Electronic Journal of Differential Equations, Vol. 2007(2007), No. 137, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
ON THE SOLVABILITY OF A FOURTH ORDER OPERATOR
DIFFERENTIAL EQUATIONS
AHMED NOUAR
Abstract. We establish sufficient conditions for the solvability of a fourth
order operator differential equation. These conditions are expressed in terms
of operator coefficients.
1. Preliminaries
In this paper we consider the problem
P (d/dt)u =
4
X
d4 u
d4−j u
4
+
A
u
+
Aj 4−j = f,
4
dt
dt
j=0
u(si ) (0) = 0, i = 0, 1,
si ∈ {0, 1, 2, 3},
(1.1)
e1
(1.2)
e2
where f ∈ L2 (R+ , H), and A0 , A1 , A2 , A3 , A4 and A are unbounded linear operators
in a separable Hilbert space H. By L2 (R+ , H) we denote the space of H-valued
R +∞
vector functions such that 0 kf (t)k2H dt < +∞, endowed with the inner product
Z +∞
hf, giL2 (R+ ,H) =
hf (t), g(t)iH dt,
0
and R+ = [0, +∞). Let
P (λ) = λ4 I + A4 +
4
X
λ4−j Aj ,
(1.3)
j=0
be the operator pencil associated with the operator differential equation (1.1).
Our aim is to establish sufficient conditions for the solvability of problem (1.1),
(1.2), expressed in terms of the norms of the operator coefficients of the pencil (1.3).
For the general theory of operator pencils see [9, 11]. In particular, unbounded
operator pencils of second and fourth orders are treated in [12, 13].
Interest for such equations is justified not only by the relation which they have
with the practical problems described by ordinary or partial differential equations
but also by their important applications in many theoretical problems. Thus, the
study of such equations is connected with the multicompleteness of the eigenvectors
and the generalized eigenvectors of the corresponding operator pencil. This relation
2000 Mathematics Subject Classification. 47E05, 35K20.
Key words and phrases. Operator Pencil; operator differential equation; normal operator.
c
2007
Texas State University - San Marcos.
Submitted February 24, 2007. Published October 17, 2007.
1
e3
2
A. NOUAR
EJDE-2007/137
has been proved in [1] and permits to answer some questions arising while studying
non self-adjoint differential operators. The notion of multicompleteness has been
introduced by Keldysh in [2, 3] and many interesting results have been obtained
during the last decades by Gasymov [1], Yakubov [9, 10, 11, 13], Markus [5], Mirzoev
[6, 7] and others.
2. Definitions
def1
Definition 2.1. Let A be a normal operator in a Hilbert space H such that σ(A)
is contained in the sector
π
Sε = {λ : | arg λ| < ε <
}, inf Re σ(A) = σ0 > 0.
12
We assume also that the operators Bj = Aj A−j , j = 0, 4 are bounded. Under these
conditions we will say that the pencil (1.3) belongs to the class Γ.
It is clear that if the operator A satisfies the conditions of the above definition it
admits a polar decomposition in the form: A = U C , where C is a positively defined
self-adjoint operator and U is a unitary operator, therefore C and U commute and
we have for any rational k
D(C k ) = D(Ak ),
kAk xkH = kC k xkH ,
x ∈ D(Ak ).
In D(C k ) we define the inner product
hx, yiHk = hC k x, C k yiH ,
x, y ∈ D(Ak ).
Then, we obtain a Hilbert space which we denote by Hk . By D(R+ , H) we denote
the pre-Hilbertian space of the infinitely differentiable H4 -valued functions with a
compact support in R+ , relative to the inner product
hu, viW = hu(4) , v (4) iL2 (R+ ,H) + hC 4 u, C 4 viL2 (R+ ,H) .
By W (R+ , H) we denote the completion of D(R+ , H) with respect to the above
inner product. We introduce also the following notation
W (R+ , H, s0 , s1 ) = {u ∈ W (R+ , H), u(s0 ) (0) = u(s1 ) (0) = 0},
M4,j (s0 , s1 ) =
kC 4−j u(j) kL2
.
kukW
u∈W (R+ ,H,s0 ,s1 )
sup
def2
Definition 2.2. We say that the couple (s0 , s1 ) belongs to the class S if it takes
one of these values: (0, 1), (0, 2), (1, 3), (2, 3).
def3
Definition 2.3. We say that the problem (1.1), (1.2) is regular if for any function
f ∈ L2 (R+ , H) there exists a unique function u ∈ W (R+ , H, s0 , s1 ) which satisfies
the equation (1.1) a.e. and the conditions (1.2). We call such a function u a regular
solution of the problem (1.1), (1.2).
3. Main Results
We consider now the pencil operator
Sj (λ, γ, C) = λ8 I + C 8 − γ(iλ)2j C 8−2j ,
EJDE-2007/137
ON THE SOLVABILITY
3
defined in H8 and where γ is a real parameter. We use the notation
(
1,
j = 0, 4,
d4,j =
4−j
4
( 4j )j/4 ( 4−j
) 4 , j = 1, 2, 3.
thm1
Theorem 3.1. For γ ∈ (0, d4,j ), j ∈ {0, . . . , 4}, the pencil operator Sj (λ, γ, C) can
be represented in the form
Sj (λ, γ, C) = Sj− (λ, γ, C)Sj+ (λ, γ, C),
where
Sj− (λ, γ, C) =
4
X
αi,j (γ)λi C 4−i ,
i=0
Sj+ (λ, γ, C) = Sj− (−λ, γ, C).
Moreover, the coefficients αi,j are real and satisfy the following relations
p
α0,0 = 1 − γ,
α4,0 = 1,
2
α1,0
− 2α0,0 α2,0 = 0,
2
α2,0
(3.1)
e4
(3.2)
e5
(3.3)
e6
− 2α1,0 α3,0 + 2α0,0 α4,0 = 0,
2
α3,0
− 2α2,0 α4,0 = 0.
and
α0,4 = 1
p
α4,4 = 1 − γ
2
α1,4
− 2α0,4 α2,4 = 0
2
α2,4
− 2α1,4 α3,4 + 2α0,4 α4,4 = 0
2
α3,4
− 2α2,4 α4,4 = 0,
and for j = 1, 2, 3
α0,j = α4,j = 1,
(
−γ, j = 1,
2
α1,j
− 2α2,j =
0,
j 6= 1.
(
−γ, j = 2,
2
α2,j
− 2α1,j α3,j + 2 =
0,
j 6= 2.
(
−γ, j = 3,
2
α3,j
− 2α2,j =
0,
j 6= 3.
Proof. For γ ∈ (0, d4,j ), j ∈ {0, . . . 4} the polynomial Qj (λ, γ) = λ8 + 1 − γ(iλ)2j
has not purely imaginary roots, its roots are simple and symmetrically situated
relatively to the real axis and the origin. So it can be represented in the form
Qj (λ, γ) = Sj (λ, γ)Sj (−λ, γ).
(3.4)
The roots of Sj (λ, γ) belong to the left half-plane and we can denote them by ω1 (γ),
ω2 (γ), ω1 (γ), ω2 (γ) where ω1 (γ) and ω2 (γ) are the conjugate of ω1 (γ) and ω2 (γ)
e7
4
A. NOUAR
EJDE-2007/137
respectively. In this case
Sj (λ, γ) = (λ − ω1 (γ))(λ − ω2 (γ))(λ − ω1 (γ))(λ − ω2 (γ)).
P4
This proves that if we write Sj (λ, γ) = i=0 αi,j λi , then the coefficients αi,j are
real.
The relations (3.1), (3.2) and (3.3) can be obtained by using the Viete theorem
and identifying the coefficients in the left and right parts of the relation (3.4). Let
Eσ be the spectral decomposition of the operator C. We have
Sj (λ, γ, C) = λ8 I + C 8 − γ(iλ)2j C 8−2j
Z ∞
(λ8 + σ 8 − γ(iλ)2j σ 8−2j )dEσ
=
σ0
∞
λ8
(iλ)2j
+
1
−
γ
)dEσ
σ8
σ 2j
σ0
Z ∞
λ
=
σ 8 Qj ( , γ)dEσ
σ
σ0
Z ∞
λ
λ
=
σ 8 Sj ( , γ)Sj (− , γ)dEσ
σ
σ
σ0
Z ∞
Z ∞
λ
λ
4
σ 4 Sj (− , γ)dEσ
=
σ Sj ( , γ)dEσ
σ
σ
σ0
σ0
Z
=
=(
4
X
σ8 (
4
X
αi,j (−λ)i C 4−i ).
αi,j λi C 4−i )(
i=0
i=0
Then taking
Sj− (λ, γ, C) =
4
X
αi,j λi C 4−i ,
i=0
we obtain Sj (λ, γ, C) = Sj− (λ, γ, C)Sj+ (λ, γ, C).
thm2
Theorem 3.2. Let γ ∈ (0, d4,j ) and u ∈ W (R+ , H). Then for j ∈ {0, . . . 4},
kSj− (d/dt, γ, C)uk2L2 = kuk2W − γkC 4−j u(j) k2L2 − hGj fe, feiH 4 ,
where
000
fe = C 7/2 u(0), C 5/2 u0 (0), C 3/2 u00 (0), C 1/2 u (0) ∈ H ⊕ H ⊕ H ⊕ H = H 4 ,


α1 α0
α2 α0
α3 α0
α4 α0
α2 α0 α2 α1 − α3 α0 α3 α1 − α4 α0 α4 α1 

Gj = 
α3 α0 α3 α1 − α4 α0 α3 α2 − α4 α1 α4 α2  ,
α4 α0
α4 α1
α4 α2
α4 α3
and αi ≡ αi,j (γ).
EJDE-2007/137
ON THE SOLVABILITY
5
Proof. By a theorem of the intermediate derivatives [4, pp.19-25], C 4−j u(j) (·) ∈
1
L2 (R+ , H) and u(j) (0) ∈ D(C 4−j− 2 ). We have
kSj− (d/dt, γ, C)uk2L2 (R+ ,H) = k
4
X
αi,j C 4−i u(i) k2L2 (R+ ,H)
i=0
=
4
X
2
αi,j
kC 4−i u(i) k2L2 (R+ ,H)
i=0
+2
4 X
i−1
X
αi,j αs,j RehC 4−i u(i) , C 4−s u(s) iL2 (R+ ,H) .
i=1 s=0
4−i (i)
Calculating the expressions hC u , C 4−s u(s) iL2 (R+ ,H) by using successive integrations by parts and taking into account that C is a selfadjoint operator we obtain
kSj− (d/dt, γ, C)uk2L2 (R+ ,H)
2
2
2
= α0,j
kC 4 uk2L2 (R+ ,H) + α4,j
ku(4) k2L2 (R+ ,H) + (α3,j
− 2α2,j α4,j )kCu(3) k2L2 (R+ ,H)
2
+ (α2,j
− 2α1,j α3,j + 2α0,j α4,j )kC 2 u(2) k2L2 (R+ ,H)
2
+ (α1,j
− 2α2,j α0,j )kC 3 u(1) k2L2 (R+ ,H) − hGj fe, feiH 4 .
Then using the relations (3.1), (3.2) and (3.3) we obtain
kSj− (d/dt, γ, C)uk2L2 (R+ ,H) = kuk2W − γkC 4−j u(j) k2L2 (R+ ,H) − hGj fe, feiH 4 .
rmk1
Remark 3.3. Theorem 3.2 shows that M4,j (s0 , s1 ) can be written with respect to
Sj− (d/dt, γ, C) and Gj . So, according to Theorem 3.1 and the definition of Gj , it
can be written with respect to γ and αi,j . In fact, the pencil Sj (λ, γ, C) from which
we have obtained the pencil Sj− (d/dt, γ, C) has been especially defined to this end.
Let Gj (s0 , s1 , γ) be the matrix obtained from the matrix Gj by deleting the lines
s0 + 1 and s1 + 1 and the columns s0 + 1 and s1 + 1.
thm3
Theorem 3.4 ([8]). Let j ∈ {0, . . . 4}, then
(
−1/2
d4,j ,
if det Gj (s0 , s1 , γ) 6= 0, for any γ ∈ (0, d4,j )
M4,j (s0 , s1 ) =
−1/2
µ4,j (s0 , s1 ), otherwise,
where µ4,j (s0 , s1 ) is the smallest positive root from (0, d4,j ) of the equation
det Gj (s0 , s1 , γ) = 0
with respect to γ.
rmk2
Remark 3.5. Theorem 3.4 implies that to calculate M4,j (s0 , s1 ), one must determine the matrix Gj (s0 , s1 , γ) from the matrix Gj obtained in Theorem 3.2 and
resolve the equation det Gj (s0 , s1 , γ) = 0 with respect to γ by using the relations
(3.1), (3.2) and (3.3) from Theorem 3.1.
lemme1
Lemma 3.6. Let the operator A be as in Definition 2.1 and u ∈ W (R+ , H, s0 , s1 )
with (s0 , s1 ) ∈ S, then
kukW ≤ (1 − 2 sin 2ε)−1/2 kP0 ukL2 (R+ ,H) ,
6
A. NOUAR
EJDE-2007/137
where
P0 u =
d4 u
+ A4 u.
dt4
Proof. If u ∈ W (R+ , H, s0 , s1 ), we have
kP0 uk2L2 (R+ ,H) = kA4 uk2L2 (R+ ,H) + ku(4) k2L2 (R+ ,H) + 2 RehA4 u, u(4) iL2 (R+ ,H) .
Using integration by parts, we have
hA4 u, u(4) iL2 (R+ ,H)
∞
∞
1
3
7
5
= hA 2 u, A∗ 2 u(3) i − hA 2 u0 , A∗ 2 u(2) i + hA2 u(2) , A∗2 u(2) iL
0
0
2 (R+ ,H)
.
So, for (s0 , s1 ) ∈ S, we obtain
kP0 uk2L2 (R+ ,H) = kuk2W + 2 RehA2 u(2) , A∗2 u(2) iL2 (R+ ,H) .
We have
hA2 u(2) , A∗2 u(2) iL2 (R+ ,H)
= hA2 u(2) , (A∗2 A−2 − I)A2 u(2) iL2 (R+ ,H) + kA2 u(2) k2L2 (R+ ,H)
≥ kA2 u(2) k2L2 (R+ ,H) − kA∗2 A−2 − IkkA2 u(2) k2L2 (R+ ,H) .
Now, we have also
∗2
−2
kA A
Z
− Ik = k
σ(A)
Z
=k
σ(A)
Z
=k
σ(A)
σ2
− 1 dEσ k
σ2
σ2 − σ2
dEσ k
σ2
e−2iθ − e2iθ dEσ k
e2iθ
where θ = arg σ. The spectrum σ(A) of the operator A is contained in the sector
Sε defined by Definition 2.1 and this means that
e−2iθ − e2iθ ≤ 2 sin 2ε,
e2iθ
this proves that
hA2 u(2) , A∗2 u(2) iL2 (R+ ,H) ≥ (1 − 2 sin 2ε)kA2 u(2) k2L2 (R+ ,H) .
Hence
kP0 uk2L2 (R+ ,H) ≥ kuk2W + 2(1 − 2 sin 2ε)kA2 u(2) k2L2 (R+ ,H) ,
and so
kP0 uk2L2 (R+ ,H) ≥ kuk2W + 2kC 2 u(2) k2L2 (R+ ,H) − 4 sin 2εkC 2 u(2) k2L2 (R+ ,H) .
EJDE-2007/137
ON THE SOLVABILITY
7
Since
kC 2 u(2) k2L2 (R+ ,H) =
Z
∞
hC 2 u(2) , C 2 u(2) idt
0
∞
∞
5
1
3
7
= hC 2 u0 , C 2 u(2) i0 − hC 2 u, C 2 u(3) i0 +
Z ∞
=
hC 4 u, u(4) idt
Z
∞
hC 4 u, u(4) idt
0
0
≤ kC 4 ukL2 (R+ ,H) ku(4) kL2 (R+ ,H)
1
≤ kC 4 uk2L2 (R+ ,H) + ku(4) k2L2 (R+ ,H)
2
1
= kuk2W
2
we conclude that
kP0 uk2L2 (R+ ,H) ≥ (1 − 2 sin 2ε)kuk2W + 2kC 2 u(2) k2L2 (R+ ,H)
≥ (1 − 2 sin 2ε)kuk2W ,
and finally, we have
1
kukW ≤ (1 − 2 sin 2ε)− 2 kP0 ukL2 (R+ ,H) .
lemme2
Lemma 3.7. Let the operator A be as in Definition 2.1, then for (s0 , s1 ) ∈ S, the
operator P0 is a bijection between W (R+ , H, s0 , s1 ) and L2 (R+ , H).
Proof. Consider the problem
P0 u = g,
u
(s0 )
(0) = u(s1 ) (0) = 0,
where g ∈ L2 (R+ , H) and (s0 , s1 ) ∈ S. The unique solution of this problem is given
by (see [1])
Z ∞
u(t) = V1 (t)Ψ0 + U1 (t)Ψ1 +
G(t, x)g(x)dx,
0
where G(t, x) is the Green function of the problem
P0 u = g,
0
u(0) = u (0) = 0,
and V1 (t) and U1 (t) are analytic semigroups generated by the operators w1 A and
w2 A respectively, where w1 and w2 are the roots on the left half plane of the equation
w4 = −1. The elements Ψ0 and Ψ1 can be obtained by solving the system
Z ∞
(s )
(s )
V1 0 (t)Ψ0 + U1 0 (t)Ψ1 +
G(s0 ) (t, x)g(x)dx = 0,
Z0 ∞
(s1 )
(s1 )
V1 (t)Ψ0 + U1 (t)Ψ1 +
G(s1 ) (t, x)g(x)dx = 0.
0
8
A. NOUAR
EJDE-2007/137
Theorem 3.8. Let pencil (1.3) belongs to the class Γ and the couple (s0 , s1 ) be in
the class S and
4
X
βj kA4−j Aj−4 k < 1,
j=0
where βj = M4,j (s0 , s1 )(1 − 2 sin 2ε)−1/2 . Then, problem (1.1), (1.2) admits a
unique regular solution.
Proof. According to the Lemma 3.7 we can look for the solution of (1.1), (1.2) in
the form u = P0−1 v, where v ∈ L2 (R+ , H). In this case, problem (1.1), (1.2) takes
the form
(I + P1 P0−1 )v = f,
(3.5)
where v ∈ L2 (R+ , H),
d4
+ A4 ,
dt4
D(P0 ) = W (R+ , H, s0 , s1 ),
P0 =
P1 =
4
X
j=0
A4−j
dj
.
dtj
A sufficient condition for the existence of a unique solution of (3.5) is that
kP1 P0−1 k < 1.
Let h ∈ L2 (R+ , H), we have
kP1 P0−1 hkL2 (R+ ,H) = k
4
X
j=0
=k
4
X
A4−j
dj
(P −1 h)kL2 (R+ ,H)
dtj 0
A4−j Aj−4 A4−j
j=0
≤
4
X
dj
(P −1 h)kL2 (R+ ,H)
dtj 0
kA4−j Aj−4 kkA4−j
dj
(P −1 h)kL2 (R+ ,H)
dtj 0
kA4−j Aj−4 kkC 4−j
dj
(P −1 h)kL2 (R+ ,H) .
dtj 0
j=0
=
4
X
j=0
Since P0−1 h ∈ W (R+ , H, s0 , s1 ),
dj
(P −1 h)kL2 (R+ ,H) ≤ M4,j (s0 , s1 )kP0−1 hkW .
dtj 0
Moreover, according to Lemma 3.6, we have
kC 4−j
1
kP0−1 hkW ≤ (1 − 2 sin 2ε)− 2 khkL2 (R+ ,H) .
In this case
kP1 P0−1 hkL2 (R+ ,H) ≤
4
X
j=0
1
kA4−j Aj−4 kM4,j (s0 , s1 )(1 − 2 sin 2ε)− 2 khkL2 (R+ ,H) .
e8
EJDE-2007/137
Then
kP1 P0−1 k ≤
ON THE SOLVABILITY
4
X
9
M4,j (s0 , s1 )(1 − 2 sin 2ε)−1/2 kA4−j Aj−4 k.
j=0
In view of the corresponding condition of the theorem, the above inequality implies
kP1 P0−1 k < 1.
rmk3
Remark 3.9. The value of M4,j (s0 , s1 ) obtained in Theorem 3.4 gives us the best
estimation for the norm of the operator P1 P0−1 in the sense that the coefficients βj
are optimal in the inequality
4
X
βj kA4−j Aj−4 k < 1.
j=0
Acknowledgements. We would like to thank the anonymous referees for their
constructive and helpful comments which improved the quality of this paper.
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Gas1
keld1
keld2
Lion
markus
mirz1
mirz2
mirz3
yak3
yak2
yak1
yak5
yak4
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Département des Sciences Fondamentales, Faculté des Sciences et des Sciences de
L’Ingenieur, Université de Skikda, 21000 Skikda, Algeria
E-mail address: hamdensk21@Yahoo.fr