D(A) and ∀ > 0, kA(·)ukH ≤ kAukH + C() kukH . u ∈ D(A). Let A be
the operator defined by (Au) (x) = A(x)u(x), D (A) = L2 (0, 1; H). Then the system of root
functions of L + A is complete in L2 (0, 1; H).
Proof. It is clear that
kAukL2 (0,1;H) ≤ kAukL2 (0,1;H) + C() kukL2 (0,1;H) .
Since by Theorem 4.1, we have
kAukL2 (0,1;H) ≤ C kf kL2 (0,1;H) = C k(L − λI) ukL2 (0,1;H) ,
hence
kAukL2 (0,1;H) ≤ k(L − λI) ukL2 (0,1;H) + C() kukL2 (0,1;H)
and so, for |λ| sufficiently large and |arg λ| ≤ ϕ < π,
R(λ, L + A) ∈ σp (L2 (0, 1; H)) ,
and from Theorem 4.1 we have
kR(λ, L + A)k ≤ C |λ|−1
for |λ| sufficiently large and |arg λ| ≤ ϕ < π. Then the system of root functions is complete in
L2 (0, 1; H).
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
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10
A ISSA A IBECHE
6. A PPLICATION
Let us consider, in the cylindrical domain Ω = [0, 1] × G the non local boundary value
problem for the Laplace equation with a parameter

L(λ)u = λu(x, y) − ∆u(x, y) + b(x, y)u(x, y) = f (x, y), (x, y) ∈ Ω;






 L1 u =
δu(0, y) = f1 (y),
y ∈ G;
(6.1)
(P )
∂


L2 u =
u(1, y) + Bu(0, y) = f2 (y),
y ∈ G;

∂x




Pu =
u(x, y 0 ) = 0,
(x, y 0 ) ∈ Γ,
where Γ = [0, 1] × ∂G and ∂G is the boundary of G.
A number λ0 is called an eigenvalue of (P ) if the problem

L(λ0 )u = 0, (x, y) ∈ Ω;






 L1 u = 0,
y ∈ G;
0
(6.2)
(P )


L2 u = 0,
y ∈ G;





P u = 0,
(x, y 0 ) ∈ Γ,
has a non trivial solution that belongs to W22 (Ω). The non trivial solution u0 of (P 0 ) that belongs
to W22 (Ω) is called the eigenfunction of (P ) corresponding to the eigenvalue λ0. Solutions uk
of

L(λ0 )uk + uk−1 = 0, (x, y) ∈ Ω;






 L1 uk = 0,
y ∈ G;
00
(6.3)
(P )


L2 uk = 0,
y ∈ G;





P uk = 0,
(x, y 0 ) ∈ Γ,
belonging to W22 (Ω) are associated functions of the k − th rank to the eigenvalue u0 of (P ) .
Eigenfunctions and associated functions of (P ) are gathered under the general name of root
functions of (P ) .
0,1
Theorem 6.1. Let b(x, y) ∈ W∞
(Ω), δ 6= 0, ∂G ∈ C 2 then
−
mk
+3
(1) (P ) , for f ∈ W20,1 (Ω, P u = 0), fk ∈ W2 2 4 (G, P u = 0) and for λ such that |λ|
sufficiently large and |arg λ| ≤ ϕ < π, has a unique solution that belongs to the space
W22 (Ω), and for this solution we have the coercive estimate
|λ| ku00 kL2 (Ω) + kukW 2 (Ω)
2
≤C
kf kL2 (Ω) +
2
X
kfk k
k=1
m
− 2k + 3
4 (G,P u=0)
W2
+
2
X
m
− 2k
|λ|
!
+ 34
kfk kH
,
k=1
where C does not depend on u and λ.
(2) Root functions of (P ) are complete in L2 (Ω).
Proof. Consider in H = L2 (Ω) the operators A and A(x) defined by
Au = −∆u(y) + λ0 u(y),
A (x) u = b(x, y)u(y) − λ0 u(y),
J. Inequal. Pure and Appl. Math., 4(2) Art. 43, 2003
D(A) = W22 (G, P u = 0)
D(A (x)) = W21 (G, P u = 0, m = 0).
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C OERCIVENESS I NEQUALITY
11
Then the problem (P ) can be written in the form

λu(x) − u00 (x) + Au(x) + A(x)u(x) = 0 x ∈ (0, 1),



δu(0)
= f1



u0 (1) + Bu(0)
= f2 .
We have the compact imbedding W22 (Ω) ⊂ L2 (Ω). On the other hand
2
sj I, W22 (Ω), L2 (Ω) ' j − r+1 .
By virtue of Lemma 3.1 in [21, p. 60] we have
sj (I, H(A), L2 (Ω)) ' sj A−1 , L2 (Ω), L2 (Ω) .
Since H(A) ⊂ W22 (Ω), then it follows that A−1 ∈ σp (L2 (Ω), L2 (Ω)) , then kR(λ, A)k ≤
C |λ|−1 for |arg λ| ≤ ϕ < π and |λ| sufficiently large. Hence, all conditions of Theorem 5.3
has been checked.
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