Bisectorial operator pencils and the problem of bounded solutions

ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2012, Vol. 56, No. 3, pp. 26–35. c© Allerton Press, Inc., 2012.Original Russian Text c© A.V. Pechkurov, 2012, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2012, No. 3, pp. 31–41.

Bisectorial Operator Pencils and the Problem of Bounded Solutions

A. V. Pechkurov1*

1Voronezh State University, Universitetskaya pl., 1, Voronezh, 394000 RussiaReceived March 5, 2011

Abstract—We consider a linear differential equation unresolved with respect to the derivative. Weassume that the spectrum of the corresponding pencil is contained in two sectors. We study theunique existence of a bounded solution with any bounded free term.

DOI: 10.3103/S1066369X1203005X

Keywords and phrases: bisectorial operator pencil, problem of bounded solutions, Greenfunction.

It is well-known [1–3] that the equation u′ − Au = f , where A is a linear bounded operator, has aunique solution bounded on the axis with any bounded right-hand side f if and only if the spectrum of Adoes not intersect the imaginary axis.

One can easily extend a particular case of this assertion, when the whole spectrum lies in the left half-plane, to the case when the coefficient A is unbounded and satisfies the sectorial condition [4–8]. Thiscondition means that the spectrum lies in some sector in the left half-plane, and the resolvent satisfiessome growth estimate at infinity.

In this paper we consider the case of a bisectorial pencil, i.e., the case, when the spectrum belongsto two sectors which lie in the left and right half-planes, correspondingly (see the left Fig. 1a)). Thenotion of a bisectorial pencil (in the original terminology, a bisemigroup) λ �→ λ1−A (here the symbol 1denotes the identity operator) generated by an unbounded operator A was introduced in [9] and studiedin [10]. Here we consider a more general bisectorial pencil λ �→ λF − G. We assume that operators Fand G act from a Banach space X to a Banach space Y and are bounded; usually one can reduce thecase of unbounded F and G to the case of bounded ones [11].

Let us mention two specific features of the problem under consideration. In the case of a sectorialpencil and the equation u′ −Au = f , when the coefficient A acts from its domain of definition D(A) ⊂ Xto X, the corresponding semigroup of operators T (t), t ≥ 0, acts from X to X (rather than to D(A)).

That is why values of the solution u(t) =+∞∫

0

T (s)f(t − s) ds belong to X (rather than to D(A)). In the

case of a bisectorial pencil λ �→ λF − G and the equation Fu′ − Gu = f , the space X, where F and Gare defined, appears to be an analog of D(A); therefore, if values of the function f belong to Y , thenthe range of the solution is wider than X. For the equation Fu′ − Gu = f the determination of a widersuitable space containing X is a separate problem. In this connection we consider here functions fwhose values belong to some subspace Y 1 of the space Y . This indirectly means that X turns into aspace that includes the domain of definition of F and G. Auxiliary considerations (see Example 2 inSection 4) allow us to apply the theorem proved in this paper even if values of f run over the wholespace Y ; in this case the set of values of u is wider than X.

The next specific feature is the following property. If for ordinary differential equations and thosewith an unbounded sectorial operator the Green function t �→ G(t) has a discontinuity of the first kind atzero, then for an arbitrary bisectorial pencil there occurs an integrable discontinuity of the second kind(Proposition 5).

The subject studied in [12] is most close to that of our paper. One proves in [12] that if there existsan integrable Green function, then with any bounded right-hand side there exists a unique bounded

*E-mail: [email protected].

26

BISECTORIAL OPERATOR PENCILS AND THE PROBLEM OF BOUNDED SOLUTIONS 27

solution; however, no conditions for the existence of such a function are obtained. Close questions werealso studied in [13]; however, as distinct from this paper, in [13] the part of the spectrum that lies in heright half-plane is not necessarily bounded.

Let us mention one more class of problems [4–7] and [13–15] among considered ones, namely, thecase when the point at infinity is a pole of the resolvent of the pencil or, equivalently, the semigroup orbisemigroup generated by the pencil is degenerate. In this paper we do not study this case.

In Section 1 we define the notion of a bisectorial pencil. In Section 2 we study properties of the Greenfunction. In Section 3 we prove the unique existence of a bounded on the axis solution to the differentialequation. Finally, in Section 4 we give some examples.

1. Y 1-BISECTORIAL PENCILS

Let X and Y be Banach spaces. The symbol B(X,Y ) denotes the space of all linear boundedoperators from X to Y . Let F,G ∈ B(X,Y ). We understand a (linear) pencil as a function λ �→λF − G, λ ∈ C. We understand the resolvent set of a pencil as the set ρ(F,G) consisting of all λ ∈ C

with which the operator λF − G is invertible, and we do the resolvent of the pencil as the function (thefamily) Rλ = (λF − G)−1, λ ∈ ρ(F,G). The complement σ(F,G) = C \ ρ(F,G) to the resolvent set iscalled the spectrum of the pencil.

We call a pencil λ �→ λF − G (cf. [10], P. 16) bisectorial if there exist δ0 ∈ (0, π/2] and h0 > 0 suchthat the set of complex numbers (see Fig. 1a))

Ωδ0,h0 ={

λ : −π

2− δ0 < arg λ <

π

2+ δ0,

π

2− δ0 < arg λ < −3π

2+ δ0

}

∪ {λ : |Re λ| < h0}

belongs to the resolvent set of the pencil, and for each δ ∈ (0, δ0) and h ∈ (0, h0) there exist M ∈ R andm ∈ Z such that

‖(λF − G)−1‖Y →X ≤ M(1 + |λ|)m, λ ∈ Ωδ,h. (1)

Assume that the space Y has a linear subspace Y 1 complete with respect to its norm ‖ · ‖1 such that‖y‖ ≤ ‖y‖1 for y ∈ Y 1. Evidently, ‖T‖Y 1→X ≤ ‖T‖Y →X for any linear bounded operator T : Y → X.As the simplest example, we put Y 1 = Y ; in Section 4 we give a more complex example.

A pencil λ �→ λF − G is called Y 1-bisectorial if there exist δ0 ∈ (0, π/2] and h0 > 0 such that theset Ωδ0,h0 belongs to the resolvent set of the pencil, and for each δ ∈ (0, δ0) and h ∈ (0, h0) there existsM ∈ R satisfying the bound

‖(λF − G)−1‖Y 1→X ≤ M

1 + |λ| , λ ∈ Ωδ,h. (2)

In what follows we consider a fixed Y 1-bisectorial pencil λ �→ λF − G.

Remark. In the statement of the main theorem of this paper the space Y is not mentioned. Evidently,everywhere in proofs one can replace inequality (1) with that (2). Nevertheless, we keep inequality (1),because it guarantees the existence of the Green function G(t) on a wider space Y .

For each function f analytic in a neighborhood of some set �δ,h = C \ Ωδ,h, where δ ∈ (0, δ0) andh ∈ (0, h0), we set (assuming that the integral converges)

ϕ(f) =1

2πi

∫

Γδ,h

f(λ)(λF − G)−1 dλ;

here the contour Γδ,h is the boundary of the set �δ,h, its orientation is shown in Fig. 1b). Note that Γδ,h

consists of two parts Γ+δ,h and Γ−

δ,h.

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 56 No. 3 2012

28 PECHKUROV

Fig. 1. a) the set Ωδ0,h0 ; b) boundaries of sets Ωδ0,h0 and Ωδ,h (the arrows indicate the orientation); c) the correctedcurve Γ+

δ,h.

2. THE GREEN FUNCTION

Consider functions

exp+t (λ) =

{eλt, if Re λ < 0;0, if Re λ > 0,

exp−t (λ) =

{0, if Reλ < 0;eλt, if Reλ > 0,

gt(λ) =

{exp+

t (λ), if t > 0;− exp−

t (λ), if t < 0.

Let

X+(t) = ϕ(exp+t ), t > 0,

X−(t) = ϕ(exp−t ), t < 0,

G(t) = ϕ(gt), t = 0.

We call G the Green function. According to formula (4) below, G is really the Green function for theproblem of bounded solutions.

Proposition 1. Integrals

X±(t) =1

2πi

∫

Γδ,h

exp±t (λ)(λF − G)−1 dλ

are independent of the choice of the contour Γδ,h with δ ∈ (0, δ0) and h ∈ (0, h0).

The proof follows from the exponential decrease of functions exp±t on Γδ,h.

Proposition 2. With t = 0 the function t �→ G(t) is differentiable with respect to the norm of thespace B(Y,X) (and, consequently, with respect to the norm of the space B(Y 1,X)) and satisfiesthe differential equation F G(t) − GG(t) = 0.

Proof. Let us show that functions X±(t) = ϕ(exp±t ) with t > 0 and t < 0, correspondingly, satisfy the

equation FX(t) − GX(t) = 0. For the derivatives we have the representation

X±(t) = limΔt→0

12πi

∫

Γ±δ,h

eλΔt − 1Δt

eλt(λF − G)−1 dλ =1

2πi

∫

Γ±δ,h

λeλt(λF − G)−1 dλ.

(The passage to the limit under the integral sign is possible, because the latter integral convergesuniformly with respect to t ∈ K for any compact set K ⊂ R \ {0}.) Then



FX±(t) − GX±(t) = F1

2πi

∫

Γ±δ,h

λ exp±t (λ)(λF − G)−1dλ − G

12πi

∫

Γ±δ,h

exp±t (λ)(λF − G)−1dλ

=1

2πi

∫

Γ±δ,h

exp±t (λ)(λF − G)(λF − G)−1 dλ =

12πi

∫

Γ±δ,h

exp±t (λ) dλ = 0.

The latter equality follows from the Cauchy theorem.

Lemma. For λ ∈ ρ(F,G) \ {0} the following identity is valid (here 1 is the identity operator):

F (λF − G)−1 =1λ1 +

1λ

G(λF − G)−1.

The proof consists in the direct verification of the desired assertion.

Proposition 3. The product FG(t) : Y 1 → Y has a limit with respect to the norm of the spaceB(Y 1, Y ) as t → ±0, and

limt→+0

FG(t) − limt→−0

FG(t) = 1.

Proof. In accordance with the Jordan lemma ([16], P. 436) we transform the contour Γ+δ,h (Fig. 1b)) in

the integral (hereinafter (λF − G)−1 : Y 1 → X and F,G : X → Y )

FG(t) = F1

2πi

∫

Γδ,h

gt(λ)(λF − G)−1 dλ =1

2πi

∫

Γ+δ,h

eλtF (λF − G)−1 dλ,

which defines FG(t) for t > 0, to the vertical line

FG(t) =1

2πi

∫ +i∞

−i∞eλtF (λF − G)−1 dλ.

In accordance with the above lemma we reduce this integral to the form (tracing to the left of zero)1)

FG(t) =1

2πi

∫ −0+i∞

−0−i∞eλtF (λF − G)−1 dλ

=1

2πi

∫ −0+i∞

−0−i∞eλt 1

λ1 dλ +

12πi

∫ −0+i∞

−0−i∞eλt 1

λG(λF − G)−1 dλ.

In view of the Jordan lemma and the Cauchy theorem the first integral vanishes with t > 0. Therefore,

FG(t) =1

2πi

∫ −0+i∞

−0−i∞eλt 1

λG(λF − G)−1 dλ.

In this representation we proceed to the limit as t → +0 (this is possible, because due to bound (2) theintegral converges uniformly):

FG(+0) =1

2πi

∫ −0+i∞

−0−i∞

1λ

G(λF − G)−1 dλ.

Analogously we obtain

FG(−0) =1

2πi

∫ +0+i∞

+0−i∞

1λ

G(λF − G)−1 dλ,

where the zero point is located to the left of the vertical line.

1)Since the residue vanishes at zero, the direction of bypass does not matter.


30 PECHKUROV

Subtracting integrals that define FG(+0) and FG(−0), we obtain an integral over a circumference γof a small radius that encircles zero and is traced clockwise:

FG(+0) − FG(−0) =1

2πi

∫

γ

1λ

G(λF − G)−1 dλ.

Applying the residue theory, we obtain FG(+0) − FG(−0) = −G(0F − G)−1 = GG−1 = 1.

Proposition 4. The function G and its derivative decrease exponentially as t → ∞ in the normof the space B(Y,X) (and, consequently, in the norm of the space B(Y 1,X)), i.e., there exist Mand γ > 0 such that with sufficiently large |t| it holds

‖G(t)‖Y →X + ‖G(t)‖Y →X ≤ Me−γ|t|.

Proof. By definition, X+ is representable in the form

X+(t) =1

2πi

∫

Γ+δ,h

eλt(λF − G)−1 dλ, t > 0.

We transform the curve Γ+δ,h as is shown in Fig. 1c), namely, for Γ+

δ,h we choose two rays arg(λ − a) =±

(π2 + δ

), where a ∈ (−h, 0), and δ > 0 is sufficiently small.

We parameterize the upgoing ray Γ++δ,h of the curve Γ+

δ,h as follows: λ = a − s sin δ + is cos δ, s ∈[0,+∞). Let us estimate the integral along this ray with t ≥ 1. We have

∥∥∥∥

12πi

∫

Γ++δ,h

eλt(λF − G)−1 dλ

∥∥∥∥

Y 1→X

≤ 12π

∫ +∞

0e(a−s sin δ)tM

(1 + |a + sei(π/2+δ)|

)mds

≤ M1eat

∫ +∞

0e−ts sin δ(1 + s)m ds ≤ M1e

at

∫ +∞

0e−s sin δ(1 + s)m ds.

The integral at the right, evidently, converges. It remains to recall that a < 0.One can estimate the rest integrals analogously.

Proposition 5. The function t �→ G(t) : Y 1 → X is integrable at zero.

Proof. Let t > 0. Then

G(t) =1

2πi

∫

Γ+δ,h

eλt(λF − G)−1 dλ,

where Γ+δ,h is the left boundary of the set Ωδ,h (see Fig. 1c)). Let us represent Γ+

δ,h as the union of two rays

going to infinity. We parameterize the upgoing ray Γ++δ,h as follows: λ = −s sin δ + is cos δ, s ∈ [0,+∞).

Let us estimate the integral along Γ++δ,h with the help of bound (2):

∥∥∥∥

12πi

∫

Γ++δ,h


∥∥∥∥

Y 1→X

≤ 12π

∫ +∞

0

Me−ts sin δ

1 + |sei(π/2+δ)|ds = M1

∫ +∞

0

e−ts sin δ ds

1 + s.

We change the variable ts = σ and obtain

M1

∫ +∞

0e−ts sin δ ds

1 + s= M1

∫ +∞

0e−σ sin δ dσ

t + σ≤ M1

(∫ +∞

1e−σ sin δ dσ

σ+

∫ 1

0

dσ

t + σ

)

.

The first integral is independent of t, and the second one is calculated as follows:1∫

0

dσt+σ = ln(1 + t)− ln t.

Therefore, for t close to zero we finally obtain the following summable bound:∥∥∥∥

12πi

∫

Γ++δ,h


∥∥∥∥

Y 1→X

≤ M2 + M3| ln t|.

One can estimate analogously the rest integrals that define t �→ G(t).



3. THE EXISTENCE OF BOUNDED SOLUTIONS

Let the symbol C = C(R, Y ) stand for the space of all bounded continuous functions f : R → Y .Analogously, let the symbol C1 = C1(R,X) denote the space of all differentiable functions u : R → Xbounded and continuous together with their derivative.

Theorem. Let a pencil be Y 1-bisectorial. Then for any f ∈ C(R, Y 1) the equation(Fu

)′(t) − Gu(t) = f(t), t ∈ R, (3)

has a unique solution u ∈ C1(R,X) representable in the form

u(t) =∫ +∞

−∞G(t − s)f(s) ds, t ∈ R. (4)

Proof. Let us prove that function (4) satisfies Eq. (3) (since the function f belongs to C(R, Y 1), andthe kernel G(·) is integrable (Propositions refp:exp decreasing:bi and 5), this integral converges). UsingProposition 2 and rules of differentiation of an improper integral depending on a parameter, we obtain

(∫ t

−∞FG(t − s)f(s) ds

)′= FG(+0)f(t) +

∫ t

−∞

(FG(t − s)

)′tf(s) ds

= FG(+0)f(t) +∫ t

−∞GG(t − s)f(s) ds = FG(+0)f(t) + G

∫ t

−∞G(t − s)f(s) ds,

(∫ +∞

tFG(t − s)f(s) ds

)′= −FG(−0)f(t) +

∫ +∞

t

(FG(t − s)

)′tf(s) ds

= −FG(−0)f(t) +∫ +∞

tGG(t − s)f(s) ds = −FG(−0)f(t) + G

∫ +∞

tG(t − s)f(s) ds.

Hence

(Fu

)′(t) =(∫ t

−∞FG(t − s)f(s) ds

)′+

(∫ +∞

tFG(t − s)f(s) ds

)′

=(FG(+0) − FG(−0)

)f(t) + G

( ∫ t

−∞G(t − s)f(s) ds +

∫ +∞

tG(t − s)f(s) − ds

)

=(FG(+0) − FG(−0)

)f(t) + Gu(t).

Evidently,(Fu

)′(t) − Gu(t) =(FG(+0) − FG(−0)

)f(t). It remains to apply Proposition 3.

Let us prove the uniqueness. Let u ∈ C1 be a solution to the equation Fu − Gu = 0. In thisequation we perform the Fourier transform (in the sense of generalized functions [17, 18]). We getiωF u(ω)−Gu(ω) = 0 or (iωF −G)u(ω) = 0, where u is the Fourier transform of the function u. Sincethe pencil is bisectorial, the operator iωF − G is invertible with all ω ∈ R. Consequently, u = 0. Thenu = 0. See [19] for details.

4. EXAMPLES

Let the symbol C2π stand for the Banach space of all continuous 2π-periodic functions x : R → C

with the norm ‖x‖ = supt

|x(t)|, and let the symbol C2π 0 denote the closed subspace of functions

x ∈ C2π such that2π∫

0

x(t)dt = 0. We denote by C12π 0 the subspace of functions x ∈ C2π 0 that belong to

C2π 0 together with their derivative, where the norm is defined as follows: ‖x‖1 = ‖x‖C2π 0 + ‖x′‖C2π 0 .Consider the operator D : C1

2π 0 → C2π 0 defined by the formula

Dx = ix′.


32 PECHKUROV

Evidently, operators D, 1 : C12π 0 → C2π 0 are bounded, and D is an isomorphism. One can easily see

that the spectrum of the operator D considered as an operator from C2π 0 to C2π 0 with the domainC1

2π 0 ⊂ C2π 0 coincides with the set Z \ {0}.One can easily make sure that the resolvent of the operator D : C2π 0 → C1

2π 0 is representable in theform

((λ1 − D)−1g

)(t) = −1

i

∫ 2π

0

e−iλs

1 − e−2πiλg(t − s) ds = −1

i

∫ t

t−2π

e−iλ(t−s)

1 − e−2πiλg(s) ds.

Therefore

‖(λ1 − D)−1‖C2π 0→C2π 0 ≤ ‖(λ1 − D)−1‖C2π→C2π ≤∫ 2π

0

|e−iλs||1 − e−2πiλ| ds

=∫ 2π

0

es Im λ

|1 − e2π Im λe−2πi Re λ| ds =|e2π Im λ − 1|

| Im λ| · |1 − e2π Im λe−2πi Re λ| .

Hence we can easily deduce that the resolvent allows the bound

‖(λ1 − D)−1‖C2π 0→C2π 0 ≤ M

1 + |λ| , λ ∈ Ωδ,h, (5)

with arbitrary δ ∈ (0, π/2) and certain M and h > 0.According to bound (5) and the identity D(λ1 − D)−1 = λ(λ1 − D)−1 − 1, we have

‖D(λ1 − D)−1‖C2π 0→C2π 0 ≤ M |λ|1 + |λ| + 1 ≤ M1, λ ∈ Ωδ,h. (6)

Consequently,

‖(λ1 − D)−1‖C2π 0→C12π 0

≤ M2, λ ∈ Ωδ,h. (7)

Example 1. Consider the equation

u′(t) −(Du

)(t) = f(t), (8)

where f : R → C2π 0, with respect to the function u : R → C2π 0.Let Y = C2π 0 and X = Y 1 = C1

2π 0. Bounds (7) and (5) mean that bounds (1) and (2) are valid,though bound (1) holds true with m = 0. Consequently, in view of the main theorem of this paper,Eq. (8) with any f ∈ C(R, C1

2π 0) has a unique solution x ∈ C1(R, C12π 0). This example illustrates the

bisectorial property in the simplest case.

Example 2. Let us discuss the way for obtaining (by means of additional considerations) with the helpof the main theorem) solutions that correspond to f ∈ C(R, C2π 0). Let the symbol 2) C−1

2π 0 stand forone more copy of the space C2π 0 with the same norm. In order to distinguish between elements of thespace C−1

2π 0 and those of the space C2π 0, we use denotations like y(1); thus, the correspondence y �→ y(1)

represents an isometric isomorphism of C2π 0 onto C−12π 0. Let us define an embedding 11 : C2π 0 → C−1

2π 0

by the rule x �→ (D−1x)(1). Then, as is seen from the commutative diagram,

C12π 0

D−−−−→ C2π 0⏐⏐�1

⏐⏐�11

C2π 0D1−−−−→ C−1

2π 0;

here 1 : C12π 0 → C2π 0 means the canonical embedding, the mapping y �→ y(1) appears to be an ex-

tension D1 : C2π 0 → C−12π 0 of the mapping D : C1

2π 0 → C2π 0, and the inverse one D−11 : C−1

2π 0 → C2π 0

obeys the rule y(1) �→ y. Let us mention once again that D1 : C2π 0 → C−12π 0 is an isometric isomorphism.

2)One can treat elements of the space C−12π 0 as generalized derivatives of functions x ∈ C−1

2π 0 (with the multiplier −i).



In accordance with accepted definitions, the operator λ11 − D1 : C2π 0 → C−12π 0 is defined by the rule

x �→ λ(D−1x)(1) − x(1) = (λD−1x − x)(1) =((λ1 − D)D−1x

)(1).

In this rule we put z = (λ1 − D)D−1x, where x, z ∈ C2π 0, or, equivalently, x = D(λ1 − D)−1z. Asa result we obtain the representation D(λ1 − D)−1z �→ z(1), whence we see that the operator (λ11 −D1)−1 : C−1

2π 0 → C2π 0 is defined by the rule

z(1) �→ D(λ1− D)−1z. (9)

Hence

‖(λ11 − D1)−1‖C−12π 0→C2π 0

≤ ‖(λ1 − D)−1‖C2π 0→C12π 0

· ‖D‖C12π 0→C2π 0

.

Therefore, taking into account bound (7), we obtain ‖(λ11 − D1)−1‖C−12π 0→C2π 0

≤ M3 with λ ∈ Ωδ,h.

By applying formula (9) we represent the operator 11(λ11 − D1)−1 = (λ11 − D1)−1 : C−12π 0 → C−1

2π 0 in

the form z(1) �→((λ1− D)−1z

)(1), which gives the bound

‖(λ11 − D1)−1‖C−12π 0→C−1

2π 0≤ ‖(λ1 − D)−1‖C2π 0→C1

2π 0.

Hence with the help of formula (5) we get

‖(λ11 − D1)−1‖C−12π 0→C−1

2π 0≤ M

1 + |λ| , λ ∈ Ωδ,h.

Let Y = C−12π 0 and X = Y 1 = C2π 0. The above bounds imply those (1) and (2), though bound (1) is

true with m = 0. Consequently, according to the main theorem, Eq. (8) with any f ∈ C(R, C2π 0) hasa unique solution x ∈ C1(R, C2π 0).

Example 3. Consider the equation(D−1u

)′(t) − u(t) = f(t), (10)

where the operator F = 1 and the function f : R → C2π 0, with respect to the function u : R → C2π 0.The operator D is assumed to be the same. The representation (λD−1 − 1)−1 =

(D−1(λ1 − D)−1

)=

D(λ1 − D)−1 and formula (6) give

‖(λD−1 − 1)−1‖C2π 0→C2π 0 ≤ M4, λ ∈ Ωδ,h.

At the same time, formula (5) and the identity (λD−1 − 1)−1 = (λ1 − D)−1D imply that

‖(λD−1 − 1)−1‖C12π 0→C2π 0

= ‖(λ1 − D)−1D‖C12π 0→C2π 0

≤ ‖(λ1 − D)−1‖C2π 0→C2π 0 · ‖D‖C12π 0→C2π 0

≤ M5

1 + |λ| , λ ∈ Ωδ,h.

Let X = Y = C2π 0 and Y 1 = C12π 0. The latter inequalities mean that bounds (1) and (2) take place,

though bound (1) is true with m = 0. Consequently, by the theorem, Eq. (10) with any f ∈ C(R, C12π 0)

has a unique solution x ∈ C1(R, C2π 0).

Example 4. Let us briefly describe (for reasons of space) the scheme of construction of a more complexexample of an equation having the bisectorial property. Consider the following system of differentialequations:

B

(

− i∂

∂x

)∂u(t, x)

∂t− A

(

− i∂

∂x

)

u(t, x) = f(t, x), t ∈ R, x ∈ Rn,

where A,B : Cn → B(Ck, Ck) are polynomials, ∂

∂x = ( ∂∂x1

, . . . , ∂∂xn

), and u is a function with values

in Ck.


34 PECHKUROV

For Y we consider the Hilbert space L2 = L2(Rn, Ck) of all square summable functions defined onR

n with values in Ck, and for X we do the subspace L2 consisting of functions v for which B

(− i ∂

∂x

)v

and A(− i ∂

∂x

)v (understood in the sense of generalized functions) belong to L2.

We assume that the polynomial B is subordinate to the polynomial A, i.e., the degree l of thepolynomial A is strictly greater than the degree of the polynomial B. In addition, we assume that thepolynomial A is elliptic, i.e., the sum Al(ω) of monomials in the polynomial A, whose degree exactlyequals l, is invertible for all ω ∈ R

n with ‖ω‖ = 1. In this case the space X appears to be the Sobolevspace W l

2.Evidently, we can write the considered equation in the operator form (3) and understand its solution

in the corresponding sense.

Let us apply the Fourier transform to the pencil λB(− i ∂

∂x

)− A

(− i ∂

∂x

), i.e., proceed to the pencil

Cλ = F(λB

(− i ∂

∂x

)− A

(− i ∂

∂x

))F−1; here F is the Fourier transform in L2. We can immediately

verify that Cλ is the multiplication operator(Cλv

)(ω) =

(λB(ω) − A(ω)

)v(ω), ω ∈ R

n.

Since the polynomial B is subordinate to the polynomial A, the coefficient c(ω) = λB(ω) − A(ω)(with any λ ∈ C) as ω → ∞ behaves in the same way as the function ω �→ A(ω). Therefore the necessaryand sufficient condition for the invertibility of the operator Cλ is the invertibility of the coefficient c(ω) =λB(ω) − A(ω) with all ω ∈ R

n. Therefore, the spectrum of the pencil λ �→ λB(− i ∂

∂x

)− A

(− i ∂

∂x

)

coincides with the set of λ ∈ C, for which the coefficient c(ω) = λB(ω)−A(ω) ∈ B(Ck) is not invertiblefor at least one ω ∈ R

n. In other words, the spectrum of the pencil λ �→ λB(− i ∂

∂x

)− A

(− i ∂

∂x

)

coincides with the set of λ ∈ C which solve the equation

det(λB(ω) − A(ω)

)= 0

with at least one ω ∈ Rn. Thus, the spectrum of the pencil is calculated efficiently.

Under natural constraints on A and B one can estimate the growth rate of the resolvent of the pencilat infinity and thus verify the bisectorial property.

Let us mention one simplest particular case. Let k = 2; assume that the operator B(− i ∂

∂x

)is

identical and A(− i ∂

∂x

)is defined by the matrix

⎛

⎝−Δ + 1 0

0 Δ − 1

⎞

⎠ ,

where Δ is the Laplace operator. Then, evidently, the pencil is bisectorial. This trivial example showsthat in the considered class of equations the bisectorial property is possible.

ACKNOWLEDGMENTS

This work was supported by the Russian Foundation for Basic Research, grant No. 10-01-000276.

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Translated by O. A. Kashina


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Bisectorial operator pencils and the problem of bounded solutions