Operator Pencils and Half-Range Problem in Operator Theory

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1 Operator pencils and Cauchy problem. The ﬁnite dimensionalcase 3

2 Theorem on holomorphic operator function 7

3 New proof of the completeness theorem in ﬁnite dimensional case. Representation of the resolvent as a meromorphic function of ﬁnite order growth 14

4 Keldysh-Lidskii theorem on the completeness 28

5 Half-range minimality and completeness problems for dissipative pencils 36

6 Mandelstam radiation principle (non-resonant case) and half-range problems 43

7 Generalized Mandelstam radiation principle (resonant case). The factorization of a quadratic pencil 52

8 Dissipativeand linearlydissipative operatorpencils 65

9 Factorizationofdissipativeoperatorpencils 82

10 Pontrjagin spaces. The proof of Azizov-Iohvidov-Langer theorem 91 10.1 Pontrjagin theorem and the formulation of Azizov-Iohvidov-Langer theorem...... 91 10.2Example...... 93 10.3 Criteria for to be Pontrjagin subspace in Pκ...... 95 10.4Rieszbasistheorem..L ...... 97

11 Operator pencils arising in elasticity and hydrodynamics: The instability index formula 106 11.1 Classesofunboundedoperatorpencils ...... 107 11.2 Root subspaces of linear dissipative pencils and their properties . . . 110 11.3 Quadratic dissipative pencils and the instability indexformula . . . . 120 11.4Applications ...... 122 2

12 Factorization of elliptic pencils and the Mandelstam hypothesis 128 12.1 Ellipticpencilsandtheirspectrum ...... 131 12.2 The real spectrum of a strongly elliptic pencil ...... 137 12.3 Factorization of positive strongly elliptic pencils ...... 141 12.4 Elliptic pencils satisfying the Keldysh-Agmon condition ...... 147 12.5 Theresolventgrowthcondition ...... 147 12.6 Half-range completeness and minimality ...... 148 12.7Factorization...... 151 12.8 TheMandelstamhypothethis...... 153 12.9 Application to the Lame system of the elasticity theory ...... 155

13 Scattering of waves by periodic gratings and factorization problems 165 13.1Introduction ...... 165 13.2 Scattered Waves and the Radiation Condition for the Two- dimensionalHelmholtzEquation ...... 167 13.3 Factorization of elliptic operator pencils and solvability of the corre- spondingequationsonthesemi-axis ...... 169 13.4 Existence and Uniqueness of the Solution of the Plane Scattering Problem ...... 175 13.5 Scattering by Two-periodic Surfaces in M3...... 181

14 On the stability of a top with a cavity ﬁlled with a viscous ﬂuid 189 14.1Introduction ...... 189 14.2 Analysis of the Operator Equation Corresponding to the Linear Equa- tionsofEvolutionoftheSystem ...... 193 14.3 A Stability Criterion and the Instability Index ...... 197 14.4 Further Properties of the Evolution Operator T of the System. The SpectrumatLargeViscosity ...... 201 14.5 TheSymmetricTop ...... 206 3 Part 1

1 Operator pencils and Cauchy problem. The ﬁ- nite dimensional case

Among key-stones which ly in the base of the topic represented in this article I would like to mention three ones. The first one is due L. S. Pontrjagin [P], who proved (1944) the existence of maximal semi-definite invariant subspaces for self- adjoint (and later for dissipative) operators in Pontrjagin space. The second one is due M. V. Keldysh [K1, K2], who began to investigate(1951) spectral properties of operator pencils and suggested an analytic approach to prove the completeness properties of the root functions of wide class of non-self-adjoint operators and operator pencils. The third one is due to M. G. Krein and H. Langer [KL] who proved (1964) the first factorization theorem for self-adjoint quadratic operator pencils in the infinite dimensional case using a generalization of Pontrjagin theorem. Since that time many mathematicians were involved in the investigation of related problems. Here we shall present some our points of view to this subject. Let us come to the subject: we deal with operator polynomials

A(λ)= A + λA + + λnA (1) 0 1 · · · n where Aj, j = 0, 1,...,n are operators in Hilbert space H. Further, we will consider the different situations: H is finite dimensional; H is infinite dimensional and Aj are bounded operators; Aj are unbounded operators (in this case H has to be infinite dimensional). I would like to mention two origins for the study of operator pencils. First the algebraic origin, by which I mean the theory of matrices. This is under- standable, because the eigenvalue problem for operator pencil A(λ) is a generalization of the eigenvalue problem for monic linear operator polynomials A(λ)= A λI, i.e. the classical eigenvalue problem for matrix A. − Second the Fourier method for solving equations with operator coefficients of the form d du dnu A( i )u(t)= A u iA + +( i)nA =0, (2) − dt 0 − 1 dt · · · − n dtn where u(t) is the function (determined for example for t > 0) with values in Hilbert space H. 4

Certainly, the problem of solvability of such equations and the problem of stability of their solutions are closely connected with some problems on factorization of operator pencil, with problems of its eigenvalue distribution, and with problems on basis properties, completeness and minimality of its eigenfunctions (the latter concept can be considered in different senses: for example, n-multiple completeness and half range completeness). Now we clarify the connection between operator pencil (1) and differential equation (2). First we can consider the operator pencil A(λ) as the characteristic polynomial of the differential equation to arrive at A(λ)y = 0 as the characteristic 0 equation. We say y = is an eigenvector corresponding to eigenvalue λ0 of the 6 ∅ 0 1 p operator pencil A(λ) if A(λ0)y =0. We say also, that y ,...,y is the sequence of vectors associated with y0 if

s 1 s 1 1 (s) 0 A(λ )y + A′(λ )y − + + A (λ )y =0 0 1! 0 · · · s! 0 for all s = 1,...,p. The sequence y0, y1,...,yp we call the chain of eigen and associated vectors (EAV) corresponding to eigenvalue λ0 and we say p +1 is the length of that chain. A simple veriﬁcation shows that if y0, y1,...,yp is a chain of EAV corresponding to eigenvalue λ0 then for all s =0, 1,...,p the functions

s s iλ t s it s 1 (it) 0 u (t)= e 0 [y + y − + + y ] 1! · · · s! are solutions of diﬀerential equation (2). These functions we call elementary solutions of the equation (2). Now assume that dim Ker A(λ )= ℓ< . Let y0,...,y0 be a basis in subspace 0 ∞ 1 ℓ Ker A(λ0) and 0 1 pk yk, yk,...,yk , j =1,...,ℓ (3) are the chains of EAV corresponding to λ0 of the maximum possible length pk +1. The number m =(p1 +1)+ +(pℓ +1) certainly depends on the choice of the basis y0 . If m< for any choice· · · of y0 then there exists a basis y0 such that the { k} ∞ { k} { k} corresponding number m has the maximum possible value, say m0. That number m0 is called the algebraic multiplicity f the eigenvalue λ0 and the chains (3) consisting of m0 elements numbered in such a way that p1 > p2 > > pℓ, are called a canonical system of eigen and associated elements. Certainly canonical· · · system is not unique. The number ℓ = dimKer A(λ0) is called the geometric multiplicity of the eigenvalue λ0. 5

Note, if A(λ)= A λI then the canonical system of EAV (3) coincides with Jor- − dan chains corresponding to eigenvalue λ0 (see [Lancaster and Tismenetsky, §6.4]). Note also that if the operator polynomial A(λ) is not linear then one can not assert s the linear independence of the elements of the system (3). Some of elements yk for 1 6 s 6 pk may even be equal to zero. Let us try to ﬁnd the solution of the equation (2) satisfying the initial conditions

( i)ju(j)(0) = ϕ , j =0, 1,...,n 1. (4) − j − The problem (2), (4) is called the Cauchy problem. Following the Fourier method we try to ﬁnd its solution in the form

s s iλkt s it s 1 (it) 0 s u(t)= c e (y + y − + + y )= c u (t) (5) k k 1! k · · · s! k k k,s Xs,k Xs,k s s where the ck are unknown coeﬃcients, yk, s = 0, 1,...,pk, are the elements of canonical systems (3) corresponding to all eigenvalues λk of the pencil A(λ). here we avoid the introduction of the third index if addition to s and k and assume that canonical system (3) corresponds to eigenvalue λk (instead of λ0) and λk is repeated in the sum (5) as many times as its geometric multiplicity. Using (5) we can rewrite the initial conditions (4) in the form

s,0 ϕ0 yk s,1 ϕ1 s yk s s   = ck   = ckyk, (6) · · · s,k s,n· · · 1 s,k ϕn 1 X y −  X  −   k      e where for r =0, 1,...,n 1 − dr (it)s ys,r =( i)ru(r)(0) = ( i)r [eiλkt(ys + + y0)] = k − k,s − dtr k · · · s! k t=0 r s d λkt s t s 1 t 0 = [e (y + y − + + y )] dtr k 1! k · · · s! k t=0 The elements ys Hn = H H H are called the Keldysh derived chains k ∈ × ×···× constructed from canonical system (3). If the eigenvalue λk is semi-simple (this is the case when theree are no associated vectors) then the Keldysh derived chains have the representation n 1 yk =(yk, λkyk,...,λk− yk).

e 6

Let us assume that dim H < and let det A(λ) 0. In this case the pencil A(λ) has a ﬁnite number of eigenvalues∞ and to establish6≡ the Fourier method for the Cauchy problem (2), (4) we have to show that the system of Keldysh derived chains ys is a basis in Hn (then the Cauchy problem will be solvable for any set of initial { k} vectors ϕ0,...,ϕn 1). − Theoreme 1.1. Let dim H < and det A(λ) 0. Then a system of Keldysh s n∞ 6≡ derived chains yk is basis in H if and only if Ker An =0. Proof. With pencil A(λ) we associate the following linear pencil in space Hn e A (λ)= A λA (7) 0 − 1 where 0 0 ... 0 An A0 A1 ... An 1 − − I 0 ... 0 0 A 0 I... 0 A   0 =  . . . .  , 1 = 0 I... 0 0 ...... ......   0 0 ...I      0 0 ...I 0      A simple veriﬁcation (see for example [Keldysh 1], [Markus 1]) shows that the EAV (3) are the chains of eigen and associated vectors corresponding to eigenvalue λk of the pencil A(λ) if and only if the Keldysh derived chains

0 1 pk yk, yk,..., yk (8) are EAV of the linear pencil (7) or linear operator A 1A acting in the space Hn e e e 1− 0 (note that A1 is invertible if Ker An = 0). But the system of EAV of any linear operator in ﬁnite dimensional space is a basis. Hence the system of Keldysh derived chains is a basis in Hn. s To show that yk is not a basis if Ker An = 0 we can assume without loss of generality that A{ is} invertible. (Otherwise we6 have to shift λ λ + λ , where 0 → 0 λ0 is a point suche that A(λ0) is invertible. We can ﬁnd such a point because det A(λ) 0.) Then the system of EAV for pencil (7) coincides with a system 6≡ A 1A A 1A of EAV for I λ 0− 1 and in turn coincides with EAV for operator 0− 1 with the exception− of a canonical system corresponding to the eigenvalue µ = 0. A 1A The operator 0− 1 is singular, hence the algebraic multiplicity of the eigenvalue µ = 0 is equal to k > 0. Then the system Keldysh derived chains has defect k. Theorem 1 is proved. 7 2 Theorem on holomorphic operator function

In attempting to generalize theorem 1 to infinite-dimensional spaces one comes up against some deep problems. Under the assumption that the spectrum of the pencil A(λ) is descrete we will show the minimality of Keldysh derived chains in the space Hn. Under some reasonable additional assumptions we will sketch the proof of its completeness. But the basis property, as a rule, does not hold. Even for the simple pencil A(λ)= I λ2C, where C is a self-adjoint positive compactoperator in H the Keldysh derived− chains do not form a basis in H2. Nevertheless, for some pencils it is possible to find the space H which is embedded in Hn and such that the system s H yk consisting of Keldysh derived chains has the basis property in . Some results of{ this} kind may be found in the recent paper [Shkalikov 1, §2.3]. They are based one eigenexpansion theorems for p-subordinate linear operators due to V. Kaznelson, A. Markus and V. Matsaev (see [Markus 1], for example). For convenience we give the definitions ot the concepts which we have mentioned.

Deﬁnition 2.1. The system yk 1∞ is complete in Hilbert space H if from the equalities { } (yk,x)=0, k =1, 2,..., it follows that x =0.

Exercise 2.1. (see [A. Kolmogorov and S.Fomin 1]). The system yk 1∞ in separable Hilbert space H is complete if and only if it is dense in H, i.e.{ for} any x H ∈ and for any ε > 0 there exists a linear combination YN = c1y1 + + cN yN such that Y x <ε. · · · k N − k Deﬁnition 2.2. The system y ∞ is minimal in Hilbert space H if there exists a { k}1 system z ∞ H, such that { j}1 ∈

(yk, zj)= δkj, j,k =1, 2,..., where δkj is the Kronecker symbol.

Exercise 2.2. The system yk 1∞ is minimal in Hilbert space H if and only if for all j =1, 2,..., { } yj Clos y1,...,yj 1, yj+2,... 6∈ { − } (by Clos x ,x ,..., we denote the closure of the linear span of the set x ∞). { 1 2 } { k}1 8

Definition 2.3. The system y ∞ is a basis in Hilbert space H if y 1 (i.e. { k}1 k kk ≍ c1 6 yk 6 c2 with some positive constants c1, c2 independent of k) and any elementk yk H can be uniquely represented by a series ∈ ∞ y = ckyk (1) Xk=1 with some coefficients c and this series strongly converges in H. If this series { k} converges unconditionally for any y H then the basis yk is called an unconditional basis or a Riesz basis. ∈ Note 2.1. It is not a simple exercise to give an example of a basis which is not a Riesz basis. It was K. Babenko who proved in 1948 that the system yk(x) 1∞ = α { } x sin kx ∞ is a basis in L [0, π] but not a Riesz basis, provided that 1/2 < α1/2. { }1 2 − The following theorem allows us to give another definition of a Riesz basis.

Theorem 2.1. The system yk 1∞ in Hilbert space H forms a Riesz basis if and only if there exists an orthogonal{ } basis

e ∞ { k}1 in H and a bounded invertible operator A such that

Aek = yk, k =1, 2,.... The proof of this theorem can be found in [Gohberg, Krein 1, Ch.6]. More precise historical comments are given in [Nikolskii 1]. If the system yk 1∞ is a basis in H then the coefficients ck = ck(y) in the representation (1){ are} linear functionals in H. Since these functionals are defined for all y H, we have by virtue of the Banach-Steinhaus theorem that they are bounded.∈ Therefore, using the Riesz theorem we can find elements z H such { k} ∈ that ck(y)=(y, zk). Since the representation (1) for element y = yj is unique we have (yj, zk)= δjk. The theorem zk is said to be adjoint to yk . It is known (see [Gohberg, Krein, Ch.6], for example){ } that if y is a basis (or{ a} Riesz basis) then { k} zk is too. { }Hence any basis y in H is a minimal and, obviously, complete system. The { k} converse assertion is certainly not true. For example, the system y ∞, where { k}1 1 1 y1 = 1, 0, 0,... , y2 = 1, 1, 0, 0,... ,...,yk = 1, 1,..., 1, 0, 0,... { } √2{ } √k{ } 9 is minimal and complete in H = ℓ , bet it is not a basis in ℓ . (Hint: if e ∞ is 2 2 { k}1 the standard basis in ℓ2 then the operator A defined by equalities Aek = yk is not invertible.) If the system is complete and minimal but is not a basis then it can have some intermediate property.

Deﬁnition 2.4. The system y ∞ is a basis with parenthesis in Hilbert space H { k}1 if there exists a sequence of integers ms 1∞ (m1 =0) such that any element y H can be uniquely represented by a series{ } ∈

ms+1 ∞ ∞ y = ckYk = Ys s=1 ! s=1 X k=Xms+1 X and the series ΣYs strongly converges in H.

Deﬁnition 2.5. Let Λ = λk 1∞ be a sequence of complex numbers such that for α { } some α > 0 Re λk > 0 for all k suﬃciently large (we take the main branch of α α λ , i.e. λ > 0 if λ > 0). The minimal system yk 1∞ H is a basis for the Abel method of summability of order a α with respect{ } to sequence∈ Λ if there exists a sequence of integers m ∞ (m =0) such that for any x H the series { s}1 1 ∈ ms+1 ∞ αt ∞ λk x(t)= e− (x, yk∗)yk = Xs(t) s=1 ! s=1 X k=Xms+1 X

(the system yk∗ 1∞ is adjoint to yk 1∞) strongly converges for any t> 0 and x(t) x 0 if t { +0} . { } k − k→ → It was V.Lidskii [1] who introduced this method for summability of Fourier series (x, yk∗)yk with respect for systems yk 1∞ of EAV of compact operators A. In this { } 1 situation Λ = λk is the sequence of eigenvalues of A− (the deﬁnition of x(t) P { } 1 slightly changes if λk are not semi-simple eigenvalues of A− ). According to a theorem of Hilbert the system yk 1∞ of eigenvectors of self- adjoint compact operator A acting in Hilbert space H{ is} an orthogonal basis in H. The system y ∞ of EAV of non-self-adjoint compact operator A corresponding to { k}1 eigenvalues λ = 0 forms a minimal system, because the system y∗ ∞ of EAV of k 6 { k}1 the operator A∗ is adjoint to yk 1∞. Certainly, for a nonself-adjoint operator, the system of its EAV is not always{ complete,} For example the compact operator x Ay(x)= y(t)dt Z0 10

in the space H = L2[0, 1] has no eigenvectors. But if the property of completeness is proved, then one can try to prove the basis property, or the basis property for the Abel method of summability. The investigations in this field were very intensive and a number of deep and refined results were obtained. The reader can make acquaintance with some of them in books [Cohberg, Krein 1], [Markus 1], [Agranovich 1]. We will touch this topic again in a subsequent lecture. The proof of the completeness theorem in Section 1 depended on the finite dimensional context. To give the new approach we have to start from an important theorem on holomorphic operator functions. First we have to recall some definitions. We say A(λ) is an analytic vector function of complex variable λ with values in Hilbert space H and defined in a domain Ω C, if at each point λ Ω the ration ⊂ ∈ A(λ + h) A(λ) − h converges in the norm of H to a limit A′(λ) if h 0. Futher by ( ,u)v we denote one dimensional operator→ V such that V =(y,u)v. · y Obviously, V ∗ =( ,v)u. By σ we denote the class of compact operators in H. The following· result is due∞ to Keldysh [1, 2] (its first part was independently proved by I. Gohberg).

Theorem 2.2. Let A(λ) = A0 + S(λ), where S(λ) is an holomorphic operator function in a domain Ω and S(λ) σ for each λ Ω. Also, let there exist a point ∈ ∞ ∈ 1 λ0 Ω such that the operator A(λ0) is invertible. Then A− (λ) is a meromorphic ∈ 1 operator function in Ω, i.e. it can be represented in the form A− (λ)= D(λ)/∆(λ), where D(λ) is a holomorphic operator function and ∆(λ) is a holomorphic scalar 1 function in Ω. The principal part of the function A− (λ) at the pole λ = c has the representation

ℓ ( , z0)y0 ( , z1)y0 +( , z0)y1 · k k + · k k · k k + (λ c)pk+1 (λ c)pk · · · k=1 − − X pk 0 1 0 pk ( , z )yk +( , zpk 1)yk + +( , zk)y + · k · − · · · · k (2) λ c − where 0 1 pk yk, yk,...,yk , k =1,...,ℓ (3) is an arbitrary canonical system of A(λ) corresponding to eigenvalue c and

0 1 pk zk, zk,...,zk , k =1,...,ℓ (4) 11 is the canonical system of operator function A∗(λ)=[A(λ)]∗ corresponding to eigenvalue c. The adjoint canonical system (4) is uniquely determined by the given canonical system (3). Proof. First, note that this theorem generalizes the well known Fredholm theorem for linear operator function A(λ)= I λA, A σ . Then note that without loss − ∈ ∞ of generality we can assume that A0 = I. Otherwise we can shift λ λ + λ0 and 1 → consider the operator function A− (λ )[A(λ )+(S(λ + λ ) S(λ ))] = I + S (λ), 0 0 0 − 0 1 where S1(λ) σ . Let em 1∞ be an orthogonal basis in H and Pm be the set ∈ ∞ { } m { } of orthogonal projectors such that Pm(H) = Span ek 1 . Obviously, Pm I and Q = I P 0 if m in the strong operator{ topology} (i.e. P x →x 0 m − m → →∞ k m − k→ for each x H). Let Ω1 be a closed domain in Ω. Since the operator S(λ) is compact we∈ have for any fixed λ Ω , Q S(λ) 0 if m . The operator ∈ 1 k m k → → ∞ function QmS(λ) is holomorphic, hence for any δ > 0 there exists an ε > 0 and m +0= m (λ) such that for all m > m and all µ for which µ λ 6 ε we have 0 0 | − | QmS(µ) 6 δ. We can cover the domain Ω1 by such discs and then choose a finite ksubcover.| Thus for all µ Ω and all m > m , we obtain ∈ 1 1 Q S(µ) 6 δ < 1 (5) k m k where m1 does not depend on µ but only on Ω1. 1 To find the inverse operator A− (λ) we have to solve the equation [I + S(λ)]x = f, f H. (6) ∈ Take any m > m1 and denote Pm = P,Qm = Q. We can rewrite (6) in the form P c + PS(λ)x = Px + PS(λ)Px + PS(λ)Qx = Pf, (7)

Qx + QS(λ)x = Qx + QS(λ)Qx + QS(λ)Px = Qf. (8) From (8) it follows that

[I + QS(λ)]Qx = Q[f S(λ)Px]. − Remembering that Q = Qm and taking into account (5) we obtain

1 Qx =[I + QS(λ)]− Q[f S(λ)Px]. − Using this equality we can rewrite (7):

1 1 P [I + S(λ) S(λ)[I + QS(λ)]− QS(λ)]Px = P [f PS(λ)[I + QS(λ)]− Qf] (9) − − 12

If x = x e then Px = x e + + x e . This means that the equation (9) k k 1 1 · · · m m represents an algebraic system of m equations with unknown variables x1, ,xm. P · · · Note, that the equation (6) has a unique solution if λ = λ0, hence the equation (9) has too. This means that the determinant d(λ) of the algebraic system (9) is non-zero at λ0. Now S(λ) is holomorphic in Ω1 and so is d(λ), moreover d(λ) 0. From (9) we obtain 6≡

1 1 Px = F (λ)P [f PS(λ)(I + QS(λ))− Qf] d(λ) − where F (λ) is a holomorphic operator function in Ω1. Then

1 1 A− (λ)f = x = Px + Qx = [(I + QS(λ))− Qf]+

1 1 [F (λ)P QS(λ)F (λ)P ][f PS(λ)(I + QS(λ))− Qf] d(λ) − −

1 Hence the operator A− (λ) exists with the exception of some finite number of poles in Ω1 Ω. Since Ω1 is an arbitrary closed sub-domain in Ω. we obtain the first assertion⊂ of the theorem. The complete proof of the second statement is technically difficult. We sketch here only the main idea, for more details we refer the reader to the original paper 1 [Keldysh 1]. Suppose that the principal part of the resolvent A− (λ) for the pole λ = c has the form R R R 0 + 1 + + m , (λ c)m 1 (λ c)m · · · λ c − − − − where R are some operators in H and R =0. Then we can write s 0 6 1 x = A(λ)A− (λ)x =

1 R0x Rmx = A(c)+ A′(c)(λ c)+ + + + R(λ)x , 1! − · · · (λ c)m 1 · · · λ c − − − where R(λ) is holomorphic at λ = c. The left side of this equality has no pole, hence the coeﬃcients of powers of (λ c) ν, ν = m +1,m,..., 1, on the right − − 13 side are equal to zero. It follows that

A(c)R0x =0, 1 A(c)R x + A′(c)R x =0, 1 1! 0 ......

1 1 (m) A(c)Rmx + A′(c)Rm 1x + + A (c)R0x =0. 1! − · · · m! This means that for each x H the sequence R x, R x, , R x is a chain of EAV ∈ 0 1 · · · m of length m+1. From the definition of a canonical system (3) it follows that m = p1 and R x = c y0 + + c y0, (10) 0 1 1 · · · ℓ ℓ because y0 ℓ is a basis in Ker A(c) = Ker[I + S(c)] (dimKer[I + S(c)] = ℓ< { k}1 ∞ since S(c) σ ). Note that, if elements y1 and y2 generate chains of EAV of length ∈ ∞ m1 and m2, then y1 + y2 generates a chain of EAV of length min(m1, m2). Thus from the definition of a canonical system; the coefficients cj in (10) are equal to zero for all j such that pj < p1 = m. Moreover cj = cj(x) are continuous linear 0 functionals on x, and by virtue of Riesz’ theorem, we can write cj(x)=(x, zj ) and

R =( , z0)y0 + +( , z0)y0, (11) 0 · 1 1 · · · · q q where q is such a number that p1 = = pq > pq+1. As before, from the equality 1 · · · [A(λ)]∗[A− (λ)]∗x = x we can conclude that R0∗x, R1∗x,...,Rm∗ x is a chain of EAV 0 0 of A∗(λ). This means that elements z1,...,zq generate chains of EAV of length m, and may be taken as the first elements of canonical system (4). The representation m 1 (11) shows that we proved (2) for the (leading) coefficient of (λ c)− − . More − detailed analysis allows us to get the necessary representations for R1, R2,...,Rm and to prove (2). 14 3 New proof of the completeness theorem in finite dimensional case. Representation of the resolvent as a meromorphic function of finite order growth

1. Now we are able to give a new approach to the proof of the theorem on completeness of Keldysh derived chains of operator polynomials. To prove it in ﬁnite 1 dimensional space we need only the representation for the principal part of A− (λ) in the neighborhood of a pole. Let A(λ)= A + λA + + λnA , (1) 0 1 · · · n An be invertible, and (for simplicity) assume that A(λ) has only simple eigenvalues λk (i.e. the algebraic multiplicity of each eigenvalue λk equals 1). In this case {Keldysh} chains have the representation

n 1 y = y , λ y ,...,λ − y , k =1, 2,..., (2) k { k k k k k} where y are corresponding eigenvectors of A(λ). Suppose the system (2) is not k e complete.{ } Then there exists a vector f = f ,...,f Hn such that { 1 n}∈ n 1 (f, y )=(f , y )+ +(f , λ − y )=0, k =1, 2,.... (3) k 1 k · · · n k k n Denote A∗(λe)=[A(λ)]∗ = A0∗ + λA1∗ + + λ An∗ . If dim H < then there is a Laurent expansion · · · ∞ 1 ( , yk)zk [A∗(λ)]− = · + R∗(λ) (4) λ λ − k where R∗(λ) is holomorphic in the neighborhood of λk and zk is the eigenvector of A∗(λ) corresponding to λk. Let us consider the meromorphic vector function

1 F (λ)=[A∗(λ)]− f(λ), (5)

n 1 where f(λ)= f + λf + + λ − f . 1 2 · · · n 15

Using (4) we obtain

1 1 F (λ)=[A∗(λ)]− [f(λ )+ (λ λ )f ′(λ )+ ] k 1! − k k · · · (f(λk), yk)zk = + F1(λ) λ λk − n 1 [(f1, yk)+ +(fn, λk− yk)]zk = · · · + F1(λ) λ λ − k where F1(λ) is a holomorphic function in the neighborhood of λk. Now the equalities (3) show that F (λ) has no poles at λ . Hence F (λ) is an entire vector function { k} (i.e. holomorphic in the whole complex plane). For suﬃciently large λ > r0 we also have the estimate | |

1 F (λ) 6 [A∗(λ)]− f(λ) k k k n 1 kk k 1 1 n 1 1 6 λ − An− (I + λ− An 1An− + + λ− A0An− )− − | | k kk · · ·n 1 k× ( f1 + + λ − fn ) 1 × k k · · · | | k k 6 M λ − , | | where the constant M does not depend on λ. Hence F (λ) is bounded in the whole complex plane and from Liouville’s theorem it follows that F (λ) const. Since F (λ) 0 when λ we have F (λ) 0. ≡ k Fromk→ (5) we have→∞ ≡ f(λ)= A∗(λ)F (λ)=0, therefore f = f ,...,f =0. Hence the system (2) is complete. { 1 n} Note 3.1. We can usually work with holomorphic vector or operator functions as well as with scalar holomorphic vector function in a disk Dε(c)= λ : λ c <ε then { | − | } ∞ f(λ)= f (λ c)k, f H, (6) k − k ∈ Xk=0 and the series converges strongly for λ D (c). Indeed, for each g H the scalar ∈ ε ∈ function (f(λ),g) is holomorphic in Dε(c). Hence

∞ (f(λ),g)= c (λ c)k, λ D (c) k − ∈ ε Xk=0 16

The coefficients c = c (g) are linear functionals defined for all g H, therefore k k ∈ ck =(fk,g). Now from the Banach-Steinhaus theorem we can deduce that the series (6) converges strongly. Also, if F (λ) = F0 + F1λ + is an entire bounded vector function then so is the scalar function (F (λ),g) for· each · · g H. From the Liouville theorem it follows ∈ that (F (λ),g) = const. Hence (Fj,g)=0, j = 1, 2,..., for each g H and F (λ)= F const≡ . (See [Hille and Fhillips], for example, for details.) ∈ 0 ≡ Note 3.2. The proof of completeness does not change significantly if the eigenvalues λk are not simple or semi-simple. It is an easy exercise to reduce from the definition{ } of EAV and representation (2.2)1 that the equalities

0 1 m (f, yk)=0, (f, yk)=0,..., (f, yk )=0 are equivalent to the following: e e e

R0∗f(λk)=0, 1 R∗f(λ )+ R∗f ′(λ )=0 1 k 1! 0 k ......

1 (m) R∗ f(λ )+ + R∗f (λ )=0. m k · · · m! 0 k

n h Therefore if vector f H is orthogonal to all derived chains yk then the function (5) is an entire vector∈ function. This observation allows us{ to} ﬁnish the proof of completeness as before. e h Note 3.3. Keldysh deﬁnes the system of EAV yk to be n-multiple complete in h { } n H if the derived chains yk form a complete system in H . But after considering some concrete operator pencils{ } we will see that we have to investigate the properties h n H of derived chains yk notein the space H , but in some space which is embedded in Hn. { }

2. The new proofe of the completeness theorem can be generalized for inﬁnite dimensional spaces H. The essence of the matter is contained in the subsequent theorem on the growth of the resolvent of operator pencils. First we have to recall some deﬁnitions.

1Here and further the notation (n.m) means the reference (m) from lecture n. 17

An entire scalar function f(λ) is said to be a function of ﬁnite order if there exists a constant p> 0 such that the inequality

λ p f(λ) < e| | | | is valid for all sufficiently large λ >r0 = r0(p). The infimum of such numbers p is called the order of the entire function| | f(λ). We say an entire function f(λ) has a finite type with order p if there exists a constant k > 0 such that k λ p f(λ) < e | | | | for all sufficiently large λ > r1 = r1(k). The infimum of such numbers k is called the type of f(λ) with order| | p. It is an easy exercise to verify that the order p and the type σ of an entire function f(λ) are determined by equalities

lnln M (r) ln M (r) p = lim f , σ = lim r , r r p →∞ ln r →∞ r where Mf (r) = max f(λ) . λ =r | | The same definition| | of growth is applied to holomorphic vector or operator function. The only difference is that instead of f(λ) we have to consider f(λ) . | | 1/k2 k If A is a compact operator (A σ ), then operator C = (A∗A) is compact too. The eigenvalues of the operator∈ C∞are called s-numbers of the operator A. We will assume that the sequence of s-numbers of A is enumerated in decreasing order, so that A = s (A) > s (A) > . k k 1 2 · · · ∞ p We will write A σp if sk(A) < . ∈ k=1 ∞ P Theorem 3.1 (Fundamental theorem on the growth of the resolvent). Let operator pencil (1) be such that A(λ0) is invertible for some λ0 C. Suppose also that there exists a number p> 0 such that one of the following conditions∈ is fulfilled:

A σ j =1, 2,...,n; (7) j ∈ p/j j/p sk(Aj)= o(k− ) j =1, 2,...,n; (8) j/p sk(Aj)= O(k− ) j =1, 2,...,n; (9)

1 Then A− (λ) is a meromorphic operator function whose order does not exceed p. 1 1 This means that A− (λ) admits the representation A− (λ) = D(λ)/∆(λ), where 18

D(λ) and ∆(λ) are the operator and scalar functions, respectively, of order p or less. Moreover, the type of D(λ) and ∆(λ) is finite if (9) holds and equal to zero if either (7) or (8) is fulfilled. In some respects this theorem is due to M. Keldysh because he was the first to prove a general result of this kind, although he took advantage of an important result due to T. Carleman estimating the Fredholm resolvent of a Hilbert-Schmidt operator. New approaches to the proof for linear pencils A(λ) = I λA were proposed by V. Lidskii, V. Matsaev, I. Gohberg and M. Krein. Some contributions− were also made by M. Gasimov and G. Radzievskii. An independent approach to the proof of such theorems for operator pencils with differential operators was developed by F. Brouder and S. Agmon (see comments and references in [Shkalikov 1, §2]). To prove this theorem we have to make a tour of some selected topics in the theory on non-self-adjoint operators. Certainly we will not be able to give proofs of all the results which we will use. Item 1. Let A be a bounded operator in H. It is well-known, and easy to see, that H = Ker A Im A = Ker A∗ Im A ⊕ ∗ ⊕ Each bounded operator A can be represented in the form A = UC, (10)

1 where C = (A∗A) 2 and a partial isometry, such that Ker U = Ker C and U : Im A∗ = Im C Im A is an isometric one-to-one map. This representation is called the polar representation→ . It is not complicated to prove (see proof in [Gohberg, Krein, Ch.1]), but very useful. In particular, if A is a compact operator, then C is too. Hence ∞ C = s (A)( , e )e , k · k k Xk=1 where ek 1∞ is the orthonormal system of eigenvectors of C. Then form (10) we get the{Schmidt} representation

∞ A = s (A)( , e )f , (11) k · k k Xk=1 where f ∞ = Ue ∞ is also an orthonormal system, because U is an isometric { k}1 { k}1 operator for x Im C and ek Im C. Item 2. Let∈ A,B be compact∈ operators and D a bounded operator in H. The following properties of s-numbers are fulﬁlled: 19

1. sk(A)= sk(A∗), k =1, 2,... ; 2. s (DA) 6 D s (A), s (AD) 6 D s (A), k =1, 2,... ; k k k k k k k k 3. sk+1(A)= min A F , k =0, 1,..., where Rk is the set of all operators F Rk k − k with range of∈ dimention k or less. In particular, if F R , then ∈ r

sk(A + F ) 6 sk r(A), k = r +1,r +2,.... −

4. sk+m+1(A + B) 6 sk(A)+ sm(B), k,m =1, 2,..., hence for a set of compact operators A1,A2,...,Aq one has the inequalities s (A + A + + A ) 6 s (A )+ s (A )+ + s (A ), k = 1, 2,..., k 1 2 · · · q k1 1 k1 2 · · · k1 q where k1 = [(k 1)/q]+1 (by [a] we denote the integer part of the number a); −

5. sk+m 1(AB) 6 sk(A)sm(B), k,m =1, 2,..., − and if A1,...,Aq are compact operators then

s (A A A ) 6 s (A )s (A ) s (A ), k 1 2 · · · q k1 1 k1 2 · · · k1 q where k =1, 2,..., k = [(k 1)/q]+1; 1 − 6. λ (A)λ (A) λ (A) 6 s (A)s (A) s (A), k =1, 2,..., | 1 2 · · · k | 1 2 · · · k where λj(A) are eigenvalues of the operator A numbered as many times as their algebraic multiplicity and in order of decreasing absolute value;

k k p p 7. λj(A) 6 sj (A), p> 0, k =1, 2,... ; j=1 | | j=1 P P k k 8. (1+ r λj(A) ) 6 (1 + rsj(A)), r> 0, k =1, 2,... ; j=1 | | j=1 Q Q k k k 9. sj(A + B) 6 sj(A)+ sj(B), k =1, 2,... ; j=1 j=1 j=1 P P P k k 10. sj(AB) 6 sj(A)sj(B), k =1, 2,..., j=1 j=1 P P 20

and if Aj σpj , j = 1,...,q, then using the H¨older inequality one can deduce ∈ 1 1 1 1 A1A2 ...Aq σp, where p− = p1− + p2− + + pq− . These results∈ on s-numbers are due· to · · H. Weyl, K. Fan, A. Horn and D.Allachverdiev. The proof of all these basic properties of s-numbers can be found in the book [Gohberg, Krein, Ch.2]. Only the proof of the property 6 (H. Weyl’s theorem) is rather complicated. We propose a new sort proof due to A. Kostyuchenko. Denote by Hk the span of EAV of the operator A corresponding to the ﬁrst k eigenvalues (counted according to their algebraic multiplicity) and by Ak denote the restriction of A to its invariant subspace H . We can choose a Jordan basis e k, k { j}1 so that either Aej = λjej or Aej = λjej + ej 1, j =1,...,k. Write Ak = PkAPk, − where Pk is the orthoprojector on Hk. Then by Schmidt’s orthogonalization of k k ej 1 we get the basis fj 1 for which operator Ak evidently has triangular form and{ }(A f ,f )= λ (A) {(the} basis f k is called the Schur basis for A ). Hence k j j j { j}1 k det A 2 = λ λ λ 2 | k| | 1 2 · · · k| = det Ak∗ det Ak = det Ak∗Ak =[s (A ) s (A )]2 6 [s (A) s (A)]2. 1 k · · · k k 1 · · · k The property 6 follows using property 2 of s-numbers. Item 3. An operator A is said to be nuclear if A σ1, i.e. sk(A) < . Using the Schmidt representation (11) the following remarkable∈ fact can be established∞ (see [Gohberg , Krein, Ch. III, Sec.8]). P

Proposition 3.1. A σ1 if and only if for any orthonormal basis ϕk 1∞ in the space H the series ∈ { } ∞ (Aϕk,ϕk) (12) Xk=1 converges. Moreover, the sum (12) does not depend on the choice of the basis ϕ ∞. { k}1 The sum (12) for operator A σ1 is denoted by Tr A. It is worth mentioning the following properties the following∈ properties of the functional Tr A (we can consider class σ1 as a normed space with norm A 1 = sk(A) and then Tr A is a linear functional in this space): k k P 1. Tr A∗ = Tr A; 2. Tr(AB) = Tr(BA), A,B σ ; ∈ 1 21

∞ 3. Tr A = λk(A). k=1 The ﬁrst propertyP is trivial, the second one can be easily obtained from the Schmidt representation. The third one is the assertion of a well-known theorem of Lidskii and is nontrivial. Further, we will need the inequality

Tr A 6 A = s (A). (13) | | k k1 k Certainly, (13) follows immediately from propertyX 7 of s-numbers, and the Lidskii theorem. But an elementary proof can also be proposed. Taking in (12) ϕk = ek and using the Schmidt representation (11) we easily get (13). Item 4. If A σ1 then the determinant of operator I A is deﬁned by the formula ∈ − ∞ det(I A)= (1 λ (A)). − − k Yk=1 This product obviously converges, because of property 7 for s-numbers

∞ λ (A) 6 A . | j | k k1 Xk=1 We consider also the characteristic determinant of the operator A

∞ det(I µA) := D (µ) := (1 µλ (A)). − A − k Yk=1 Theorem 3.2. det(I A) is a continuous functional in the space of nuclear operators with norm A −. k k1 Proof. We have to show that for any ε > 0 there exists a δ > 0 such that the inequality A B <δ implies k − k1 det(I A) det(I B) < ε. | − − − | We have (with λk = λk(A)),

∞ DA′ (µ) λk [ln det(I µA)]′ = = − DA(µ) − 1 µλk k=1 − X 1 = Tr[A(I µA)− ]. − − 22

The last equality is valid because of Lidskii’s theorem. (Evidently, for any ﬁxed 1 µ = λ the operator A(I µA)− is nuclear.) Now we get the representation 6 k − 1 D (µ) = exp Tr[A(I ζA)− ]dζ , (14) A − − ZΓ where Γ is any smoothcontour, which connects the points 0 and µ, and does not 1 contain the points λk− . Obviously, operator I ζB is invertible for all ζ Γ if B <δ and δ is{ suﬃciently} small. Hence, there− exists a constant M such∈ that k Ak 1 1 1 1 max (I ζA)− 6 M, max (I ζB)− 6 M. (15) ζ Γ k − k ζ Γ k − k ∈ ∈ Notice that

1 1 1 1 A(I µA)− B(I µB)− =(I µA)− [A(I µB) (I µA)B](I µB)− − − − − 1 − − −1 − =(I µA)− (A B)(I µB)− . − − − Now, using property 2 of s-numbers and the estimates (13), (15), we obtain

1 1 2 Tr[A(I ζA)− B(I ζB)− ]dζ 6 γM A B , − − − k − k1 ZΓ where γ is the length of Γ. From representation (14) it follows that

D (µ) D (µ) = | A − B | 1 1 = D (µ) 1 exp Tr[A(I ζA)− B(I ζB)− ]dζ A − − − − ZΓ 2 6 γM DA(µ) A B 1. | |k − k The last inequality (we can put µ =1) proves the thorem.

Proposition 3.2. Let A σ1 and Qn be a sequence of orthoprojectors such that Q 0, when n . Then∈ { } n → →∞ Q AQ 0, Q A 0, AQ 0. (16) k n nk1 → k n k1 → k nk1 → Proof. Using the Schmidt representation (11) again we get

N ∞ A = s ( , e )f + s ( , e )f)k = A + A , k · k k k · k N ε Xk=1 k=XN+1 23

∞ where Aε 1 = sk(A) < ε. Since AN is ﬁnite dimensional, we have k k k=N+1 QnAN QN 1 0, PQnAN 1 0 and (16) follows. k k → k k → The theorem on continuity of the determinant and the last proposition give us some important results.

Corollary 3.1. Let A σ1 and Pn be orthogonal projectors such that Pn I when n . Then ∈ { } → →∞

det(I A) lim det(I PnAPn). n − →∞ − Proof. We have to notice only that

A P AP = Q AP + P AQ + Q AQ 0 k − n nk1 k n n n n n nk1 →

Corollary 3.2. If A σ ,B σ then ∈ 1 ∈ 1 det(I A)(I B) = det(I A)(I B). (17) − − − − Proof. The equality (17) is known in ﬁnite dimensional space, hence if P I, then n →

det(I A)(I B)= lim det(I PnAPn PnBPn + PnABPn) n − − →∞ − − = lim det[(I PnAPn)(I PnBPn) PnAQnBPn] n →∞ − − − = lim det(I PnAPn)(I PnBPn) [since AQn 1 0] n →∞ − − k k → = lim det(I PnAPn) lim det(I PnBPn) n n →∞ − · →∞ − = det(I A) det(I B). − −

Corollary 3.3. If A σ ,B σ and I B is invertible then ∈ 1 ∈ 1 −

1 det(I A) det(I A)(I B)− = − . (18) − − det(I B) − 24

1 1 Proof. Since (I B)− = I + B(I B)− , we have − −

1 ∞ 1 ∞ 1 1 det(I B)− = (1 + λ (1 λ )− )= (1 λ )− = . − k − k − k det(I B) Yk=1 Yk=1 − Hence the equality (18) follows from (17). Item 5. Now we establish the assertion of the fundamental theorem for a linear pencil A(λ)= I λA, A σ , p 6 1. − ∈ p Theorem 3.3. If A σ , p 6 1 then ∈ p

1 D(λ) (I λA)− = − det(I λA) − and ∞ det(I λA) 6 (1 + λ s (A)), (19) | − | | | k Yk=1 1 ∞ D(λ) = (I λA)− det(I λA) 6 (1 + λ s (A)). (20) k k k − − k | | k Yk=1 Proof. The estimate (19) follows from property 8 of s-numbers

∞ ∞ ∞ (1 λλj(A)) 6 (1+ λ λj(A) ) 6 (1 + λ sj(A)). − | || | | | k=1 k=1 k=1 Y Y Y

To prove the estimate (20), choose arbitrary vectors ϕ, ψ such that ϕ = ψ =1 and consider the operator k k k k

A = A + ξ( , ψ)ϕ, ξ> 0. 1 · According to property 3 of s-numbers we have

sj+1(A1) 6 sj(A), j =1, 2,..., s (A )= A 6 A + ξ = s (A)+ ξ. 1 1 k 1k k k 1 Hence ∞ det(I λA ) 6 [1 + λ(s (A)+ ξ)] (1+ λ s (A)). (21) | − 1 | | 1 | | j j=1 Y 25

1 Since (I λA )(I λA− )= I λK, where K is a one dimensional operator − 1 − − 1 Kf = ξ((I λA)− f, ψ)ϕ, − we have from corollary 3,

1 det(I λA1) det(I λA )(I λA)− =1 λλ (K)= − . − 1 − − 1 det(I λA) −

Solving the equation Kf = λ1f, we ﬁnd d = ϕ and λ1 = λ1(K) = ξ((I 1 − λA)− ϕ, ψ). Therefore

1 det(I λA1) 1 λξ((I λA)− ϕ, ψ)= − − − det(I λA) − and taking into account (21), we obtain

∞ (1 + λ sk(A)) 1 1 1 s1 k=1 | | ((I λA)− ϕ, ψ) 6 + + +1 . | − | λξ λξ ξ Qdet(I λA) | − | Now let ξ > 0 tend to inﬁnity and notice that R = sup (Rϕ,ψ) . Then the k k ϕ = ψ =1 | | k k k k last estimate gives (20). Now we can get the assertion of the fundamental theorem by applying Borel’s theorem on growth of canonical products (see, for example, [B. Levin]). We formulate this result for simple canonical products.

Theorem 3.4 (Borel’s Theorem). Let one of the following conditions be fulﬁlled:

1. sp < , p > 1; k ∞ 1/p 2. sPk = o(k− ), p< 1;

1/p 3. sk = O(k− ), p< 1.

∞ Then ∆(λ)= (1 + skλ) is an entire function or order p or less and ﬁnite type. k=1 Moreover, the typeQ of ∆(λ) is equal to zero if either condition 1 or condition 2 holds. 26

Item 6. We start to prove the fundamental theorem for the general case using the note that, without loss of generality, one can assume A0 = I (as in the theorem on holomorphic operator functions). Denote

F (λ)= A λ A λ2 A λn. − 1 − 2 −···− n

Evidently, for each fixed λ and any ε> 0 we haveF (λ) σp+ε. Choose the smallest integer ℓ such that p/ℓ < 1. Then taking into account the∈ property 5 of s-numbers we obtain F ℓ(λ) σ . Thus for any fixed λ C the function ∈ 1 ∈ ∆(λ) = det(I F ℓ(λ)) − is well defined. Moreover, using corollary 1 and Weierstrass’ theorem on uniformly convergent sequences of holomorphic functions, we find that ∆(λ) is an entire func- ℓ tion (the uniform convergence of the functions det(I PnF (λ)Pn), Pn I, follows from the theorem on continuous dependence of− the determinant and→ from the continuity of the function F ℓ(λ) in the nuclear norm). From the simple relation

1 1 ℓ 1 ℓ 1 A− (λ)=[I F (λ)]− =[I + F (λ)+ + F − (λ)][I F (λ)]− (22) − · · · − 1 we conclude that the growth of the meromorphic function A− (λ) is the same as ℓ 1 the growth of [I F (λ)]− , because the left divisor in (22) is polynomial and does − 1 not have any inﬂuence on the order or type of A− (λ). Using the theorem on the estimate of the resolvent in the latter item, we have

ℓ 1 D(λ) [I F (λ)]− = − ∆(λ) and ∞ ∞ D(λ) 6 [1 + s (F ℓ(λ))], ∆(λ) 6 [1 + s (F ℓ(λ))] (23) k k k | k Yk=1 Yk=1 (to get these estimates we make substitution in (19), (20): λ 1, A F ℓ(λ)). Denote → →

ℓ ℓ ℓ ℓ ℓ+1 ℓ 1 ℓ 2 ℓ 1 F (λ)=( 1) λ A + λ [A − A + A − A A + + A A − ]+ − { 1 1 2 1 2 1 · · · 2 1 nℓ ℓ+2 ℓ 1 nℓ ℓ ℓ j + λ [A − A + ]+ λ A =( 1) λ B . 1 2 · · · n} − j Xj=ℓ 27

First, assume that condition (9) holds. Then taking into account the properties 4, j j/p 5 of s-numbers we ﬁnd sk(B ) = O(k− ), j = ℓ, ℓ +1,...,nℓ, k = 1, 2,.... Using the property 4 again we obtain

nℓ nℓ ℓ j j j/p s (F (λ)) 6 λ s (B ) 6 M λ k− , (24) k | | k1 j | | Xj=ℓ Xj=ℓ

k 1 where k1 = n(ℓ −1)+1 +1, M = const. Hence − h i nℓ nℓ ∞ ∞ ∞ ℓ jk j/p jk j/p [1 + s (F (λ))] 6 (1 + M λ − ) 6 (1+ M λ − ). k | | | | Yk=1 Yk=1 Xj=ℓ Yj=ℓ Yk=1 According to Borel’s theorem the function

∞ j/p fj(µ)= (1+ Mµk− ), j = ℓ, ℓ +1,...,nℓ Yk=1 has the order p/j and finite type. Then we find from the definition that the function ℓ nℓ f(λ)= fℓ(λ ) fnℓ(λ ) has order p and finite type and so do the functions D(λ) and ∆(λ). · · · The proof does not change if (9) is replaced by (8). If the condition (7) holds then using the properties 5. 10 of s-numbers we may deduce that B σ , j = j ∈ p/j ℓ,...,nℓ. Taking into account the first estimate in (24) and noting that index k1 repeats κ = n(ℓ 1)+1 times when k runs through the integers, we obtain − κ nℓ ∞ ∞ [1 + s (F ℓ(λ))] 6 [1+ λ js (B )] . k  | | k j  kY1=1 Yj=ℓ Yk=1   Recalling that B σ and applying Borel’s theorem, we find from (23) that D(λ) j ∈ p/j and ∆(λ) have the order 6 p and type 0 with order p. This proves the theorem. 28 4 Keldysh-Lidskii theorem on the completeness

To prove the subsequent theorems on completeness we need to recall some classical results of the theory of entire functions (see, for example, [Levin, Ch.1]). Theorem 4.1 (Phragmen-Lindel¨of Theorem). Let f(λ) be a holomorphic function 2 of order p in a sector Ω (ϕ)= λ : ϕ arg λ 6 α and α { | − | } f(λ) 6 M (1) | | on the sides of the sector Ωα(ϕ). If α<π/2p then the estimate (1) holds throughout all the sector Ωα(ϕ). Note 4.1. One can deduce from this theorem a slight generalization. Instead of the estimate (1), assume that the following holds:

f(λ) 6 M(1 + λ m) (2) | | | | on the sides of the sector Ωα(ϕ). Then the same estimate holds throughout all of the sector Ωα(ϕ), probably with a new constant M1 instead of M. To prove this fact we may consider a function f(λ)/p(λ), where p(λ) is a polynomial of degree in with zeros lying outside of the sector Ωα(ϕ). The next result is a corollary of the theorem giving a lower estimate of entire functions due to E. C. Titchmarsh.

Theorem 4.2 (Theorem on a Ratio of Holomorphic Functions). Let F (λ)= F1(λ) , F2(λ) where Fj(λ), j =1, 2, are entire functions of the order pj and type σj with order pj. If F (λ) is also an entire function then it is a function of order p = max(p1,p2) or less and of the type σ = σ1 + σ2 or less with order p. Now we can formulate and prove a general theorem on completeness. The main assumption of this theorem is that, on some rays in the complex plane, the growth of the resolvent of a pencil does not exceed polynomial growth. It may seem at first sight that such a condition is unnatural and is difficult to establish for some specific linear or polynomial pencils. But we will dispel any such illusion later on. Now we mention only that there are a lot of papers devoted to estimates of the resolvent for boundary value problems containing a spectral parameter for ordinary differential

2We defined the order and the type of an entire function, but the same definition is applied to functions which are holomorphic in a sector. 29 operators, as well as for partial differential operators. Some of them are due to G. Birkhoff and Ja. Tamarkin, S. Agmon and L. Nierenberg, M. Agranovich and M. Vishik (for estimates of the resolvent in non-Hilbert spaces see the book [Tribel] and references there).

Theorem 4.3 (General Theorem on Completeness). Let the operator pencil

A(λ)= A + λA + + λnA 0 1 · · · n be such that Aj σp/j, j = 1,...,n, for some p > 0 and Ker An∗ = 0. Also, let ∈ q there exist a ﬁnite set of a rays γk k=1 dividing the complex plane into q sectors with angles less than π/2p and such{ } that the resolvent of the pencil A(λ) exists on these rays for suﬃciently large λ >r and has the estimate | | 0 1 m A− (λ) 6 M(1 + λ ), (3) k k | | where M and m are constants. Then the system of Keldysh derived chains of A(λ) is complete in Hn.

n Proof. We have noticed (see Note 3.2) that if a vector f = f1,...,fn H is orthogonal to all Keldysh derived chains then { } ∈

1 n 1 F (λ)=[A∗(λ)]− (f1 + λf2 + + λ − fn 1) · · · − is an entire vector function. It follows from the fundamental theorem on the estimate of the resolvent and the theorem on a ratio of entire functions that F (λ) has the order p or less. Taking into account the equality

1 1 [A∗(λ)]− = [A− (λ)]∗ k k k k q and the estimate (3) which holds asymptotically on the rays γk 1, we obtain the estimate { } m+n 1 F (λ) 6 M (1+ λ − ) (4) k k 1 | | q on the rays γk 1. The angle between neighboring rays is less than π/2p. Hence, in virtue of the{ Phragmen-Lindel¨of} theorem we obtain the estimate (4) in each sector contained between neighboring rays. Thus the estimate (4) holds in the whole complex plane and, from Liouville’s theorem, we conclude that

F (λ)= F + λF + + λrF , r 6 m + n 1. 0 1 · · · r − 30

On the other hand

n r n 1 (A∗ + λA∗ + + λ A∗ )(F + λF + + λ F )= f + + λ − f . 0 1 · · · n 0 1 · · · r 1 · · · n The right side of this equality is a polynomial of degree n 1 but the left side − is a polynomial of degree n + r or less. Hence we find An∗ Fr = 0, An∗ Fr 1 = − 0,...,A∗ F = 0. Since the kernel of the operator A∗ is trivial we have F (λ) 0 n 0 n ≡ and then f = f1,...,fn =0. The theorem is proved. We will deduce{ some useful} corollaries from this general theorem. But first we recall some definitions and prove some auxiliary results. The numerical range θ(A) of an operator A is the set of all complex numbers (Au,u), where u takes values in the unit sphere: u = 1. A theorem due to Hausdorff asserts that θ(A) is a convex set. It is knownk alsok (see, for example, [Kato, Ch.5]) that the closure of θ(A) contains the spectrum of A and for all µ θ(A) σ(A) the following estimate holds 6∈ ∪

1 1 (A µI)− 6 . (5) k − k dist(µ, θ(A))

An operator T is called sectorial if its numerical range θ(T ) is a subset of a sector ϕ arg λ 6 α for some ϕ [0, 2π) and α [0,π/2]. The numbers ϕ and α are called| − the vertex| and semi-angle∈ of the sectorial∈ operator T . Further, when dealing with a sectorial operator, we will assume (for simplicity) that its vertex is equal to zero. An operator T is called accretive (dissipative) if its numerical range lies in the right-half plane, Re λ > 0 (in the left-half plane Re λ 6 0).

Lemma 4.1. If T is a sectorial operator with semi-angle α then outside the sector Ω = λ : arg λ 6 α + ε , (ε> 0) the following estimates hold: α+ε { | | }

1 1 (I λT )− 6 , (6) k − k sin ε 1 1 (I λT )− T 6 . (7) k − k λ sin ε | | Moreover, if Ker T ∗ =0, then for any ﬁxed vector x

1 (I λT )− x 0 (8) k − k→ when λ outside the sector Ω . →∞ α+ε 31

Proof. If λ Ω then according to (5) 6∈ α+ε

1 1 1 1 1 1 (I λT )− = λ − (λ− I T )− 6 6 k − k | | k − k λ dist(λ 1, Ω ) sin ε | | − α and the ﬁrst estimate (6) follows. Also, if λ Ω , then for all y H 6∈ α+ε ∈ (I λT )y T y > ((I λT )y,Ty) = (y,Ty) λ(Ty,Ty) k − kk k | − | | − | > T y 2 dist(λ, Ω ) > T y 2 λ sin ε k k α k k | | 1 and T y 6 ( λ sin ε)− (I λT )y . Hence for all x =(I λT )y we have k k | | k − k − 1 1 T (I λT )− x 6 ( λ sin ε)− x k − k | | k k and the second estimate (7) follows. If Ker T ∗ = 0 then Im T = H, therefore for any x H, and given any δ > 0, there exists a vector y = T z such that y x <δ. Then∈ for λ Ω we have k − k 6∈ α+ε

1 1 z (I λT )− x 6 k k + δ . k − k sin ε λ | | Since δ may be chosen arbitrary small, we obtain (8).

Lemma 4.2. Let A = (I + S)T , where S σ , Ker A∗ = 0, and let T be a 3 ∈ ∞ sectorial operator with semi-angle α. Then outside a sector Ωα+ε, ε> 0, and for suﬃciently large λ >r = r (ε), the following estimate holds: | | 0 0 1 (I λA)− 6 M = M(ε). (9) k − k

Proof. It follows from the assumption Ker A∗ = 0 that Ker(I + S∗)=0 and 1 Ker(I + S)=0 (because S σ ). Hence I + S is invertible and (I + S)− = 1 ∈ ∞ I S(I + S)− = I + S1, S1 σ . Note also, that Ker T ∗ =0. Further, we have − ∈ ∞ 1 1 1 (I λA)− = [(I + S)(I λT )(I + S1(I λT )− )]− k − k k 1 − 1 − k 1 1 (10) 6 (I + S)− (I λT )− (I + S (I λT )− )− . k kk − kk 1 − k 3An operator A, having such a representation, we call a compact perturbation of a sectorial operator. 32

Notice, we have if V =( ,ϕ)ψ is a one dimensional operator (or ﬁnite dimensional) then it follows from (8) that·

1 1 V (I λT )− = (I λT ∗)− V ∗ 0 k − k k − k→ when λ outside Ω . (Obviously, the numerical range of T ∗ lies in the → ∞ α+ε sector Ωα and Ker T = Ker T ∗ = 0, see Note 3 below.) Operator S1 is compact, hence it may be approximated in the operator norm with any accuracy by a ﬁnite 1 dimensional operator. Thus S (I λT )− 0 when λ outside Ω and k 1 − k → → ∞ α+ε the estimate (9) follows from (6) and (10).

Note 4.2. Lemmas 1 and 2 are valid if the sectorial operator T is replaced by a self- adjoint operator C (not necessarily non-negative). The only diﬀerence is that in this case all estimates hold outside the sector Λε = λ : arg λ 6 ε or π arg λ 6 ε . { | | | − | } Now we are able to present some corollaries from the general theorem on completeness.

Corollary 4.1 (Keldysh-Lidskii theorem). Let T be a sectorial operator with semi- angle α and Ker T ∗ =0. If T σp and p<π/2α then the system of EAV of T is complete. ∈

Proof. Since the estimate (6) holds, we ﬁnd that all assumptions of the general theorem are fulﬁlled for the linear operator pencil A(λ)= I λT . −

Note 4.3. Actually, one can omit the assumption Ker T ∗ =0 in the latter corollary, because for an accretive operator T (and of course, for a sectorial operator T ) we have Ker T ∗ = Ker T (this equality follows from the representation T = TR + iTI, where TR =(T + T ∗)/2,TI =(T T ∗)/2i > 0). Hence, for a sectorial operator we have the representation H = Im−T Ker T . Since the restriction of T to its ⊕ invariant subspace H1 = Im T has a complete system of EAV in H1, we ﬁnd that the assertion of Corollary 1 is valid without the assumption that Ker T ∗ =0. Corollary 4.2. Let A = (I + S)T, S σ ,T be a sectorial operator with ∈ ∞ semi-angle α, and Ker A∗ =0. If T σp and p<π/2α then the system of EAV of the operator A is complete. ∈ 33

Proof. By Lemma 2, we ﬁnd that all assumption of the general theorem are fulﬁlled for the linear pencil A(λ)= I λA. −

Corollary 4.3 (Theorem of Keldysh). Let

2 2 n 1 n 1 n n A(λ)= I + S0 + S1Cλ + S2C λ + + Sn 1C − λ − +(I + Sn)C λ , (11) · · · − where Sj, j = 0, 1,...,n, are compact operators, Ker(I + Sn) = 0, and C = C∗ > 0. If C σp for some p > 0 then the system of Keldysh derived chains of the pencil A(∈λ) is complete in Hn. In particular, the system of EAV of a compactly perturbed positive self-adjoint operator

A =(I + S)C, S σ , Ker(I + S)=0, C> 0, C σp, ∈ ∞ ∈ is complete H.

Proof. Without loss of generality we can assume that Sn = 0, otherwise we have 1 to consider the operator pencil (I + Sn)− A(λ), which has the representation (11) with Sn =0. If Sn =0 then

n 1 − 1 n n 1 k k n n 1 1 A− (λ)=(I + λ C )− (I + Skλ C (I + λ C )− )− . (12) Xk=0 If ω n are the roots of the equation ωn +1=0, then { j}1 n n n I + λ C = (1+ ωjλC) j=1 Y and for k =0, 1,...,n 1, we have − k k n n 1 S λ C (I + λ C )− k k k k n 1 1 1 =[Sk(I + ωk+1λC)− ][ λ(I + ωjλC)− ][ (I + λωjC)− ] j=1 Y j=Yk+1 = A1(λ)A2(λ)A3(λ).

n Denote by Ωε the union of n sectors in the complex plane with semi-angles ε and vertex ω , k = 1,...,n. Then, according to Lemma 2 A (λ) 0 if λ − k k 1 k → → ∞ 34

n outside of Ωε and according to Lemma 1 Aj(λ) 6 M, j = 2, 3 outside the n k k domain Ωε . Hence, n 1 − k k n n 1 1 (I + S λ C (I + λ C )− )− 6 M k k k 1 Xk=0 n for λ Ωε and λ sufficiently large. Using Lemma 1 again and the representation (12) we6∈ obtain | | 1 A− (λ) 6 M k k 2 if λ Ωn and λ > r is sufficiently large. Now we can choose ε such that 6∈ ε | | 0 ε<π/2p, and all assumptions of the general theorem are fulfilled.

Note 4.4. Taking Note 2 into account we may assume in Corollary 3 that C is an arbitrary self-adjoint operator with Ker C =0, instead of C > 0. It is also worth mentioning that we can refine the assertion of Corollary 1 by replacing the condition p<π/2α with p 6 π/2α. For this purpose we need to apply the following fact from theory of entire functions (see, for example, [Levin, Ch.1]). Theorem 4.4 (Refined Version of The Phragmen-Lindelöf Theorem). Let f(λ) be 4 a holomorphic function of order p and of minimal type in a sector Ωα = λ : arg λ 6 ε . If p 6 π/2α and { | | } f(λ) 6 M(1 + λ m) | | | | on the sides of the sector Ωα, then the same estimate holds throughout the sector Ωα, probably with a new constant M1 instead of M. Theorem 4.5 (Refined Theorem on Completeness of EAV of a Sectorial Operator5). Let T be a sectorial operator with semi-angle α and either T σp, or sk(T ) = 1/p ∈ o(k− ), k =1, 2,.... If p 6 π/2α then the system of EAV of the operator T is complete. Proof. Let Ω = λ : λ = µ ε, arg µ 6 α =Ω ε. Then for all λ Ω α,ε { − | | } α − 6∈ α,ε 1 1 sin ε sin ε dist(λ− , Ω ) > λ − sin . α | | λ ∼ λ 2 | | | | 4A holomorphic function f(λ) is said to have a minimal type with order p if it has the type zero with order p. 5For accretive operators this theorem follows from a deep theorem of M. Krein on completeness of EAV of compact accretive operators with nuclear real component. 35

Hence, using the estimate (5), we ﬁnd for all λ Ω 6∈ α,ε

1 1 1 1 M (I λT )− = λ − (λ− I T )− 6 λ , (13) k − k | | k − k sin ε| | where M does not depend on ε and λ. Taking into account the Note 2, we can assume that Ker T ∗ = 0. If f is 1 orthogonal to the EAV of T , then the vector function F (λ)=(I λT )− f is an entire function of order p and minimal type (according to the− assertion of the fundamental theorem on the estimate of the resolvent and the theorem on a ratio of entire functions). Since the estimate (13) holds outside the sector Ωα,ε with semi-angle α and p 6 π/2α, we obtain from the reﬁned version of the Phragmen-Lindel¨of Theorem, that F (λ) is a linear function. As at the end of the general theorem on completeness, we can show that F (λ) 0 and f =0. ≡ 36 5 Half-range minimality and completeness problems for dissipative pencils

Let us return to the subject of Section 1 and consider the Cauchy problem d du dnu A( i )u(t) = A u iA + ... +( i)nA =0, (1) − dt 0 − 1 dt − n dtn ( i)ju(j)(0) = ϕ , j =0, 1,...,n 1, (2) − j − where u(t) is a function with values in Hilbert space H. Note the following simple result for ﬁnite dimensional space H.

Proposition 5.1. If dim H < and Ker An∗ = 0 then the Cauchy problem (1), ∞ n 1 (2) has a unique solution for any given initial vectors ϕ − . { j}0 Proof. To prove the existence of the solution we consider two approaches. First, h according to the theorem 1.1, the system of Keldysh derived chains y˜k is a basis in Hn. Hence, there exist coeﬃcients ch such that { } { k} h h ϕ = ϕ0,ϕ1,...,ϕn 1 = cky˜k . { − } Xh,k Then the function (see formulas (1.5), (1.6))

h h iλkt h it h 1 (it) 0 u(t)= c e (y + y − + ... + y ) (3) k k 1! k h! k Xh,k is the solution of (1), (2). Second, denoting

n 1 (n 1) u˜(t)= u(t), iu′(t),..., ( i) − u − (t) , { − − } we can rewrite (1), (2) in the form

du˜ 1 i = A u˜(t), A = A − A , (4) − dt 1 0

u˜(0) = ϕ = ϕ0,ϕ1,...,ϕn 1 , (5) { − } where A0, A1 are the operators defined in (1.7). For any bounded operator L we an define the operator 1 1 eL = I + L + L2 + ..., 1! 2! 37 where the series converges in the uniform operator topology. Hence, the solution of (4), (5) can be represented by the formula u˜(t)= eiA tϕ. The first component of the function u˜(t) represents the solution of (1), (2). The uniqueness of the solution of (4), (5) (or (1), (2)) is a well-known fact in the theory of ordinary differential equations. The second approach can be applied for infinite dimensional space H, if all operators Aj, j = 0, 1,...,n, are bounded and An is invertible (in this case the operator A is bounded). But this does not cover some important practical problems of mathematical physics. As we will see soon, for most interesting equations operator A is unbounded and to define the exponent we have to recall some semigroup operator theory. As a rule, the operator A does not generate a C0 semigroup in Hn and we need to look for another space, where it has better prop−erties. But, all our attempts to prove that A is the generator of C0 semigroup in some space will A− be fruitless if there exists a subsequence λks σ( ) such that Im λks (the condition Im λ 6 const for all λ σ(A ) is∈ a necessary condition for →A ∞to be a k k ∈ generator C0 semigroup). Hence, if the spectrum of the pencil A(λ) ( recall that it has the same− spectrum as A ) does not satisfy the condition Im λ 6 const, for all λ σ(A) k k ∈ then the Cauchy problem for the equation (1) is not correctly set. It is a well-known fact from the theory of partial differential equations that the Cauchy problem is correctly set for some types of hyperbolic and parabolic equations but is not so for elliptic equations. Examining some concrete problems, one discovers that when equation (1) originates with an elliptic problem (in this case the order n =2l is even), it makes sense to impose only l conditions at t =0. For example, ( i)ju(j)(0) = ϕ , j =0, 1,...,l 1. (6) − j − If the equation (1) is considered on the finite interval t [0,T ], then one has to impose another l condition at t = T , for example ∈ ( i)ju(j)(T )= ψ , j =0, 1,...,l 1. (7) − j − j+l (j+l) (These conditions may be different. For example, ( i) u (T ) = ψj, j = 0, 1,...,l 1). The case T = is of special interest. In− this case the conditions (7) are replaced− by the condition∞ lim u(t)=0, (8) t →∞ 38 of u(t) = const, 0

iλkt u(t)= cke yk. (10) ImXλk>0 We put in this sum only the eigenvectors yk, corresponding to the eigenvalues λk with positive imaginary part, otherwise the condition (8) does not hold. Also, we have to satisfy the conditions (6). From (10) and (6) we obtain

l 1 ϕ = ϕ0,ϕ1,...,ϕl 1 = ck yk, λkyk,...,λk− yk . (11) { − } { } ImXλk>0 l 1 The vectors yˆk = yk, λkyk,...,λk− yk we call the Keldysh derived chains of length { } h l. The deﬁnition of the derived chains yˆk , corresponding to the associated vectors is h similar to the deﬁnition of y˜k given in Section 1. Denote by + ( +) the system yˆ of Keldysh derive chains of length l corre- E0 ER { k} sponding to all eigenvalues λk with Im λk > 0 (Im λk > 0) . Proposition 5.2. If dim H < then the problem (1), (6), (8) has a unique ∞l 1 + solution for all given vectors ϕj 0− if and only if the system 0 = yˆk Im λk>0 is a basis in the space H. { } E { } Proof. According to the Proposition 1, any solution u(t) of (1) has the representation (3).If all the eigenvalues are semisimple (this is not essential) and the condition (8) holds, then u(t) must have the representation (10). Hence, the conditions (6) are equivalent to (11). But a vector ϕ Hl can be uniquely represented by the series + ∈ (11) if and only if the system 0 is a basis. Obviously, for the problem (1),E (6), (9) is also valid, but we have to replace the + + system 0 by the system R . Hence we come to the problem which we call the half-rangeE basis problem. Similarly,E we can consider the problem of completeness + + of the system R (half-range completeness) and the problem of minimality of 0 (half-range minimalityE ). E Using the method of G. Radzeivskii [1] (1974) we can easily prove the following result. 39

Theorem 5.1. Let dim H < and the operator pencil A(λ) satisfy the following conditions: ∞ a) Ker An =0; b) Im(A(λ)x,x) 6 0 for all x H and λ R; ∈ ∈ c) there exists a point λ0 R such that 0 / θ(A(λ0)). Then the system + corresponding∈ to the operator∈ pencil A(λ) is complete in Hl. ER l Proof. Suppose that there exists a vector f = f1,...,fl H , which is orthogonal to the system +, i.e. { }∈ ER l 1 ¯ (f1, yk)+(f2, λkyk)+ ... +(fl, λk− yk)=(f(λk), yk)=0 . (12) for all λk with Im λk > 0, where l 1 f(λ)= f1 + λf2 + ... + λ − fl. Consider the meromorphic scalar function

1 F (λ)=([A∗(λ)]− f(λ),f(λ¯)).

1 From the representation of [A∗(λ)]− in the neighborhood of a pole λ¯k and the equalities (12) it follows that F (λ) is a holomorphic function in the lower half-plane. 1 n Since Ker A =0, we have A− (λ) = O( λ ), n || || | | n+2(l 1) 2 F (λ)= O( λ − − )= O( λ − ). | | | | | 1 Denoting g(λ)=[A∗(λ)]− f(λ), we can rewrite F (λ) in the form

F (λ)=(g(λ),A∗(λ¯)g(λ¯))=(A(λ)g(λ),g(λ¯)). Since g(λ) = g(λ¯) for all λ R, it follows from condition b) that Im F (λ) 6 0 for λ R. If Im F (λ)=0 for∈ all λ R, then it follows from the Riemann-Schwartz principle∈ that F (λ) has a symmetric∈ holomorphic continuation in the upper half- plane. Since F (λ) has no poles on the real axis, it is an entire function. Taking into account the estimate (13), we deduce from Liouville’s theorem that F (λ) 0. ≡ Suppose Im F (λ1) = 0 for some λ1 R. Then ϕ(λ) = Im F (λ) is an harmonic bounded function in6 the lower half-plane∈ and ϕ(λ) 6 0 for λ R. Since ϕ(λ) is not identically zero we have (according to the maximum princ∈iple) ϕ(λ) < 0 for Im λ < 0. Now recall the Caratheodory theorem (see [Levin, Ch1.]): If F (λ) is a holomorphic function in the open lower plane and Im F (λ) < 0 for Im λ< 0 then

1 1 F (λ) > F ( i) λ − . | | 5| − | | | 40

This contradicts the estimate (13). Hence F (λ) 0. Now we can write ≡

1 1 (l 1) (l 1) f(λ)= f(λ )+ f ′(λ )(λ λ )+ ... + f − (λ )(λ λ ) − 0 1! 0 − 0 (l 1)! 0 − 0 − (l 1) and using condition c), we can ﬁnd f(λ0)=0,f ′(λ0)=0,...,f − (λ0)=0, i.e. f(λ) 0. This proves the theorem. ≡

Note 5.1. It may seem at ﬁrst sight that one can omit condition c) of Theorem 1. But it is essential. For example, the self-adjoint quadratic operator pencil 0 1 0 3i 0 2 A(λ)= + λ − λ2 1 0 3i 0 − 2 0 i in two-dimensional space H has two eigenvalues λ1 = i and λ2 = 2 in the upper half-plane and the same eigenvector corresponds to both eigenvalues. This example was given by G. Radzievskii [2] and, independently, a similar example was given by A. Kostyuchenko. Certainly, condition c) automatically holds for monic operator polynomials.

Denote by − ( −) a system of Keldysh derived chains of length l corresponding E0 ER to the eigenvalues λk with Im λk < 0(Im λk 6 0). Corollary 5.1. If the condition a) - c) of Theorem 1 are fulﬁlled and the operator + pencil A(λ) has no real eigenvalues, then both systems 0 and 0− form bases in the space Hl. E E

Proof. Under our assumptions E− = E− and the completeness of the system − 0 R E0 can be proved using the same methods. Hence, dim(Span 0±) = κ± > lm, where + + E l m = dim H. Since κ + κ− = nm =2lm, we have κ = κ− = lm = dim H . If the operator pencil A(l) has real eigenvalues, we cannot deduce from Theorem 1 + l that the systems 0 , 0− are minimal in the space H . In fact they are minimal and we will prove evenE a moreE general fact for infinite dimensional space H. Definition 5.6. The system ek 1∞ in the Hilbert space H is called linearly independent if any finite system{ is linearly} independent.

Note 5.2. Certainly, if the system ek 1∞ is minimal then it is linearly independent. { } k The converse assertion is not true. For example, the system of functions x ∞ { }k=0 ∈ L2[0, 1] is linearly independent, but it is not minimal in L2[0, 1]. 41

Deﬁnition 5.1. The point λk σ(A) is called a point of the discrete spectrum of ∈ 1 the operator pencil A(λ) if it is an isolated point of σ(A) and the resolvent A− (λ) has the representation (2.2) in some neighborhood of λk. The set of all such points we denote by σd(A). The following theorem is also due to G. Radzievskii [2] (1987).

Theorem 5.2. Let the operator pencil A(λ) satisfy the conditions b), c) of Theorem 1 and all operators A , j = 0, 1,...,n, be bounded. Let + be the system yˆ j E0 { k} consisting of all Keldysh derived chains of length l, corresponding to λk σd(A) with Im λ > 0. Then the system + is a linearly independent system. ∈ k E0 Proof. Suppose that

N N l 1 c yˆ = c y , λ y ,...,λ − y =0, (13) k k k{ k k k k k} Xk=1 Xk=1 where λ σ (A) and Im λ > 0. Consider the meromorphic scalar function k ∈ d k N c y f(λ)=(A(λ)d(λ),d(λ¯)), d(λ)= k k . λ λk Xk=1 − Notice that is a regular point for d(λ) and at inﬁnity is the Laurent expansion ∞ 1 2 d(λ)= a1λ− + a2λ− + ..., where N j 1 aj = ckλk− yk, j =1, 2,.... Xk=1 Evidently, A(λ) 6 M λ 2l, 2l = n. Thaking into account (14), we have a = a = || || | | 1 2 ... = al =0. Hence

2l 2(l+1) 2 F (λ) = O( λ − )= O( λ − ), λ . | | | | | | →∞ N ¯ N It appears that the function F (λ) may have poles at the points λk 1 and λk 1 . N { } { } However, since A(λk)yk =0, all points λk 1 are regular and F (λ) is holomorphic function in the closed upper half-plane.{ It follows} from condition b) that Im F (λ) 6 0 for all λ R. Now we can repeat the arguments in the proof of Theorem 1 and ∈ 42

(j) deduce F (λ) 0. Using condition c), we obtain d (λ0)=0, for all j = 0, 1,... . Then it follows≡ from the uniqueness theorem for holomorphic functions, that d(λ) ≡ 0. Hence, ck =0, k =1,...,N. We can easily deduce from Theorems 1 and 2 the following assertion.

Corollary 5.2. If conditions a) - c) of Theorem 1 are fulﬁlled then for any given l vectors ϕj 0 there exists a solution of problem (1), (6), (9). Under conditions b), c) of Theorem{ } 1 the solution of problem (1), (6), (8) is unique.

Note 5.3. If the pencil A(λ) has real eigenvalues then under the same assumptions we cannot guarantee the existence of solution (1), (6), (8) and the uniqueness of the solution (1), (6), (9). Hence, in this case we have to replace condition (8) or (9) by a more reﬁned one. To do this we have to consider some concrete problems in mathematical physics. 43

6 Mandelstam radiation principle (non-resonant case) and half-range problems

The linear diﬀerential equations which arise in the theory of electromagnetic waves in elasticity theory can often be reduced to the following form:

∂2u ∂u ∂2u A iB + Cu + =0, (1) − ∂z2 − ∂z ∂t2 where the function u(t, z) = u(t,x,y,z) takes values in a Hilbert space H, the variable t denotes the time, operators A, B are symmetric and C is self-adjoint in H. Such kinds of equations arise in the wave-guide regions Q = Ω R+, where Ω is a bounded domain in the x y plane and the direction of z is orthogonal× to − this plane (see ﬁgure 1). Then the role of H is played by the space L2(Ω). Plane wave-guide regions may also be considered. In this situation Ω is an interval (see ﬁgure 2). The solutions of the wave equation (1), which are periodic in time, i.e.

u(t, z)= v(z)eiωt, v(z)= v(z,x,y) (2) are of considerable interest. The constant ω is called the angular frequency. Substi- tuting (2) into (1) we obtain the equation of stable oscillations with given frequency ω d2v dv A iB + Cv ω2Iv =0. (3) − dz2 − dz − Let wk be the eigenvectors, corresponding to the eigenvalues λk, of the related operator pencil L (λ)= λ2A + λB + C ω2I. (4) ω − iλkz The elementary solutions vk(z) = wke of equation (3) are called propagating waves if λ R and the evanescent waves if Im λ > 0. The waves with Im λ < 0 k ∈ k k have no physical meaning. The eigenvectors wk are called the amplitudes and λk are called the wave-numbers. Example. The simplest but also important equation of type (3) is Helmholtz’ equation in the semi-strip Q = [0, 1] [0, ) × ∞ ∂2 ∂2 ∆v ω2v =0, v = v(x, z), ∆= + , (5) − − ∂x2 ∂z2 v(0, z)= v(1, z)=0, 0 6 z < . (6) ∞ 44

Denote by C the operator Cw = w′′ with domain of deﬁnition − D(C)= w w W 2[0, 1], w(0) = w(1) = 0 , { | ∈ 2 } 2 where W2 [0, 1] is Sobolev space consisting of the functions w(x) such that w and w′ are absolutely continious and w′′ L [0, 1]. Then C is a positive self-adjoint ∈ 2 operator (see, for example, [Najmark]) in the space H = L2[0, 1] and the problem (5), (6) can be rewritten in the form d2v + Cv ω2v =0. (7) − dz2 − We also have to impose the initial condition

v(x, 0) = v(0) = ϕ, ϕ = ϕ(x) L [0, 1], (8) ∈ 2 and to deﬁne a restiction on the behaviour of the solution v(z) when z . The elementary solutions of equation (7) have the representation →∞

v (z)= w (x)eiλkz, k = 1, 2,..., k k ± ± 2 2 2 where λk = √ω π k and wk(x) = sin πkx are the solutions of the eigenvalue problem − (λ2I + C ω2I)w =0. − If ω < π then the pencil λ2I + C ω2I has no real eigenvalues and we can represent the soluiton of the problem (7), (8)− in the form

∞ ∞ √π2k2 ω2 z v(z)= ckvk(z)= cke− − sin πkx, Xk=1 Xk=1 1 where ck = π (ϕ(x), sin πkx)L2. Obviously, this solution satisfies the condition v(z) 0 when z . (9) → →∞ If ω > π then the equation (7) has a finite set of real wave-numbers λk (see figure 3). Now we can not find for any ϕ H = L2[0, 1] the solution of the problem (7), (8) satisfying the condition (9). But∈ if we replace the condition (9) by the condition v(z) 6 const, we can not guarantee the uniqueness of the solution. For example, if|| ϕ(x||) = sin πx then both functions

+ i√ω2 π2 z i√ω2 π2 z v (z)= e − sin πx, v−(z)= e− − sin πx 45 are bounded and satisfy (7), (8). Hence, to select the unique solution, we consider not all propagating waves of the equation (7) but only half of them. How does one choose this half? In our particular case the answer is easy: for example, we can choose the half of propagating waves corresponding to all positive wave-numbers λk. The real wave-number λk characterizes the phase velocity of the corresponding wave. The wave with λk > 0 runs to positive inﬁnity, but the wave with λk < 0 runs to negative inﬁnity. Thus we can select the propagating waves according to their phase velocities and claim that the waves with positive phase velocity have physical meaning, but the waves with negative phase velocity do not. Then we come to the following principle.

Sommerfeld Radiation Princliple. The solution of equation (7) must have the representation v(z)= v0(z)+ v1(z), where v0(z) 0 if z and v1(z) is a ﬁnite superposition of propagating waves with|| positive|| → phase→ velocity, ∞ i.e.

iλkz v1(z)= ckwke , λXk>0 where wk are the amplitudes and ck are some coefficients. It is not{ difficult} to prove the uniqueness{ } and existence of the solution of (7), (8), satisfying the Sommerfeld radiation principle. But more than 50 years ago physicists discovered that the Sommerfeld radiation principle does not work for more complicated equations type (3). To establish a new principle we have to introduce the group velocity of the wave. Using Rellich’s theorem (see Ch. 7.2 of Kato [1]) we find that the real eigenvalues λk = λk(ω) of the operator pencil (4) are holomorphic functions of ω with the exception of some exclusive frequencies ω = ξ , ξ , which are called the resonance frequencies (ω is called a resonance k k →∞ frequency if there exists at least one real eigenvalue λk which is not semisimple). Hence, for all non-resonant frequencies the functions λk′ (ω) are well defined. The 1 iλkz number ′ is called the group velocit y of the propagating wave vk(z)= wke . λk(ω) Now we can formulate the other principle of wave selection.

Mandelstam Radiation Principle. The solution of equation (3) must have the representation v(z)= v0(z)+ v1(z), 46

where v0(z) 0 when z and v1(z) is a finite superposition of propagating waves with|| positive|| → group velocity.→∞ Our next goal is to obtain the different representations for the group velocity 1/λk′ (ω). Further, we suppose that C is a self-adjoint operator and the domains of definition of the symmetric operators A,B contain the domain of C. Proposition 6.1. If ω > 0 is not a resonant frequency of the self-adjoint operator pencil (4) then 1 (L (λ )w ,w ) = ω′ k k k (10) λk′ (ω) 2ω(wk,wk) where Lω′ (λk)=2λkA + B and wk is the corresponding amplitude. 6 Proof. Let λk be a real eigenvalue corresponding to the amplitude wk of the pencil Lω(λ) with a non-resonant frequency ω > 0. For fixed λ in the real neighborhood of the point λk consider the eigenvalue problem (L (λ) ξ2I)y(λ)=0 L (λ)= Aλ2 + Bλ + C, (11) 0 − 0 viewing ξ as a spectral parameter. According to Rellich’s theorem (see Ch. 7.2 of Kato [1]) the eigenvalue ξk = ξ(λ), such that ξ(λk) = ω > 0, is a holomorphic function of λ in the real neighborhood of λk and there is a corresponding holomorphic eigenvector yk = y(λ), y(λk)= wk. It follows from (11) that

2 [L′ (λ) 2ξ(λ)ξ′(λ)]y(λ)+[L (λ) ξ (λ)I]y′(λ)=0. 0 − 0 −

Substituting λ = λk, ξ(λk)= ω, y(λk)= wk, we obtain from this equation

([L′ (λ )w 2ωξ′(λ )w + L (λ )y′(λ )],w )= 0 k k − k k ω k k k =(L′ (λ )w ,w ) 2ωξ′(λ )(w ,w )=0. ω k k k − k k k Taking into account that

1 dλk(ξ) dξ(λ) − 1 λ′ (ω)= = =[ξ′(λ )]− , k dξ dλ k ξ=ω λ=λk

we obtain the relation (10).

6Similar assertions were discovered for matrix polynomials by I. Gohberg, P. Lancaster and L. Rodman [1] (1979), by I. Vorovich and V. Babeshko [1] (1979), by A. Kostyuchenko and M. Orazov [1] (1981), by A. Zilbergleit and Ju. Kopilevich [1] (1983). 47

Proposition 6.2 (Kostyuchenko and Shkalikov [1], 1983). If λk is a simple real eigenvalue of the operator pencil Lω(λ) then the principal part of the resolvent 1 Lω− (λ) at the pole λk has the representation ε ( ,w )w k · k k , λ λ − k where 1 εk = . (12) (L′(λk)wk,wk)

Proof. Assuming that λk is a simple eigenvalue of Lω(λ), we imply that it is also a 1 point of the discrete spectrum. Hence the principal part of Lω− (λ) has the representation ( , z ) w · k k , λ λ − k where z Ker L∗ (λ ). Since L∗ (λ )= L (λ ), we can ﬁnd a number ε such that k ∈ ω k ω k ω k k zk = εkwk. From the identity

1 wk = Lω(λ)Lω− (λ)wk εk(wk,wk)wk = L (λ )+(λ λ )L′ (λ )+ ... + R(λ )w + ... ω k − k ω k λ λ k k − k = ε (w ,w )L′ (λ )w +(λ λ )[...]+ ... k k k ω k k − k we ﬁnd (wk,wk)= ε(wk,wk)(Lω′ (λk)wk,wk) and the equality (12) follows.

Note 6.1. The assertion of Proposition 2 is valid not only for seladjoint operator pencil Lω(λ) but for arbitrary operator pencil A(λ), satisfying the condition Im(A(λ)x,x) 6 0 for all x H and all λ R. Indeed, according to note 4.2, we ∈ ∈ have in this case Ker A(λk) = Ker A∗(λk), hence we can repeat the arguments in the proof of Proposition 2.

For the simple eigenvalue λk the number εk = sign(L′(λk)wk,wk) is called the sign charachteristic of the corresponding eigenvector wk (see, for example, Ch. 10 of Gohberg, Lancaster and Rodman [2]). According to the deﬁnition given in the 48

papers Daffin [1] and Langer and Krein [1], the simple eigenvalue λk is called an eigenvalue of positive (negative) type if the corresponding number εk > 0(< 0). Taking into account Note 1 and the equality sign λk′ (ω) sign εk, we can reformulate the Mandelstam radiation principle for differential −equations of an arbitrary order, if the corresponding operator pencil is dissipative. As we have mentioned, the concrete problems of mathematical physics involve unbounded operators. In this lecture we will consider the problem on solvability of equations of arbitrary order, but only in finite dimensional space H. As in Section 5, let us consider the problem (n =2l) d dv dnv A( i )v(z)= A v iA + ... +( i)nA =0, (13) − dz 0 − 1 dz − n dzn ( i)jv(j)(0) = ϕ , j =0, 1,...,l 1, (14) − j − where A0,...,An are operators acting in finite dimensional space H. Assume, that the related operator pencil n n A(λ)= A0 + λA1 + ... + λ A is dissipative, i.e. Im(A(λ)x,x) 6 0 for all x H and all λ R. (15) ∈ ∈ Assume also that all real eigenvalues of A(λ) are simple. Now, let us introduce the systems E± which we call the first and the second part of eigen and associated vectors of A(λ) respectively: + h E = wk Im λk>0 wk λk R,εk>0 , { } ∪{ } ∈ h E− = wk Im λk<0 wk λk R,εk<0 . { } ∪{ } ∈ + Define also systems the systems ( −) consisting of Keldysh derived chains of E h E + length l, corresponding to vectors w E (E−). Remembering the definition of k ∈ the systems 0±, R± given in Section 5, we notice that 0± ± R±. The followingE problemE is of our interest: to find a solutionE of⊂equation E ⊂ E (13) satisfying the initial conditions (14) and the Mandelstam radiation principle at infinity. We say this is the half-range Cauchy problem. Observe that (in finite dimensional case only!) a solution v(z) of (13) satisfies the Mandelstam radiation principle if and only if v(j)(0) Span E+, j =0, 1,...,n 1. ∈ − Repeating the arguments in the proof of proposition 5.2 we obtain the following result. 49

l 1 Proposition 6.3. For any given initial vectors ϕj − there exists a unique solution of the problem (13), (14), satisfying the Mandelstam{ } radiation principle at inﬁnity if and only if the system + is a basis in Hl. E Theorem 6.1 (Shkalikov [2], 1985). Let dim H < , the pencil A(λ) satisﬁes the condition (15), Ker A =0 and let there exist a point∞ λ R, such that n 0 ∈ 0 / θ(A(λ )). (16) ∈ 0 + l then the systems and − are complete in H . E E Proof. Consider, for example, the system +. Suppose there exists a vector f = f ,...,f Hl, which is orthogonal to theE system +, i.e. { 1 l}∈ E l 1 λk(f1, yk)+(f2, λkyk)+ ... +(fl, λn− yk)=0 for all λk with Im λk > 0 and real λk is of positive type. Consider the meromorphic scalar function

1 l 1 F (λ)=([A∗(λ)]− f(λ),f(λ¯)), f(λ)= f1 + λf2 + ... + λ − fl 1, − where A∗(λ)=[A(λ¯)]∗. Repeating the arguments in the proof of Theorem 5.1, we conclude that F (λ) has no poles in the lower half-plane as well as in the real points λk of positive type. Hence, on the real axis the function F (λ) may have poles only at points λk of negative type and according to Proposition 2 the principal part of F (λ) at pole λ = λk is equal to ε (f(λ ),w )(w ,f(λ )) ε (f(λ ),w ) 2 k k k k k = | k k | . (17) λ λ λ λ − k − k 1 q Consider the contour Γ = CR I1 Cε ... Iq Cε Iq+1, which is depicted in ∪ ∪ ∪ ∪ k ∪ ∪ Figure 4 (CR is a large semi-circle of radius R, Cε ,k =1,...,q, are small semicircles of radii ε with centers in poles λ1, λ1,...,λq and I1,...,Iq+1 are intervals on the real axis). Then

q+1 q F (λ)dλ = + + F (λ)dλ =0 (18) k ZΓ ZCR k=1 ZIk k=1 ZCε X X 2 Since Ker An =0, we deduce (see the estimate (5.13)) that F (λ) = O( λ − ) when λ , hence | | | | →∞ F (λ)dλ 0, when R . (19) → →∞ ZCR 50

According to (17), we have

q q 2 F (λ) πi εk (f(λk),wk) if ε 0, (20) k Cε → | | → Xk=1 Z Xk=1 and q ∞ F (λ)dλ V.P. F (λ)dλ if ε 0, R . (21) Ik → → →∞ Xk=1 Z Z−∞ Hence, from (18) - (21) we have

q ∞ πi ε (f(λ ),w ) 2 + V.P. F (λ)dλ =0. (22) k| k k | Xk=1 Z−∞ Now notice that ε < 0 in (22) for all k and Im F (λ) 6 0 if λ R (see Theorem k ∈ 5.1). Then we immediately obtain from (22) that (f(λk),wk)=0, k = 1,...,q, and Im F (λ) 0 for λ R. Hence, F (λ) is a real fuction on the real axis and has no real poles.≡ Using the∈ Riemann-Schwartz symmetry principal and Liouville’s theorem, we obtain F (λ) 0. Then condition (16) allows us to conclude f(λ) 0. ≡ ≡

Theorem 6.2 (Radzievskii [2], 1987). Let the conditions of Theorem 1 hold with the exception of conditions Ker A =0 and dim H < . Then the system + and n ∞ E − are linearly independent. E Proof. Following the proof of Theorem 5.2, consider the meromorphic function

N c w F (λ)=(A(λ)d(λ),d(λ¯)), d(λ)= k k , λ λk Xk=1 − where wk are the eigenvectors corresponding to eigenvalues λk with Im λk > 0 and real λk of positive type. It was proved in Theorem 5.2 that F (λ) is holomorphic in the open upper half-plane. Obviously, the principal part of F (λ) at the real pole λk is equal to 2 (A′(λ )w ,w ) c k k k | k| . λ λ − k By virtue of Note 1 it follows that

sign(A′(λk)wk,wk)= sign εk. 51

x Figure 1

Hence, all residues of the function F (λ) corresponding to real poles are positive, 2 Im F (λ) 6 0 for all λ R and F (λ) = O( λ − ) if λ . Taking the contour Γ depicted in Figure 5 and∈ repeating| the| arguments| | of Theorem→∞ 1, we conclude that F (λ) 0. Now condition (16) shows that d(λ) 0, hence the system + is linearly ≡ ≡ E independent. The same arguments apply to the system −. As a corollary of Theorems 1 and 2 and Proposition 1 weE obtain the following result.

Theorem 6.3. Let dim H < and the pencil A(λ) satisfy the conditions of The- ∞ l 1 orem 5.1. Then for any given initial vectors ϕj 0− there exists a unique solution of the problem (13), (14) satisfying the Mandelstam{ } radiation principle at inﬁnity.

Note 6.2. If Ker An =0 and dim H < then the duality principle is valid: if the + l ∞ systems and − are complete in H then they are linearly independent and vice E E + + versa. Indeed, let x (x−) denote the number of vectors of the system ( −) and E E let dim H = m. It follows from the deﬁnition of the systems ± and Theorem 1.1 that E + x + x− = mn =2ml. (23) + + + If and − are complete then x > ml and x− > ml. Now (23) implies x = E E + + x− = ml, hence the system ( −) is a basis. On the contrary if and − are +E E E E linearly independent then x 6 ml, x− 6 ml and we obtain again from (23) that + ( −) is a basis. E E 52

x v(1, z)=0 1 v(x, 0) = ϕ ∆v ω2v =0 − 0 v(0, z)=0 z

Figure 2

Figure 3 (wave-number distribution of Helmholtz equation.)

7 Generalized Mandelstam radiation principle (resonant case). The factorization of a quadratic pencil

We have assumed in proving theorems on half-range completeness and half-range minimality that all eigenvalues of the operator pencil A(λ) are simple. Certainly, the simplicity of the non-real eigenvalues is not essential. For example, if vector f = f1,...,fℓ is orthogonal to all Keldysh derived chains of length l corresponding to eigenvalues λ with Im λ > 0 then the vector function k k 1 ℓ 1 [A∗(λ)]− (f + λf + + λ − f ) 1 2 · · · ℓ is holomorphic in the lower half-plane (see Note 3.2). Hence, the proof of Theo- rem 6.1 does not change. Similarly, in Theorem 6.2 we have only to replace the function d(λ) by the function p k c wh c w0 d(λ)= k,h k + k k , (1) pk+1 h (λ λk) − λ λk Im λk>0 h=0 λk R,εk>0 X X − ∈X − 0 1 pk where wk,wk,...,wk are the chains of canonical systems corresponding to the eigenvalues λk (λk is repeated as many times as its geometric multiplicity). But if the 53

I1 I2 Iq+1 1 2 ε Cε Cε Cq

Figure 4

CR λ-plane

1 1 Cε Cq

I1 Iq+1

Figure 5 real eigenvalues of A(λ) are not simple, then we come to a new problem, which is serious even in the ﬁnite dimensional case. In this situation we have to select the proper subset of elements from the canonical system

0 1 pk wk,wk,...,wk , k = N1,...,N2 (2) corresponding to the real eigenvalue λk = c (the number N2 N1 +1 is the geometric multiplicity of c). − Actually, we can not select the proper subset from any canonical system (1). First we have to choose a special canonical system and then to divide it. The problem on selection of elements from the canonical system (1) was originated in the paper of Kostyuchenko and Orazov [2] (1975). In this paper an important supplement was made to the remarkable theorem of Krein and Langer [1], which asserts: if 2 L(λ) = λ I + λB + C, where B = B∗ is bounded and C is a positive compact operator then L(λ) admits a factorization L(λ)=(λI + B Z)(λI + Z), − such that the spectrum of the operator Z lies in the closed upper-half plane and coincides with the spectrum of L(λ) in the open upper-half plane. A natural 54 question arises: How to divide the Jordan chains of L(λ), corresponding to the real eigenvalues λk σ(L), to obtain the Jordan chains of Z? This problem can be solved by using a geometrical∈ approach, because the problem on factorization of L(λ) is equivalent to the existence of the maximal invariant subspace for the linearization Z in Klein space with indeﬁnite metric G, where

1 B C 2 I 0 Z = 1 , G = . C 2 0 0 I − − Investigations in this ﬁeld have a long history. Important ideas on connection between factorization and existence of the maximal invariant endspace were developed in the paper Langer [3]. For ﬁnite dimensional space a comprehensive treatment of this theory can be found in the books of Gohberg, Lancaster and Rodman [2-4]. However, we will use an analytic, rather than a geometric approach. First we will prove the existence of a special canonical system and then we will be able to divide it and to select the proper part. The following result is due to Kostyuchenko and Shkalikov [1] (1983). Theorem 7.1 (Theorem on the Existence of a Normal Canonical System). Let A(λ) be holomorphic self-adjoint7 operator function in a neighborhood of the real point 1 c and c be the point of discrete spectrum of A(λ), i.e., the resolvent A− (λ) has a 1 pole at λ == c and the principal part of A− (λ) at this pole has the representation (cf.(2.2))

N 2 ( , z0)w0 ( , z1)w0 +( , z0)w1 · k k + · k k · k k + ... (λ c)pk+1 λ c)pk kX=N1 − − pk 0 pk 1 1 0 pk ( , z )w +( , z − )w + +( , z )w + · k k · k k · · · · k k (3) λ c) − where 0 1 pk zk, zk,...,zk , k = N1,...,N2 (4) is the adjoint canonical system to the canonical system (2) of EAV of the operator function A(λ). Then a canonical system (2) can be chosen in such a way that

h h zk = εkwk , k = N1,...,N2; h =0, 1,...,pk, (5)

7An operator function A(λ) is called self-adjoint in a neighborhood of point c if there exist ε > 0 such that A(λ) = [A(λ)]∗ for all λ : c ε<λ pq+1 pN . Denote by Lh the operators of finite range coinciding with ≥···≥ p 1 h the coefficients of the powers (λ c)− 1− − , h =0, 1,...,p , in the representation − 1 (3). Evidently, the operators Lh are self-adjoint. In particular, the operator L0 is 1 0 self-adjoint, hence we can find the vectors e1,...,eq such that q L = ε ( , e )e , ε = 1. 0 k · k k k ± Xk=0 0 q 0 q 0 q Evidently, the vectors zk 1 and wk 1 lie in Span ek 1. Then we obtain from (3) the following representation{ } { } { }

q q+q1 L = ε ( ,w1)e0 +( , e0)w1 +( ,f )e0 + ( , z0)w0, (6) 1 k · k k · k k · k k · k k Xk=1 kX=q+1 1 1 where fk = zk wk and q1 is the number of chains with length equal to p1 1. − 0 q 0 q+q1 0 q+q1 − Denote H1 = Span ek 1,H2 = Span wk q+1 . Since wk 1 are linearly independent (this follows{ from} the deﬁnition{ } of a canonical{ system),} we have H H = . Hence, we obtain the unique representation z0 = ϕ + ψ , where 1 2 ∅ j j j ϕj H1, ψj H2. Consider the operators ∈T ∈ q+q1 q q+q1 B = ( , ψ )w0, C = ε ( ,f )e0 + ( ,ϕ )w0. 1 · k k 1 k · k k · k k kX=q+1 Xk=1 kX=q+1 The operator L1 is self-adjoint and from the representation (6) we ﬁnd that B1 + C1 is self-adjoint. Now notice, that in the orthogonal basis consisting of elements 8 0 0 0 0 e1,...,eq, wq+1,...,wq+q1 we have the matrix representation B1 = bjk and bjk may be not equal to zero only in the right lower quadrant, i.e. if min j,{ k }>q. On the contrary, all elements of the matrix C = c in the same basis{ are} equal to 1 { jk} zero if min j, k > q. Since B1 + C1 is self-adjoint, we have in this situation that both operators{ B} and C are self-adjoint. Hence there exists a basis e0 q+q1 in the 1 1 { k}q+1 space H2 such that q+q1 B = ε ( , e0)e0 1 k · k k kX=q+1 8 0 N We may choose a canonical system (2) so that wk 1 is an orthogonal system. Then the system 0 0 0 0 { } e1,...,eq, wq+1,...,wq+q1 is orthogonal. 56

Taking into account the matrix representation of the operator C1 = C1∗, we can choose the elements x1,...,xq such that q C = ( ,x )e +( , e )x . 1 · k k · k k Xk=1 1 1 Then denoting ek = wk + εkxk, we obtain from representation (6) the following one

q q+q1 L = ε [( , e1)e0 +( , e0)e1]+ ε ( , e0)e0. 1 k · k k · k k k · k k Xk=1 kX=q+1 Next we can represent the operator L2 in the form q L = ε [( ,w2)e0 +( , e1)e1 +( , e0)w2 +( ,f )e0]+ 2 k · k k · k k · k k · k k Xk=1 q+q1 q+q1+q2 (7) + ε [( ,w1)e0 +( , e0)w1 +( ,f )e0]+ ( , z0)w0 k · k k · k k · k k · k k kX=q+1 k=qX+q1+1 where

2 2 zk wk, if k =1,...,q, fk = − z1 w1, if k = q +1,...,q + q . ( k − k 1 Now we can see from representation (7) that we can apply the same arguments as before and obtain the representation

q L = ε [( , e2)e0 +( , e1)e1 +( , e0)e2]+ 2 k · k k · k k · k k Xk=1 q+q1 q+q1+q2 + ε [( , e1)e0 +( , e0)e1]+ ε ( , e0)e0. k · k k · k k k · k k kX=q+1 k=Xq+q1+1

The same arguments can be repeated for operators L3,...,Lp1. Then we obtain the assertion of the theorem.

Deﬁnition 7.1. A canonical system (2) satisfying to the relations (5) is called a normal canonical system. The numbers εk appearing in (4) are called the sign characteristics of the chains of the normal canonical system. 57

Now we would like to establish a similar result for dissipative operator functions. The operator functions. The operator function A(λ) is called dissipative in the neighborhood of the real point c if there exists an ε> 0 such that Im(A(λ)x,x) 6 0 for all x H and λ : c ε<λ

Theorem 7.2 (Theorem on the Existence of the Regular Canonical System). Let a real point c of discrete spectrum of the operator pencil A(λ) and A(λ) be dissipative in the neighborhood of c. Let (2) be a canonical system corresponding to the eigenvalue c of A(λ) and (4) be the adjoint canonical system. Then

Span wh N2 αk = Span zh N2 αk := S0 (8) { k }k=N1, h=0 { k }k=N1, h=0

pk 1 0 where αk =[ 2− ] (if pk =0 then αk = 1 and we assume that the vector wk does not belong to S0). Moreover, a canonical− system (2) can be chosen in such a way that for all indices k satisfying the conditions pk = 2ℓk (i.e. for all chains of odd length) a representation

zℓk = ε wℓk + w, whereε = 1, w S0 (9) k k k k ± ∈ is valid.

Proof. The proof of this theorem is rather complicated and can be found in the paper Shkalikov [3]. Here we omit it. For the linear operator pencils this Theorem follows from Propositions 8.6, 8.7 of the next Section.

ℓk ℓk Note 7.1. It may happen, that wk = 0, then zk = 0 to (it is possible only if the order n of the pencil A(λ) is greater than 2). In this case the equality (8) does not determine the sign εk and it has to be determined from the equality

ℓk ℓk z˜k = εkw˜k +w, ˜

ℓk where z˜k (= 0!) denotes the Keldysh derived chain of length n corresponding to the ℓk6 element zk . Thus, in addition to (9) the following equality holds (see Shkalikov [3])

(Gz˜ℓk , zℓk )=(Gw˜ℓk ,wℓk )= λnε , (10) k k k k − k k 58

where n is the order of A(λ), εk = 1, λk = c =0 (otherwise we have to shift the spectral parameter) and ± 6

0 0 ... 0 A0 0 0 ... A A  0 1  G = ......  0 A0 ... An 3 An 2  − −  A0 A1 ... An 2 An 1  − −    Deﬁnition 7.2. A canonical system (2) satisfying to the relations (9), (10) is called a regular canonical system. The numbers εk appearing in (9), (10) are called the sign characteristics of the corresponding chains. Note 7.2. Hence for dissipative operator pencil we cannot introduce the sign characteristics for all chains, but only for chains of odd length. A natural question arises: Do the sing characteristics exist for the chains of even length? A simple example shows that the answer is negative. Consider the dissipative operator pencil

A(λ)= I P iCλ2, P =( , e )e , e =1, C> 0. − 0 − 0 · 0 0 0 Obviously, the point λ =0 is a point of discrete spectrum of A(λ) and the principal 1 part of A− (λ) at this point is equal to i( , e )e · 0 0 . λ2 Hence, the direct and the adjoint canonical systems coincide in this case with chains e , 0 and ie , 0 respectively. 0 − 0 The theorem on existence of the regular canonical system allows us to formulate the Mandelstam radiation principle for resonant frequencies, i.e., for the case when A(λ) has real eigenvalues λk which are not semi-simple. For this case we have not met the formulation of this principle in physical literature. Deﬁnition 7.3. Let (2) be a regular canonical system corresponding to real eigenvalue µ(= λk).Let Ek = 1 be sign characteristics corresponding to Jordan chains of odd length and let E =0±for the Jordan chains of even length. We say the solution V(Z) of equation (6.13) satisﬁes the generalized Mandelstam radiation principle at if V (Z) admits the representation ∞

V (Z)= V1(Z)+ V0(Z), 59

(j) where V0 0, j = 0, 1,..., (n 1), when Z and V1(Z) is a superposition of elementary→ solutions of the following− type: → ∞

h h iλkZ h Z h 1 Z 0 pk + εk V (Z)= ℓ (W + W − + + W ), 0 6 h 6 (11) k k 1! k · · · h! k 2 If p =0 and E = 1 then [1/2] = 1 and we assume that no elementary solutions k k − − corresponding to this index k is involved in a superposition V1(Z). We save the same definition for the generalized Mandelstam radiation principle at ; the only difference is that we have to replace the inequalities for h in (11) −∞ 6 6 (pk εk) by 0 h −2 . Denote by E+ the first half of the eigen and associated vectors of A(λ), namely,

+ h h E = wk Im λk>0 wk . { } ∪{ } (pk+εk) λk R, 06h6 ∈ 2 Thus, the system E+ consists of all vectors from the canonical system (2), corresponding to all eigenvalue λk with Im λk > 0 and of selected vectors from the regular canonical system, corresponding to real eigenvalue (this selection is produced according to the sign characteristics). Similarly denote

h h E− = wk Im λk<0 wk . − { } ∪{ } (pk εk) λk R, 06h6 ∈ 2 E + E h h Now we can determine ( −) = wˆk , where wˆk are the Keldysh derived chains { } h + h h of length ℓ corresponding to the vectors w E (E−). Obviously, wˆ = Tv where k ∈ k k ℓ 1 (ℓ 1) Tv(Z)= V (0), iv′(0),..., ( i) − V − { − − } h and Vk (Z) are deﬁned by (11). Theorem 7.3 (Theorem on Completeness). If the conditions of Theorem 6.1 hold + ℓ then the systems E and E − are complete in the space H . Proof. First, for simplicity we suppose that A(λ) is a quadratic operator pencil, hence ℓ =1. In this case E + = E+. Assume that there exists a vector f, which is orthogonal to all elements of the system E+. Then the scalar function

1 F (λ)=([A∗(λ)]− f,f), A∗(λ)=[A(λ¯)]∗, 60 is holomorphic in the lower half-plane (see the proof of Theorem 6.1) and according to (3) its principal part is equal to

N 2 (f, z0)(w0,f) (f, z1)(w0,f)+(f, z0)(w1,f) k k + k k k k + + (λ c)pk+1 (λ c)pk · · · kX=N1 − − (12) (f, zpk )(w0,f)+ +(f, z0)(wpk ,f) + k k · · · k k . (λ c) −

h (pk+εk) h Since (f,wk )=0 for h =0,..., 2 and (8), (9) hold, we have (f, zk )=0 for (pk+εk) s h 6 . Now it follows from (10) that all coeﬃcients at powers (λ c)− , s > 2 − 1, are equal to zero if pk is odd (i.e. the length of the corresponding chain is even) or pk is even but εk =1. Hence, taking into account (9), we ﬁnd that the expression (12) is equal to

(f, zℓk )(wℓk ,f) (wℓk ,f) k k = k . λ c − λ c pk=2ℓk,εk= 1 pk=2ℓk,εk= 1 X − − X − − Thus the function F (λ) may have only simple real poles with non-positive residues. Repeating the arguments in the proof of Theorem 6.1 we obtain F (λ) 0 and f =0. ≡ If ℓ> 1 then we have to consider the function 1 − ℓ 1 F (λ)=( A∗(λ) f(λ),f(λ¯)), f(λ)= f0 + f1λ + + fℓ 1λ − , · · · − + where a vector f0,f1,...,fℓ 1 is orthogonal to all vectors belonging to E . It is an easy exercise{ to show that− } this assumption as before implies the analyticity of F (λ) in the closed lower half-plane with possible exception of a simple real pole with non-positive residues. This gives F (λ) 0 and f(λ) 0. ≡ ≡

Theorem 7.4 (Theorem on Linear Independence). If the conditions of Theorem + ℓ 6.2 hold then the system E and E − are linear independent in the space H . Proof. As in the previous theorem we assume for simplicity that A(λ) is a quadratic pencil. Consider the scalar function F (λ)=(A(λ)d(λ),d(λ¯)), 61 where d(λ) is determined by (1), but the second term in (1) is replaced by

h ck,h wk (pk + εk) d1(λ)= , βk = . βk+1 h R (λ λk) − 2 λk 06h6βk X∈ X − Then A(λ) is holomorphic in the upper half-plane and may have poles on the real axis. Obviously the principal part of the function F (λ) at a real pole λ = λk(= c) coincides with the principal part of the function

F1(λ)=(A(λ)d1(λ),d1(λ)) at this pole. After some technical calculations (they are not simple; see Shkalikov [3], lemma 4) we ﬁnd that the principal part of F1(λ) at the real pole λ = c is equal to

2 ℓk ℓk ℓk ℓk ck,ℓ (Gw˜ , w˜ )(w ,w ) k k k k k . λ c λk=Xc,εk>0 − It follows from (10) that the function F (λ) may have poles on the real axis only with negative residues. Repeating the arguments in the proof of Theorem 6.2 we obtain F (λ) 0 and d(λ) 0. ≡ ≡

Theorem 7.5 (Theorem on solvability of half-range Cauchy Problem). Let the pencil A(λ) satisfy the conditions of Theorem 6.1. Then for given initial vectors ℓ 1 ϕj 0− there exists a unique solution of problem (6.13), (6.14) satisfying the generalized{ } Mandelstam radiation principle at . ∞ Proof. This theorem is a corollary of Proposition 6.3 and two previous theorems.

Note 7.3. For the case when dim H < and Ker An =0 the duality principle is valid as well as in non-resonant case (see∞ Note 6.8). Namely,6 if both systems E + and ℓ E − are complete (or linearly independent) then they are basis’ in H . Indeed, one can easily check that the equality x+ + x = 2ℓm holds in general case as well as in non-resonant case. Therefore, the same− arguments can be applied to prove this fact. 62

The ideas presented in the last two lectures can be extended to obtain similar results for dissipative pencils of odd order as well as for pencils satisfying the condition Im(λA(λ)x,x) 6 0 x H, λ R. (13) ∀ ∈ ∀ ∈ First, suppose that A(λ) satisﬁes condition(6.15) and n =2ℓ +1. Observe that + in this case (6.15) implies An = An∗ . We can represent An = An An− where An± > + + + − + + 0,An An− = An−An =0. Let P and P − be orthoprojectors onto Im An := H and Im An− := H− respectively. Let V(Z) be a function with values in H. Deﬁne "the trace" operators T by formula ± ℓ 1 (ℓ 1) ℓ (ℓ) T V (Z)= V (0), iV ′(0),..., ( i) − V − (0), ( i) P ±V (0) . ± { − − − }

Now, deﬁne the system E ± in such a way that E h ± = T Vk (Z) , { ± } h h where Vk (Z) are elementary solutions of (6.13) such that Vk (0) E±. For example, + + E + E h h ∈ if An > 0 (i.e. p = I) then ( −)= wˆk , where wˆk are Keldysh derived chains { } h + of length ℓ +1(ℓ) corresponding to vectors w E (E−). k ∈ Theorem 7.6. Let the pencil A(λ) satisfy the conditions of Theorem 6.1 and let + ℓ 1 + n =2ℓ +1. Then the system E is a basis in the space H − H while the system ℓ 1 × E − is a basis in H − H−. × + Proof. First, let us prove the completeness of the systems E and E −. Consider, for example the system E +. Assume, that this system is not complete. In this case + there exists a vector f0,f1,...,fℓ 1, P fℓ such that the function { − } 1 F (λ)=([A∗(λ)]− f(λ),f(λ¯)) where ℓ 1 ℓ + f(λ)= f0 + λf1 + + λ − fℓ 1 + λ P fℓ · · · − is holomorphic in the lower half plane and may have only simple poles on the real axis with non-positive residues (see the proof of Theorem 6.1 and the theorem on completeness from this lecture). Let us compute the residue at inﬁnity. Since An is invertible, we have

1 1 + + 1 F (λ)= (A− P f , P f )+ O , λ . λ n ℓ ℓ λ2 →∞ 63

Hence the residue of F (λ) at inﬁnity is also non-positive. Now, we can repeat the arguments which we applied in Theorem 6.1. Then we obtain F (λ) 0 and ≡ f(λ) 0. To prove the basisness we can apply the duality principle (see Note 3). ≡

Corollary 7.1. Let A(λ) satisfy condition (6.15), dim H < ,n =2ℓ +1 and let + ∞ An > 0. Then the system E (E −) consisting of Keldysh derived chains of length h + ℓ + 1(ℓ) constructed from eigen and associated vectors wk E (E−) form a basis in Hℓ+1(Hℓ). ∈

Proof. We have to note only that condition (6.16) holds for sufficiently large λ R, since A > 0. 0 ∈ n Now, we consider the pencils A(λ) satisfying condition (13). Notice that conditions (6.15) and (13) are different - each of them does not imply the other. First, consider the case n =2ℓ. In this case (13) implies A0 = A0∗, An = An∗ . Let (2) be a regular canonical system corresponding to real eigenvalue λk and let δk = sgnλkEk, where Ek are the sign characteristics of the corresponding Jordan chains (we define Ek =0 if the length of the corresponding Jordan chains is even). Let us introduce the systems

+ h h Y = wk Im λk>0 wk { } { } (pk+δk) λk R, 06h6 [ ∈ 2 h h Y − = wk Im λk<0 wk − { } { } (pk δk) λk R, 06h6 [ ∈ 2 + + Consider the spectral decompositions A = A A−,A = A A− (A± > 0 0 − 0 n n − n 0 0,An± > 0) and denote by Q± and P ± the orthoprojectors onto Im A0± and Im An± respectively. For a vector valued function V (Z) deﬁne the "trace" operator

ℓ 1 ℓ 1 ℓ (ℓ) T V (Z)= Q±V (0), iv′(0),..., ( i) − V − (0), ( i) P ±V (0) ± { − − − } and introduce the systems h Y = T Vk (Z) (14) ± { ± } h h where Vk (Z) are elementary solutions of (6.13) such that Vk (0) Y ±. For example, h h ∈ if A0 > 0 and An > 0 then Y = wˆk where wˆk are Keldysh derived chains of − h{ } length ℓ corresponding to vectors wˆ Y −. k ∈ 64

Theorem 7.7. Let the pencil A(λ) satisfy condition (13), n = 2ℓ, Ker A0 = 0 and Ker An = 0. Let also condition (6.16) hold and dim H < . Then the ℓ 1 + + ∞ ℓ 1 system Y+(Y ) forms a basis in the space HQ− H − HP (HQ H − HP−) − × × × × where HQ± = Im Q± and HP± = Im P ±.

Proof. Let us prove the completeness of the systems Y+ and Y . Using the duality principle we obtain the basisness. − + Suppose there exists a vector Q−f0,f1,...,fℓ 1, P fℓ which is orthogonal to the system Y +. Consider the function{ − } 1 1 − F (λ)= ( A∗(λ) f(λ),f(λ¯)), λ 1 ℓ 1 ℓ + − where f(λ)= Q−f0+λf1+ +λ − fℓ 1+λ P fℓ. Denoting g(λ)= A∗(λ) f(λ) · · · − we can write 2 F (λ)= λ− (λA(λ)g(λ),g(λ¯)). Now, it follows from condition (13) that Im F (λ) 6 0 for λ R. Repeating the arguments of Theorem 5.1 we obtain that F (λ) is holomorphic∈ in the lower half plane. We also have

1 1 + + 2 F (λ)= λ− (An− P fℓ, P fℓ)+ O(λ− ) when λ , i.e., the residue of F (λ) at is non-positive too. The function F (λ→) may ∞ have other real poles which∞ are simple and the corresponding residues are non-positive (see Theorem 6.1 and the theorem on completeness in this lecture). Applying the arguments of Theorem 6.1 we ﬁnd F (λ) 0 and f(λ) 0. ≡ ≡

Theorem 8 is most interesting in the case when A0 and An are deﬁnite operators. We oﬀer the reader to formulate a Corollary from Theorem 5 for this case. Finally, if condition (13) holds and n =2ℓ+1 we should introduce the operators

ℓ (ℓ) T V (Z)= Q∓V (0), iV ′(0),..., ( i) V (0) ± { − − } and thence deﬁne the systems Y by equality (14). The following result is valid. ± Theorem 7.8. Let the conditions of Theorem 5 hold but n = 2ℓ +1. Then the ℓ + ℓ system Y+(Y ) forms a basis in HQ− H (HQ H ). − × × Proof. Repeat the arguments of Theorem 7. 65 8 Dissipative and linearly dissipative operator pencils

In this section we continue the study of dissipative operator pencils satisfying the condition Im(A(λ)x,x) 0 for all λ R and x H (1) ≤ ∈ ∈ or the condition

Im(λA(λ)x,x) < 0 for all λ R and x H. (2) ∈ ∈ But now we deal with inﬁnite dimensional space H. Further we will use the notation

A + A∗ A A∗ Aλ = ,AI = − 2 2i Obviously, Aλ and AI are self-adjoint operators and A = Aλ + iAI . Notice, that if A(λ) is a linear pencil A(λ)= A0 + λA1 (3) I then (2) holds if and only if A0 = A0∗ and A1 0. Assume that A0 is an invertible operator. Then the spectral problem for linear≤ pencil (3) is equivalent to the spectral 1 problem for the linear operator A = A0− A1. Obviously, condition (2) holds for linear pencil (3) if and only if A is an A0-dissipative operator, i.e. A is dissipative in the space H with regular indefinite metric (A0x,x). The main goal of this lecture is to show that some properties of dissipative operators in a space with indefinite metric are similar to properties of self-adjoint operators. First we recall some definitions from the theory of operators in Hilbert space withe indefinite metric. Let W be a symmetric operator (it can be unbounded, but we will consider only bounded operators). Denote by [x, y]=(Wx,y) the new scalar product in H which is indefinite if the operator W is indefinite. This metric is called regular if W is bounded and invertible. The Hilbert space with regular indefinite metric generated by the operator W is called a Pontrjagin space if either operator W +W W W W+ = | |2 or W = | |−2 is finite dimensional. The space with− regular indefinite metric is called a Krein space if both operators W+ and W are infinite dimensional. The subspace H1 H is called W -non-positive (nonnegative,− neutral) if [x,x] 0 ( 0, = 0) for all⊂x H . A W -non-positive ≤ ≥ ∈ 1 66

subspace H1 is called maximal non-positive if it is not contained in any other W - 9 non-positive subspace H2. The operator A is called dissipative in the space with indeﬁnite metric [ ] (or W -dissipative) if · Im[Ax,x] 0 for all x H. ≤ ∈ Now we establish some useful properties of W -dissipative operators. 0 h Proposition 8.1. Let A be a W -dissipative operator and S+ = span yk , where h { } yk are EAV of a linear pencil A(λ)= I + λA

0 corresponding to the eigenvalues λk with Im λk > 0. Then S+ is a W-non-positive subspace.

Proof. If yk is an eigenvector of a pencil A(λ) corresponding to an eigenvalue λk iλkt with Im λk > 0, then uk(t)= e yk is a solution of equation

iAu′(t)+ u(t)=0 (4) − and u (t) 0 when t . Similarly, the function k → →∞ h u(t)= ck,huk(t),

h X h 0 where uk(t) are elementary solutions corresponding to EAV yk S+, satisﬁes the equation (4) and u(t) 0 when t . Using (4), we obtain ∈ → →∞ ′ =[u′(ξ),u(ξ)]+[u(ξ),u′(ξ)] =[u′(ξ),iAu′(ξ)]+[iAu′(ξ),u′(ξ)] = (Vu′(ξ),u′(ξ)) − =(u′(ξ),Wu(ξ))+(Wu(ξ),u′(ξ)) (5) =(u′(ξ),iTu′(ξ))+(iTu′(ξ),u′(ξ)) = (T u′(ξ),u′(ξ)), − j I V = 2(WA) = [WA (WA)∗]/i 0. Integrating the equality (5) from t to we obtain − ≤ ∞ ∞ [u(t),u(t)] = (Vu (ξ),u (ξ))dξ. (6) ′ ′ Zt In particular, [u(0),u(0)] = [y, y] 0 for all y S0 . ≤ ∈ +

9In Section 3 we called an operator A dissipative if Aλ 0. But in the mathematical literature the word "dissipative" is also used to denote operators satisfying the condition≤ AI 0. ≤ 67

Note 8.1. Equality (6) has a physical sense. For some particular equations (4) de- scribing physical processes the form [u(t),u(t)] plays the role of an energy functional and operator V = 2(WA)I 0 is responsible for damping. Thus the equality (6) shows that, for systems with≤ damping,the energy functional decreases monotonically when t . →∞ Proposition 8.2. Let A be a W -dissipative operator and

0 1 pk yk, yk, . . ., yk , k = N1,...,N2 (7) be the canonical system of EAV of the pencil W + λW A corresponding to a real eigenvalue λ = c =0. Let k 6 0 1 pk xk,xk,...,xk , k = N1,...,N2 (8) be the adjoint canonical system of EAV of the pencil W + λA∗W corresponding to the same eigenvalue c. If [γ] is the integer part of a number γ and ak =[pk/2] then the elements 0 1 αk yk, yk, . . ., yk , k = N1,...,N2, (9) 0 1 αk xk, xk, ..., xk , k = N1,...,N2, (10) belong to Ker(WA)I . Proof. Denote T = WA. From the deﬁnition of EAV we have

h h 1 1 (cT + W )y = T y − , 0 h p (y− := 0). k − k ≤ ≤ k In particular, 0 0 I 0 0 Im((cT + W )yk, yk)= c(T yk, yk)=0. I 0 I Since T 0 we have yk KerT . Suppose we have proved that 0 h 1 ≥ I ∈ y ,...,y − Ker T for h 1 < α . Then 2h p and k k ∈ − k ≤ k h 1 h h 1 h h 1 h+1 (T yk− , yk )=(T ∗yk− , yk )= (yk− , (cT + W )yk ) h 2 h+1 − 0 2h (11) =(T y − , y )= = (y , (cT + W )y )=0. k k · · · − k k Therefore, h h 1 h I h h 0=Im (cT + W )yk + T yk− , yk = c(T yk , yk ). From this equality we deduce as before that yh Ker T I . Similarly, we can prove k ∈ xh KerT I if h α . k ∈ ≤ k 68

Proposition 8.3. Let the assumption of Proposition 8.2 hold. If pj

h r (Wxk,xj)=0 (12) for all r [p /2], h [p /2]. ≤ j ≤ k r r Proof. According to Proposition 8.2 we have Txj = T ∗xj, T = WA. Since h + i +1 p , we ﬁnd ((cf. (11)) ≤ k h r h+1 r h+1 r 1 (x ,Tx )= ((cT ∗ + W ) x ,x )=(x ,Tx − ) k j − k j k j (13) = =(xh+s+1, (cT + W )x0)=0. · · · k j From the deﬁnition of EAV we obtain

h h h h 1 W x = cT ∗x T ∗x − k k − k − k Then (12) follows from (13).

Denote by σd(A) the discrete spectrum of a pencil A(λ), i.e. the set of isolated eigenvalues of A(λ) of ﬁnite algebraic multiplicity

Proposition 8.4. Let A(λ)= W + λT and eigenvalues λk σd(A) be enumerated according to their geometric multiplicity. If (7), (8) are mutually∈ adjoint canonical systems corresponding to eigenvalues λk then the following biorthogonality relations hold: h s (T yk ,xj)= δk,jδh,ps h (14) − If λ =0 then j 6 h s ¯ 1 s 1 s¯ s (W yk ,xj λj− xj− + ... +( 1) λj− xj)= λjδk,jδh,ps h (15) − − − − Proof. We have

h h 1 h A(λ)=[A(λ )+ T (λ λ )]y = T y − +(λ λ )T y , 0 h p k − k k − k − k k ≤ ≤ k 1 (if h =0, then we assume yk− = 0). Using the representation (2.2) we obtain

h 1 h yk = A− (λ)A(λ)yk N2 pj s 0 0 s ( ,xj)yj + ... +( ,xj )yj h h 1 (16) = · · + R(λ) (λ λk)T y T y − pj+1 s k k " (λ λj) − # − − j=N1 s=0 − X X 69

where R(λ) is a holomorphic operator function at the point λj. We may assume that N =1, N = N, p p ... p 1 2 1 ≥ 2 ≥ ≥ N Let λk = λj. Taking h = 0 and comparing coefficients of the powers (λ pj 1+s 6 − λ )− − , 0 s p , we find j ≤ ≤ j 0 0 0 (T yk,xj )yj =0, (17) p =p Xj 1 0 1 0 0 0 1 0 0 0 (T yk,xj)yj + (T yk,xj)yj + (T yk,xj )yj =0. (18) pj =p1 pj =p1 pj =p 0 X X X1− 0 N It follows from the definition of a canonical system that elements yj 1 are linearly independent. Hence, it follows from 17, that { }

0 0 (T yk,xj )=0, for all pj = p1. (19) Now, it follows from (18), (19), that

(T y0,x0)=0 if p = p 1; (T y0,x1)=0 if p = p . k j j 1 − k j j 1 Repeating the argument we ﬁnd subsequently

0 s 1 s pk s (T yk,xj)=0 (T yk.xj),..., (T yk ,xj)=0 for all 0 s p . ≤ ≤ j The same arguments can be applied in the case λk = λj. Comparing the coeﬃ- cients of the powers (λ λ )ν in (18) it is found that, for h =0, 1,...,p , − j k h s (T yj ,xj)= δh,pj h, − and (14) follows. Noticing that

s 1 s 1 s 1 s 0 T ∗x = λ¯− W ∗(x λ¯− x − + +( 1) x ), j − j j − j j · · · − j we obtain (2.8).

Let (7) be a canonical system of EAV of a dissipative pencil W +λT correspond- 0 ing to a real eigenvalue λk = c. Denote by S the span of elements

y0, y1,...,yβk , k = N ,...,N , β = [(p 1)/2] (20) k k k 1 2 k k − 70

(if p = 0, we assume that β = 1 and the element y0 does not belong to S0). k k − k Let us ﬁx an index k, N1 k N2. If a number pk +1 is even we set Sk = 0 ≤ ≤ S . If pk +1 is odd, we denote by Sk the span of elements (20) combined with αj elements yj , αj =[pj/2], where index j runs through all values such that pj = pk. Similarly, by replacing the chains (7) with adjoint chains (8) we construct subspaces 0 (S )∗andSk∗. Proposition 8.5. If A(λ) = W + λT is a dissipative operator pencil then for all nonzero real λ σ (A) k ∈ d 0 0 S =(S )∗ and S = S∗ for all N k N . k k 1 ≤ ≤ 2 h h Proof. Suppose that yk Sk and xk / Sk. It follows from Proposition 3 that (10) are chains of EAV of a pencil∈ W + λT∈. Since (7) is a canonical system, we have a representations N2 h xh = c ys, 0 h α =[p /2]. (21) k j,s j ≤ ≤ j j s=0 jX=N1 X We have assumed that xh / S , therefore, at least one of the numbers c in k ∈ k j,s (2.13) is not equal to zero for s > βk = [(pk 1)/2], pj

(Wxh, g)= λ c . j − j j,s h 0 0 As λ = 0, we have c . Hence, x / S is not valid. The equality S = (S )∗ is j 6 j,s k ∈ k proved in a similar way.

Proposition 8.6. Let the assumption of Proposition (2.2) hold. Then a canonical system (2.1), corresponding to a real eigenvalue µ can be chosen in such a way that

α (W y j , yαs )= ε δ , ε = 1, α =[p /2], (22) j s j js j ± j j for all indices 1 j, s N such that p +1 or p +1 are odd. ≤ ≤ j s 71

Proof. Assume that p = p = = p >p p , p p p , 1 2 · · · q q+1 ≥···≥ r r+1 ≥ r+2 ≥···≥ N pj =2αj if 1 j r and pj =2αj +1 if r < j N. Let P1 be the orthoprojector ≤ ≤ ≤ 0 onto subspace S = S∗ (see the Proposition 6). Obviously, dim S S = q. It 1 1 1 ◦ follows from the biorthogonality relations (15) that self-adjoint operator P1W P1 q has exactly q nonzero eigenvalues which correspond to an orthogonal basis ϕs 1. { } q Obviously, a canonical system (7) can be chosen in such way that the system ϕs 1 will coincide with yα1 q. Then after a proper norming the relations (22) will{ hold} { k }1 for k = 1, 2,...,q. Considering orthoprojector Pq+1 onto subspace Sq+1 = Sq∗+1 and self-adjoint operator Pq+1W Pq+1, we can repeat the arguments and choose the chains of length 1+ pq+1 (not changing the ﬁrst chains) so that relations (22) will hold for all pk3Dpq+1. The next step is evident. Hence the proof of Proposition 7 can be completed by induction.

Note 8.2. Let a canonical system (7) satisfy condition (22). Then for all indices k αk such that pk = 2αk the elements xk of the adjoint canonical system (8) have the representation xαk = λ δ yαk = y, α = p /2, (23) k − k k k k k where y S0. This representation follows from Proposition 6 and relations (15). ∈ Let yk = c be real eigenvalue of a dissipative pencil A(λ) = W + λT , and + let canonical system (7) satisfy the condition (22). Denote by Sc (Sc−) the span of αk elements (20) combined with yk satisfying the relations (22) with δk = 1 (δk =1). + 0 h − Let S0 (S ) be the span of all EAV yk of A(λ) corresponding to eigenvalues − { } + λk σd(A) with Im λk > 0 (Im λk < 0). Denote by S (S−) the minimal subspace containing∈ S0 (S0 ) and all subspaces S+ (S ) corresponding to real eigenvalues + λk λ−k − h λk σd(A). Finally, by S we denote the minimal subspace containing all EAV yk corresponding∈ to eigenvalues λ σ (A). k ∈ d + Proposition 8.7. Let the assumption of Proposition 3 hold. Then Sc is a W - + non-positive subspace. If Sc is the span of the elements (7) then Sc is a maximal W -non-positive subspace in Sc. + Proof. It follows from Propositions 4, 5 and deﬁnitions that Sc is a W -non-positive + 1 1 subspace. Assume that Sc S Sc, where S is also W -non-positive, and ⊂ ⊂1 + that there exists an element y S such that y / Sc . Obviously, y / Sc−, ∈ ∈+ ∈ because it follows from assumptions y Sc−, y / Sc that (Wy,y) > 0. Hence, + ∈ ∈ h 0 y / Sc and y / Sc−. Then, using (15) we can ﬁnd an element yk S such that∈ (W yh, y) =∈ γ = 0. We may assume that γ > 0, otherwise we∈ have to k 6 72

h replace y by γy. Denote z = ρyk + y. Then from Proposition 4 we obtain (Wz,z)=2ργ +(Wy,y) if ρ + . This is the contradiction. →∞ → ∞

Theorem 8.1. Let W be an invertible self-adjoint operator in Hilbert space H and + A be a W -dissipative operator. Then S (S−) constructed from EAV of as pencil + I + λA is a W -non-positive (nonnegative) subsoace. Moreover S (S−) is maximal W -non-positive (nonnegative) subspace in S. Proof. It follows from biorthogonality relations (15) and the deﬁnitions that S+ is W -nonpositive. Repeating the arguments of Proposition 9 we ﬁnd that S+ is maximal W -non-positive in the subspace S.

Now we give application of Theorem 10 to the problem of half-range minimality for dissipative operator pencils. Consider an operator pencil

A(λ)= A + λA + + λnA , (24) 0 1 · · · n where Aj, j =1,...,n are bounded operators in Hilbert space H and condition (1) holds. We assume that A is invertible (if σ(A) = R then we can shift λ λ + λ , 0 6 → 0 λ0 R, λ0 / σ(A); after this translation condition (1) will also hold). With pencil (24)∈ we associate∈ the linear operators

1 1 1 1 A0− A1 A0− A2 ... A0− An 2 A0− An 1 I 0 ... 0− 0 −   A = −0 I... 0 0 , −  ......     0 0 ... I 0   −    0 0 ... 0 A0 0 0 ... A A  0 1  G = ......  0 A0 ... An 3 An 2  − −  A0 A1 ... An 2 An 1  − −  Consider also the operators 

T0 0 Gq = GAq = , 0 q n, 0 T1 ≤ ≤ 73 where 0 0 ... 0 A0 0 0 ... A A  0 1  T =( 1)q ...... , 0 −  0 A0 ... An q 3 An q 2  − − − −  A0 A1 ... An q 2 An q 1  − − − −    An q+1 An q+2 ... An 1 An − − − An q+2 An q+3 ... An 0 q 1  − −  T =( 1) − ...... 1 −  An 1 An ... 0 0   −   An 0 ... 0 0      We assume that at q =0 (q = m) the block T1 (T0) is absent in the represantion of the matrix Gq. Now we set T 0 F = 0∗ q 0 T 1 Denote by W , 0 q n, the self-adjoint operator in Hn, such that its matrix q ≤ ≤ coincides with Fq over the main diagonal, with matrix Fq∗ under the main diagonal and with (Fq + Fq∗)/2 on the main diagonal. For example, if n =2, then

R R 0 A0∗ A0 0 A1 An W0 = R ,W1 = − R ,W2 = A A 0 A A∗ 0 0 1 2 n and if n =4, then

0 0 0 A0∗ 0 A0∗ 0 0 R 0 0 A0∗ A1∗ A0 A1 0 0 W0 =  R  ,W2 =  R  . 0 A0 A1 A2∗ 0 0 A3 A4 R − − A0 A1 A2 A   0 0 A∗ 0   3   − 4      Further we assume n = 2l. A similar result can be obtained for the case n = 2l +1. One can check easily that the following important equalities hold (l = n/2)

0l 1 q 0 0 − − W A (W A)∗ = i 0 V 0 := iJ (25) 2q − 2q  0  q 0 00q   74 for q =0, 1,...,l 1, where 0 denotes a square zero matrix of order q, and − q I I 2A0 A1 0 ... 0 0 AI 2AI AI ... 0 0  1 2 3  0 AI 2AI ... 0 0 V = 3 4 (26) 0  ......   I I   0 0 0 ... 2An 2 An 1  I − −I   0 0 0 ... An 1 2An   −    Analogiously, for q =1, 2,...,l we have

0l q 0 0 − W2q 1A (W2q 1A)∗ = i 0 V1 0 := iJ2q 1, (27) − − − −   − − 0 00q 1 −   where Υ Υ 2A1 A2 0 ... 0 0 AΥ 2AΥ AΥ ... 0 0  2 3 4  0 AΥ 2AΥ ... 0 0 V = 4 5 (28) 1  ......   Υ Υ   0 0 0 ... 2Am 1 Am  −   0 0 0 ... AΥ 0   m  The following result is a generalization of Proposition 5 for polynomial operator pencils. It is proved in the paper Shkalikov [3]. Theorem 8.2. Let a polynomial pencil A(λ) be deﬁned by (24) and 0 / σ(A). Let ∈ 0 1 pk yk, yk, . . ., yk , k = N1,...,N2 (29)

0 1 pk zk, zk, . . ., zk , k = N1,...,N2 (30) be mutually adjoint canonical systems corresponding to the eigenvalues λk σd(A). Then the following biorthogonality relations hold: ∈

h s 1 n q s 1 s n + s q 1 0 (Gqy˜ , z˜ λ¯− − z˜ − + ... +( λ¯j) − − z˜ ) j j − j 1 j − s j (31) q+1 n q =( 1) λ − δs,pj s, − k − r h s where s are binomial coeﬃcients and y˜j , z˜k are Keldysh derived chains constructed from canonical systems (29), (30) respectively. 75

Proof. Let the operator Aˇ be deﬁned by the same matrix as A, except that the operators Aj are replaced by operators Aj∗, j =1,...,m. Then the following equalities can be easily veriﬁed by induction q GAq =(Aˇ )∗G, q =0, 1,....

r s r r S 1 r S 1 Aˇ z˜ =( 1) λ¯− [˜z λ¯− z˜ − + ... + j − j j − j 1 j r + S 1 +( λ¯ )S − z˜0], r = 1, 2,.... − j S j ± ± These equalities enable us to prove (31) only for some ﬁxed q, for instance, q = n. h It is known (see Section 1) that Keldysh derived chains y˜k are EAV of a linear pencil I + λA. If x˜S is the adjoint system then according to Proposition 6 { j } h S (Ay˜k , x˜j )= δk,jδh,ps h. − Hence, the relations (31) are equivalent to the following equalities S n+1 S S n+1 S Gn∗ z˜j =( 1) A∗x˜j or Gn∗ 1z˜j =( 1) x˜j (32) − − − 1 For the result (I + λA)− we have the following representation (which can be veriﬁed by multiplication of (I + λA))

1 1 L− (λ)T1 ...L− Tn 0 0 0 ... 0 λL 1T ... λL 1T 0 I 0 ... 0 (I + λA) 1 = − 1 − n + , −  ......  ......  n 1 1 n 1 1 n 2 n 3 λ − L− (λ)T1 ... λ − L− (λ)Tn  0 λ − I λ − I...I        (33) where 1 1 L− (λ)= A− A0, 2 n j+1 T1 = I,Tj(λ)= (λLj + λ Lj+1 + ... + λ − Ln), j =2,...,n, − 1 Lj = A0− Aj. On other hand according to Theorem on holomorphic operator function (Section 2) 1 the principal part of (I + λA)− in a neighborhood of the pole λk has the form

N2 pk h 0 h 1 1 0 h ( , x˜ )˜y +( , x˜ − )˜y + ... +( , x˜ )˜y · k k · k k · k k (34) pk+1 h (λ λk) − kX=N1 Xh=0 − 76

(pk+1 h) 1 Let us compute the coefficient of (λ λ )− − of the vector (I + λA)− f˜, − k f˜ = f1,...,fn, in a neighborhood of the pole λk. Using the representation (33) and (2.2) (for λk = c) after simple algebra, we find that this coefficient is given

m h 1 (q) h q 0 ( T (λ )f , z − )˜y + ... (35) q! j k j k k j=1 q=0 X X (we have not written out the remaining terms, which happen to be linear combi- 1 h nations of the elements y˜k,..., y˜k ). On the other hand from (34) we find that this coefficient equals ˜ h 0 (f, x˜k)˜yk + .... (36) h h h ˜ n Let x˜k = xk,1,xk,2,...,xk,n . Since f = f1,...,fn is any vector in H , we find comparing{ (35) and (36) }

h h 1 (q) h q x = T ∗ (λ )z − k,j q! j k k q=0 X h n j q s+1 − 1 d (λ ) h q = L∗ x − (37) ¯ j+s k − q! dλ λ=λk q=0 s=0 X X h n j − 1 s q+1 s +1 h q = λ − L∗ z − . − q! k q j+s k q=0 s=0 X X It follows from the distinction of Keldysh derived chains that

h h h h q 0 h q 1 q 1 h q n 1 q n 1 h q z˜ = λ− z − , λ − z − ,..., λ − − − z − . (38) k k q k k q k k q k q=0 q=0 q=0 n X X X o

Writing out the matrix Gn∗ 1 and using (38) we find that the first coordinate of the h − n 1 h n 1 h vector Gn∗ 1z˜k is equal to ( 1) − zk =( 1) − xk,1 (the last equality is valid according to (37)).− Hence (32) is− satisfied for the− first coordinate. Using (37), (38) we can also check the equality (32) for the subsequent coordinates. It proves Theorem 11.

Note 8.3. Theorem 11 is valid for arbitrary operator pencil A(λ). The dissipative condition (1) or (2) is not required. 77

We say a pencil A(λ) is linearly dissipative if either the operator V0, which is defined by (26), satisfies the condition V0 0 or the operator V1, which is defined by (28), satisfies the condition V 0. Since≤ conditions (25), (27) hold, a pencil A(λ) 1 ≤ is linearly dissipative if and only if a linearization A is W2q-dissipative operator for all q =0, 1,...,l 1, or A is W2q 1-dissipative for all q =1, 2,...,l. − − Proposition 8.8. If A(λ) is linearly dissipative, i.e. V0 0 (V1 0), then it is dissipative, i.e. condition (1), (condition (2)) holds. ≤ ≤

Proof. Let, for example, V0 0. Consider the function u(t)= ϕ(t)x, where x H and ϕ(t) is a smooth rapidly≤ decreasing function when t (ϕ(t) S). If∈ → ±∞ ∈ n 1 (n 1) u˜(t)= u(t), iu′,..., ( i) − u − (t) { − − } then

l ∞ ∞ I (2s) (2s) (V u˜(t), u˜(t))dt =2 =(A2su (t),u (t))dt Z−∞ s=0 Z−∞ Xl 1 − ∞ = i (A u(2s+1)(t),u(2s)(t))+ − 2s+1 (39) s=0 X Z−∞h (2s) (2s+1) +(A2s+1u (t),u (t)) dt

∞ i = 2Im (A(λ)ˆϕ(λ)x, ϕˆ(λ)x)dλ 0. ≤ Z−∞ Here we denoted ∞ iλt uˆ(λ)= e− u(t)dt =ϕ ˆ(λ)x Z−∞ and took into account that

s ∞ iλt (s) ∞ ∞ (iλ) uˆ(λ)= e− u (t)dt, (ˆu(λ), vˆ(λ))dλ = (u(t),v(t))dt. Z−∞ Z−∞ Z−∞ Hence the inequality (39) holds for all x H and all ϕ S. It is known (see, for example, Yosida [1]) that the Fourier transform∈ maps S onto∈ S continuously in both directions. Using this fact we easily obtain (1) from (39). 78

Obviously, the converse assertion is not true. For example, the pencil A(λ) = i(1 αλ2+λ4)I, 0 < α 2, satisﬁes condition (1) but it is not linearly dissipative. − ≤ Further we will consider linearly dissipative pencils, satisfying the condition V0 0. A similar result can be obtained for the case V 0. In lecture 7 we introduced≤ 1 ≤ the systems E± consisting of half of EAV of a dissipative pencil A(λ) and the systems ξ± consisting of the Keldysh derived chains of length l = n/2 corresponding to the h vector y E±. k ∈ + Proposition 8.9. Let A(λ) be a linear dissipative pencil, i.e. V0 0, ( −) be h ≤ L L the minimal subspace containing all elements y˜k (Keldysh derived chains of length h + + n) constructed from elements yk E (E−). Then ( −) is a W2q-non-positive (nonnegative) subspace for all q =0∈ , 1,...,l 1. L L − h Proof. Let λk = c be a real eigenvalue of A(λ). Keldysh derived chains y˜k coincide with EAV of the linearization A of pencil A(λ). It follows from (25) that A is W2q- dissipative, q = 0, 1,...,l 1. According to the deﬁnition of the systems E+ the elements y˜αk E+ satisfy the− equalities k ∈ y˜αk = ε z˜αk +˜y, y˜ S0, ε =1, α =[p /2]. (40) k k k ∈ c k k k Using proposition 3 we can show10 that (W y,˜ y˜)=(G y,˜ y˜) for all y S . 2q 2q ∈ c In particular, αk αk αk αk (W2qy˜k , y˜k )=(G2qy˜k , y˜k ). Now it follows from Theorem 11 and (40) that

αk αk αk αk 2q+1 n 2q (W y˜ , y˜ )=(G y˜ , y˜ )=( 1) λ − < 0 (41) 2q k k 2q k k − k + Recalling the deﬁnition of the space S for W2q-dissipative operator A we ﬁnd from (41) that + = S+. Then the assertion of Proposition 14 follows from Theorem 10. L

+ Proposition 8.10. Let A(λ) be a linearly dissipative operator pencil and ( −) L L be the same subspace as in proposition 14, and 0 / Θ(A0). Deﬁne the operator P : Hn Hl by the equality ∈ → P x˜ = P x ,x ,...,x = x ,...,x . { 1 2 n} { l+1 n} 10The proof depends on direct calculations, but these calculations are rather complicated if n> 2. Here we omit them. 79

Then + P x˜ ε x˜ , ε> 0, for all x˜ ( −) (42) k k≥ k k ∈ L L i.e. the operator P : + P ( +) has a bounded inverse. L → L Proof. Denote x˜ = Q x ,...,x = x ,...,x . { 1 n} { 1 l} Obviously, the estimate (42) holds if and only if

+ Qx˜ M P x˜ for all x˜ ( −), (43) k k≤ k k ∈ L L where the constant M does not depend on x˜. Assume that the estimate (43) does not hold. Then there exists an element x˜ + such that ∈ L Qx˜ =1, P x˜ = o(1), i.e. x = o(1), j = l +1 ...,n. (44) k k k k k jk Recalling the matrix representations of the operators W2q and A we ﬁnd from (44) that (W x,˜ x˜) = o(1), (W Ax,A˜ x˜) W A . | 2q | | 2q |≤k 2qkk k Obviously, A( +) + and according Proposition 14 + is non-positive. Hence we can apply Schwartz’L ⊂ L inequality and obtain L (W Ax,˜ x˜) 2 (W Ax,˜ Ax˜) = o(1). (45) | 2q | ≤ | 2q | On the other hand, we have

(Ax,˜ W0x˜)= 1 A0− A1x1 + ... A0∗xn x1 A0∗xn 1 + A1∗xn    −  −......  x  ,  A∗xl + A∗xl+1 + ... = (A x ,x )+ o(1).  − l   0 1  − 0 l l  xl+1   A0xl 1 + ...   −   −   ...   ......       xn 1   A0x1 + ...   −        Since 0 / Θ(A0), we ﬁnd from (45) that xl = o(1). Using this relation and (44) we obtain∈ k k (Ax,˜ W2x˜)= (A0xl 1,xl 1)+ o(1) xl 1 = o(1). − − − ⇒k − k Repeating the arguments, we ﬁnd x = o(1), j = l,l 1,..., 1. This contradicts k jk − the assumptions Qx˜ =1. Hence estimate (42) is valid. k k 80

Theorem 8.3. Let A(λ) be a linear dissipative operator pencil (V0 0) and 0 / + l ≤ ∈ Θ(A0), Then system ξ (ξ−) is minimal in the space H , l = n/2. Proof. Let an operator H : h H be bounded and have bounded inverse. 1 → 2 Obviously, if a system ek is minimal in H1 then system Kek is minimal in H2. { } h + { } n It follows from Proposition 5 that the system y˜k is minimal in H . Hence, h { } ∈ L l from Proposition 15 we ﬁnd that P y˜k is minimal in H . If all eigenvalues of A(λ) { l } 0 + + are semi-simple then the system λk− P y˜k coincides with ξ , hence ξ is minimal. In the general case one has to check{ the equalities}

h h l n r h r h + y˜ = λ− λ− P y˜ − , y˜ = ξ . (46) k k r k k { k } r=0 X h This can be easily done by using the formula (38) for Keldysh derived chains y˜k (see details in Shkalikov [1]). The equalities (46) imply that the systems ξ+ and P y˜h are connected by a triangular transformation and from this fact one can { k } easily deduce that ξ+ is a minimal.

We proved in Section 7 that the systems ξ+ is linearly independent if A(λ) satisﬁes condition (1). The result of minimality of ξ+ is much sharper and it has been proved only in the case when A(λ) is a linearly dissipative pencil. In this connection the following natural question arises. Open Problem Does condition (1) and 0 / Θ(A0) imply the minimality of the system ξ+ in Hl? ∈ Note 8.4. Theorems on minimality can be obtained for the case V 0 and n = 0 ≤ 2l +1 as well as for linearly dissipative pencils satisfying the condition V1 0 (see Section 7). ≤ Comments W -dissipative operators were introduced in the book of Dalezki, Krein [1] and were studied by Kuzhel, Azizov, Iohidov, M. Krein, Langer, Gomilko, Radzievski and many other authors. References can be found in the recent book Azizov, Iohidov [1]. In this book the proof of Proposition 1 is given, although here we proposed a new proof. Proposition 3 is due to Radzievski. Proposition 5 is proved by Keldysh [1, 2]. Proposition 6 and 7 are proved in the paper Shkalikov [3]. Theorem 10 seems to be new. Theorem 11 is proved in the papers Shkalikov [3, 6]. The second part of this lecture, connected with application to the half-range minimality problem is based on the paper Shkaliov [3]. Here a new version of the proof of Theorem 16 is given. To prove Proposition 5, we borrowed ideas from the 81 paper Langer [3], where a similar assertion is proved for self-adjoint monic operator pencils. + + o The existense of a maximal W -non-positive subspace Sc such that Sc Sc was proved by Gomilko [1] (1983) (this is related to our Proposition 9). But here⊃ we get more, in particular, an important information on connection of direct and adjoint Jordan chains is established. 82 9 Factorization of dissipative operator pencils

In this lecture we solve the problem of factorization of dissipative pencils in ﬁnite dimensional space and obtain some results on factorization of linearly dissipative pencils in inﬁnite dimensional space. First we prove one important result due to H. Langer. Let

n 1 n A(λ)= A0 + λA1 + ... + λ − An 1 + λ An (1) − be a pencil of bounded operators in Hilbert space H and let

1 1 1 1 An− An 1 An− An 2 ... An− A1 An− A0 − I − − 0 − ... − 0− 0   A˜ = 0 I... 0 0 (2)  ......     0 0 ...I 0      This is well-known and can be easily checked that A is a linearization of A(λ), i.e. the spectra of A(λ) and λI A˜ coincide and the corresponding Jordan chains of operator A˜ coincide with Keldysh− derived chains of A(λ). Notice, that operator A˜ diﬀers from linearization A used in Section 8. The choice (2) in this lecture is not incidental. One comes with serious technical diﬃculties to prove the subsequent theorem in terms of the old linearization A. k n k k n k For any operator K : H H − (acting from H into H − ) we call the subspace → Kxˆ M = xˆ Hk Hn (3) xˆ ∈ ⊂ the graph subspace of K. Obviously, the graph subspace of any bounded operator

K is closed subspace in Hn.

Theorem 9.1. (Langer [3] (1976)). The pencil (1) admits factorization

A(λ)= L(λ)K(λ) (4)

k k 1 with a pencil K(λ) = λ I λ − Kk 1 ... λK1 J0 of degree k (< n) and a pencil L(λ) of degree n −k if and− only− if the− linearization− A has an invariant − k n k subspace of a bounded operator K : H H − → 83

If M is such a subspace and

K11 K12 ...K1k K K ...K K = 21 22 2k (5)  ......  Kn k,1 Kn k,2 ...Kn k,k  − − −    then the operator Kij : H H are uniquely determined by the coeﬃcients of the right divisor, in particular,→

Kn k,j = Kk j, j =1, 2,...,k. − −

Moreover, the spectra of K(λ) and of A˜ M coincide (A˜ M is the restriction of A˜ onto M). | |

Proof. Step 1. Obviously, this is enough to prove theorem for monic pencil with An = I, therefore further we assume An = I. Let a subspace M has the representation (3) and K is deﬁned by (5). Denote

Kn k,j = Kk j, j =1, 2,...,k. − −

Observe, that if M is invariant with respect to A˜ then the operators Kij are uniquely determined by the operators Kj. Indeed, if

x˜ = K11x,K21x,...,Kn k,1x,x, 0,..., 0 (6) { − } then the (n k)-th component of A˜x˜ is Kn k 1,1x. On the other hand, it has to −2 ˜ − − 2 be equal to Kk 1x + Kk 2x, since Ax M. Therefore, Kn k 1,1 = Kk 1 + Kk 2. − s − ∈ − − − − Using the inclusions A˜ x˜ M we may determine the operators Kn k s,1. Taking in (3) x˜ = 0,x, 0,..., 0 ∈and applying the same arguments we may− determine− the { } operators Ki, 2 by Kj, and thence in a similar way all the other operator. Step 2. Let us show that the existence of an invariant subspace (3) implies the 84

1 factorization (4). For the resolvent (A λI)− we have the following representation − n 1 λ − I An 1 ... A2 A1 n 2 − λ − 0 I ...A3 A3 1   1 n 2 n 1   (A λI)− = ... A− (λ)[1,λ,...,λ − , λ − ] ...... − −  λ   0 0 ... I An 1    −   1   0 0 ... 0 I      n 3 n 2  0 I λI ... λ − I λ − I   n 4 n 3 0 0 I ... λ − I λ − I ......  +  0 0 0 ... I λI     0 0 0 ... 0 I     0 0 0 ... 0 0      which can be veriﬁed by multiplication of A˜ λI from the left. − 1 Let a vector x˜ be deﬁned by (6). Then the (n k)-th component of (A˜ λI)− x˜ equals − − k 1 λ A− (λ)L(λ)x + x (7) − where

K11 K21 I An 1 ... A2 A1   − ... 0 I ...A3 A3 n 2 n 1   Kn k,1 L(λ)=[1,λ,...,λ − , λ − ] ......  −   I   0 0 ... I An 1    −   0   0 0 ... 0 I       ...       0      1 is the pencil of degree n k. By the invariance of M we have (A˜ λI)− x˜ M. Therefore, the vector (7)− can be also represented in the form − ∈

Kk 1y1 + Kk 2y2 + ... + K0yk − − where k j 1 y = λ − A− (λ)L(λ)x, j =1, 2,...,k, j − 1 coincide with (n k + j))-th component of the vector (A˜ λI)− x˜. Hence − − k 1 k k 2 1 λ A− (λ)L(λ)x + x =( λ Kk 1 λ − Kk 2 ... K0)A− (λ)L(λ)x − − − − − − − 85 or 1 K(λ)A− (λ)L(λ)= I and factorization (4) follows. Step 3. Suppose, conversely, that the factorization (4) holds. Deﬁne the following matrices (k j n 1) ≤ ≤ −

Kk 1 Kk 2 ...K1 K0 0 ... 0 I− 0− ... 0 0 0 ... 0   Kj = 0 I... 0 0 0 ... 0  ......     0 0 ... 0 0 0 ...I    with j +1 rows and j columns and also 

Kk 1 ...K1 K0 I...− 0 0 K = (8)  ......   0 ...I 0      Suppose we have proved the equality

Kk 1 ...K0 0 ... 0 0 I...− 0 0 ... 0 0 AK˜ n 1 ...Kk =   Kn 1 ...Kk = Kn 1 ...KkK − ...... − −  0 ... 0 0 ...I 0      (9) k Then the subspace M = Kn 1 ...KkH is invariant with respect to A˜ and it is easy to see that this subspace has− the representation (3). To prove (8), observe that the ﬁrst step in the partial division of a polynomial l l 1 λ Bl + λ − Bl 1 + ... + B0 by K(λ) (l k) from the right gives a remainder whose coeﬃcients are− the entries of the product≥

[Bl,Bl 1,...,B0]Kl. − Therefore the factorization (4) yields

[An 1 Kk 1,An 2 Kk 2,...,An k K0,An k 1,...,A0]Kn 1 ...Kk =0. − − − − − − − − − − − Now it is easy to see that the last equality is equivalent to (9). 86

Step 4. Evidently, the spectrum of K(λ) coincide with spectrum of the operator K deﬁned by (8). Moreover, it follows from (9) that

(A˜ λI)Kn 1 ...Kk = Kn 1 ...Kk(K λI). − − − − k This means that σ(K) coincides with spectrum of A˜ M , where M = Kn 1 ...KkH , | − and the last assertion of Theorem 1 follows.

Now we can easily obtain the results on factorization of dissipative matrix polynomial. We consider both cases of the dissipativity condition, i.e. Im(A(λ)x,x) 0 for all x H and λ R (10) ≤ ∈ ∈ and Im(λA(λ)x,x) 0 for all x H and λ R. (11) ≤ ∈ ∈ Theorem 9.2. Let the condition (10) hold, n = 2l, dim H < , Ker A = 0 ∞ n { } and there exists λ0 R such that 0 / Θ(A(λ0)). Then A(λ) admits factorization ∈ l l 1 ∈ (4) with a pencil K(λ)= λ I λ − Kl 1 ... K0 such that the system of eigen − − − − + and associate vectors of K(λ) coincides with the system E (E−) of pencil A(λ). Proof. Without loss of generality we may assume that 0 / σ(A). Otherwise we can ∈ shift λ λ + λ0, obtain the factorization (4) and then shift back λ λ λ0. →+ h → − Let be the minimal subspace containing all elements y˜k (Keldysh derived chains ofL length n) constructed from elements yh E+ of pencil A(λ). Obviously, k ∈ + is invariant with respect to linearization A˜ of A(λ). Let us prove that + has Lrepresentation (3) with k = l. Consider the operator P : Hn Hl deﬁnedL by the equality → P x˜ = P x ,x ,...,x = x ,...,x . { 1 2 n} { l+1 n} + Denote P = P + . If all eigenvalues of A(λ) are semi-simple then the system l 0 |L + λk− P yk coincides with the system ξ which is basis according to Theorems on completness{ } and linear independence from Section 7.The completeness of ξ+ implies Im P + = Hl while the linear independence implies Ker P + = 0 . Then it follows immediately that + has the representation (3). { } In general caseL (when non-semi-simple eigenvalues exist) make use from + h h + representation (8.40) which shows that the systems ξ and P y˜k (yk E ) are { } h ∈ connected by a triangular transformation. Hence, the system P y˜k is basis and this yields again that + has the representation (3). Now apply{ Theorem} 1 to L complete proof. 87

Theorem 9.3. Let the condition (11) hold, dim H < , n =2l, A0 > 0 and An > ∞ l l 1 0. Then A(λ) admits factorization (4) with a pencil K(λ)= λ I λ − Kl 1 ... K0 such that the system of eigen and associate vectors of K(λ) coincides− with− the− system− + Y − (Y ) of pencil A(λ).

Proof. Apply Theorem 7.8 and repeat the arguments of Theorem 2.

Note 9.1. A similar results can be obtained in the case n =2l +1. Namely, if A(λ) satisfies the condition (10) and An > 0 (An < 0) then A(λ) admits the factorization (4) with K(λ) of degree l +1 (l) and the system if EAV of K(λ) coincides with E+. If n =2l +1, A(λ) satisfies the condition (11), A0 < 0, An < 0 (> 0) then A(λ) admits the factorization (4) with K(λ) of degree l (l +1) and the system of EAV of K(λ) coincides with Y +. Another versions of factorization theorems involving the + systems E− and Y can be also formulated. These assertions follow from Theorems 7.6 and 7.9. The problem on factorization of self-adjoint or dissipative operators in Hilbert space is much more deep. Langer [3] proved that each maximal G-non-positive (G-nonnegative) subspace M Hn which is invariant under the linearization A˜ of monic self-adjoint pencil A(λ)⊂has the form (3) with k = [(n + 1)/2] (k = [n/2]), where 0 0 ... 0 I 0 0 ... I An 1  −  G = ......  0 I ...A3 A2     I An 1 ... A2 A1   −  is the simmetrization of A˜. Hence the factorization problem is reduced to the problem on existence of maximal G-semi-definite subspaces invariant with respect to G-self-adjoint operator A˜. Using the results of Section 8 we can obtain a similar assertion for linearly dissipative operator pencils. But the problem on existence of maximal semi-definite subspace invariant with respect to self-adjoint operator in Krein space is still open. The deepest results of operator theory in spaces with indefinite metric are connected with this problem. It is solved for some particular classes of operators in Krein space and these results generate the corresponding fac- torizaion theorems. To make acquaintance with these results we refer the reader to the remarkable paper Langer [3]. In this connection the subsequent result on factorization of dissipative pencils in Hilbert space is of interest. It is formulated in terms of solvability of the half-range 88

Cauchy problem. But the last problem can be solved for some classes of differential equations associated with pencil A(λ) (see Note 8 in the end of this Section). We say v(z) is a regular solution of the equation d dv dnv A( i )v(z)= A v iA + ... +( i)nA =0 (12) − dz 0 − 1 dz − n dzn on interval (a, b) if v(z) has n continuous derivatives as a function with values in H on (a, b) and v(z) satisfies the equation (12). Let A(λ) be dissipative operator pencil satisfying the condition (10). Let also the real spectrum of A(λ) be discrete. Denote by S+(0, ) the linear manifold of all solutions v(z) of equation (12) on the semi-axis (0, )∞satisfying the Mandelstam radiation principle at (in sense of Definition 7.4).∞ Assume also n = 2l and consider the trace operator∞ J : S+(0, ) Hl defined by the equation ∞ → l 1 (l 1) J v(z)= v(ξ), iv′(ξ),..., ( i) − v − (z) . (13) ξ { − − } For v(z) S+(0, ) we denote by v˜(z) the following function with values in Hn ∈ ∞ n 1 (n 1) v˜(z)= v(z), iv′(z),..., ( i) − v − . (14) { − − } Proposition 9.1. Ley A(λ) be linearly dissipative pencil (V 0) and let the 0 ≤ operators Wq, q = 0, 2,...,n, be defined as in Section 8. Suppose v(z) = v1(z)+ + j v0(z) S (0, ) where v0(z) 0 for j =0, 1,...,n 1 when z , and v1(z) is a finite∈ superposition∞ of elementary→ solutions correspon− ding to→∞ real eigenvalues. Then for q =0, 2,...,n the following relations hold (W v˜ (z), v˜ (z)) 0, (15) q 0 0 ≤ (W v˜ (z), v˜ (z)) 0, (16) q 1 1 ≤ (Wqv˜0(z), v˜1(z))=0, (17) (W v˜(z), v˜(z)) 0, (18) q ≤ Proof. Let the linearization A of A(λ) be defined as in Section 8. Then the function v(z) satisfies the equation Av˜′(z)+ iv˜(z)=0, since v(z) satisfies (12). Using (8.25) we obtain

(Wqv˜0(z), v˜0(z))′ =(Wqv˜0′ , v˜0(z))+(Wqv˜0(z),v0′ (z)) = i(W v˜′ (z),Av′ (z)) + i(W Av˜′(z), v˜′(z)) = (J v˜′ (z),v′ (z)). − r 0 0 r − q 0 0 89

Integrating this equality from z to and taking into account that v0(z) vanishes at we obtain ∞ ∞ ∞ (W v˜ (z), v˜ (z)) = (J v˜′ (ξ), v˜′ (ξ))dξ 0, q 0 0 q 0 0 ≤ Zz since V 0. Hence the inequality (15) holds. 0 ≤ By our assumption the operator A is Wq-dissipative for q = 0, 1,...,n. Then the inequality (16) follows from Theorem 8.10. According to Proposition 8.3 v (z) Ker J and v′ (z) Ker J . Therefore 1 ∈ q 1 ∈ q (W v˜ (z), v˜ (z)) = (J v˜′ (z), v˜′ (z))=0. q 0 1 − q 0 1 Now (15)-(17) give the inequality (18).

Proposition 9.2. Let A(λ) be linearly dissipative (V0 0), n =2l and 0 / Θ(A0). Then for v(z) S+(0, ) the following estimate holds≤ ∈ ∈ ∞ (l 1) v(z) + v′(z) + ... + v − k k k k k k (l) (l+1) (n 1) (19) M( v (z) + v + ... + v − (z) ). ≤ k k k k k k Proof. The estimate (19) is equivalent to the following estimate

Qv˜(z) M P v˜(z) , v˜(z) S+(0, ), (20) k k≤ k k ∈ ∞ where the operators Q and P are deﬁned in Proposition 8.15. Taking into account Proposition 4 we may repeat all arguments from Proposition 8.15 and prove (20) as well as (8.37).

Note 9.2. Denote by S+ the minimal subspace in H containing all vectors v˜(ξ) for fixed ξ > 0, such that v(z) S+(0, ). Obviously, + S+ where + is defined in Proposition 8.14. It may∈ happen∞ that + = S+, henceL ⊂ PropositionL 5 and Proposition 8.15 are not identical. L 6 Theorem 9.4. Let the conditions of Proposition 5 hold. If for some ξ 0 the l ≥ image of the trace operator Jξ defined by (13) is dense in H then A(λ) admits the factorization (4) with pencil K(λ) of degree l. Moreover, if the whole spectrum of A(λ) is discrete then σ(K) lie in the upper half-plane and the system of eigen and associated vectors of K(λ) coincide with E+. 90

Proof. If v(z) S+(0, ) then the function v˜(z) defined by (14) satisfies the equation ∈ ∞ A˜v˜′ iv˜ =0, (21) − where A˜ is the linearization (2). Let S+ be a minimal subspace in Hn containing all vectors v˜(ξ) for fixed ξ 0 such that v(z) S+(0, ). Obviously, S+ is invariant with respect to A˜. According≥ to Proposition∈ 5 we have∞ representation

Kxˆ S+ = , xˆ Hl xˆ ∈L⊂ where K : Hl Hl is a bounded operator, and is a linear manifold in Hl which has to be subspace,→ since S+ is subspace. Notice,L that

l l 1 (l 1) + A v˜(ξ)= v (ξ),...,v (ξ),v(ξ), iv′(ξ),..., ( i) − v − (ξ) S { 1 l − − }∈ + l if v˜(ξ) S . Therefore Im Jξ. By our assumption Im Jξ = H , hence = Hl∈and S+ is the graphL ⊃subspace of K. Now Theorem 1 implies the factor- L ization (4) with k = l and σ(K) σ(A˜ + ). This yields the assertion of Theorem 7. − |S

l Note 9.3. The condition Im Jξ = H can be established for some classes of operator pencils. Such results are proved in ch. 8 of the paper Shkalikov [6]. In particular, l Im Jξ = H for selfadjoint operator pencils of Keldysh type (see Theorem 8.9 of Shkalikov [6]).

Comments. The problem on factorization of self-adjoint polynomials has a long history and takes the origin from the paper of Krein and Langer [1]. Impor- tant factorization theorem was proved by Rosenblum and Rovnyak [1]. The paper Langer [3] became a millstone in factorization problems for pencils of degree n> 2. Kostyuchenko and Ozarov [1, 2] classiﬁed the real spectrum of the right divisor for selfadjoint quadratic pencils. Gohberg, Lancaster and Rodman [1-3] established theorems on factorization of selfadjoint matrix polynomial with classiﬁcation of real spectrum. Nontrivial factorization theorems were proved by Markus and Matcaev (see details and comments in the book of Markus [1]). Interesting results on factorization of matrix and operator functions are contained in the books Bart, Gohberg and Kaashoek [1] and Litvinchuck and Spitkovskii [1]. Factorization theorems of this lecture for dissipative operator pencils seem to be new and are based on the paper Shkalikov [3]. 91 10 Pontrjagin spaces. The proof of Azizov- Iohvidov-Langer theorem

A classical Hilbert theorem asserts that any self-adjoint compact operator in Hilbert space H has a complete orthonormal system of eigenvectors in H. Does this result admit a generalization on Pontrjagin space? This problem is the main subject of this section. There are a number of books on operator theory in Pontjagin and in Krein spaces. We point out the books of Bognar [B], Ando [A], Iohvidov, Krein and Langer [IKL], Azizov and Iohvidov [AI] and seveys of Iohvidov and Krein [IK] and Langer [L]. Nevertheless the proof of the subsequent theorem on Riesz basis property of eigenfunctions of self-adjoint operator in Pontrjagin space (which is due to Azizov and Iohvidov) readers can ﬁnd in the only book [AI] (Theorem 4.2.12 of [AI]). Nowever, it is not easy to restore the proof from the text since if uses a of foregoing material. In our lectures we will try to elucidate the situation. We notice that such an attempt has been undertaken already in the paper of Binding and Seddighi [BS] although the latter paper dealt only with completeness, the problem on minimality and basisness had not been considered there. We hope also that this material will help readers in understanding some important concepts in the theory of operators in spaces with in deﬁnite metric.

10.1 Pontrjagin theorem and the formulation of Azizov- Iohvidov-Langer theorem.

Let Pκ =(H, G) be Pontrjagin space, i.e. H is also supplied with the scalar product (x, y) but H is also supplied by the indeﬁnite metric [x, y]=(Gx, y) where G is self-adjoint bounded and invertible operator having κ negative eigenvalues counting with multiplicities. According to spectral theorem for self-adjoint operators we can represent G = G+ G where G+G = G G+ = 0, G+ > 0, G > 0 and − − − − − our assumptions on G are equivalent to the following: G := G+ + G 0, rank G = κ < . | | − ≫ − ∞ An operator A is said to be self-adgoint in Pκ if [Ax,y]=[x,Ay] for all x, y H and this is equivalent that GA is self-adjoint in H. We present without proof∈ the following fundamental result.

Theorem 10.1 (Pontrjagin [P]). A self-adjoint operator A in Pκ has a maximal + nonnegative and maximal non-positive subspaces L and L − respectively which are 92

11 invariant under A. Moreover, dim L − = κ . The proof of this theorem is not trivial and can be found in the books mentioned above. Moreover, this theorem is also valid for bounded dissipative operators in Pκ and even for maximal dissipative operators (unbounded). See the book of Azizov and Iohvidov [AI]. We should say that the most diﬃcult results of the theory of operators in space with indeﬁnite metric are connected with this Pontrjagin theorem.

[ ] Deﬁnition 10.1. Let L be a subspace in Pκ. The subspace L ⊥ = y [y,x] = { | 0 x L is said to be orthogonal to L in Pκ (or G - orthogonal in H). The ∀ ∈ 0} [ ] subspace L =L L ⊥ is called isotropic subspace of L . ∩ [ ] [ ] It follows from Deﬁnition 1.1 that for any subspace L we have (L ⊥ ) ⊥ =L (saying L to be a subspace we always suppose that L is closed). Hence, the isotropic [ ] subspaces of L and L ⊥ coincide.

Deﬁnition 10.2. A subspace L in Pκ is called non-degenerated if its isotropic subspace L 0 = 0 . Otherwise L is called degenerated. { } Let c C be an eigenvalue of an operator A. We denote by L the subspace ∈ c consisting of all eigen and associated vectors of A corresponding to c. We call L c the root subspace corresponding to c.

Theorem 10.2. Let A be a self-adjoint compact operator in Pκ. Then there exists a Riesz basis composed of eigen and associated vectors of A if and only if the root subspace L 0 corresponding to the point 0 is non-degenerated. Moreover, if L 0 is non-degenerated then such a Riesz basis can be chosen almost G-orthogonal (i.e. all but finitely many vectors of this basis are mutually G-orthogonal). To prove this theorem is a basis goal of our lectures. Here we recall the definition of a Riesz basis, the concepts of completeness and minimality which will be used in the sequel. Definition 10.3. A system e of Hilbert space H is said to be a Riesz basis if { k} there exists such bounded and invertible operator T in H that T ek is the complete orthonormal system. { }

Deﬁnition 10.4. A system ek in H is said to be complete if any vector x H can be approximated with arbitrary{ } accuracy by a ﬁnite linear combination of elements∈ from e . { k} 11 Each maximal non-positive subspace in Pκ has the dimensional κ. This fact is trivial. 93

Deﬁnition 10.5. A system ek in H is said to be minimal if there exists a system f such that (e ,f )= δ {where} δ is the Kronecker symbol. { k} k j kj kj Exercise 10.1. A system e ∞ is complete in H if and only if the equalities { k}1 (f, ek)=0, k =1, 2,..., imply f =0.

Exercise 10.2. A system ek is minimal in H if and only if any its vector can not be approximated with any{ accuracy} by a linear combination of the other elements. Exercise 10.3. A system e is a Riesz basis in H if and only if there exists a { k} new scalar product ( , )1 in H such that the norms 1 and are equivalent and the system e ·is· complete and orthonormal in k·k( , ) . k·k { k} · · 1 10.2 Example. Let us consider one concrete example in order to see what is happening when the root subspace L 0 is degenerated. This example will help to understand the situation in general. In the space H = ℓ2 we consider the operators

000 0 0 .. 1 0 α3 α4 α5 ..  α¯ 0 β 0 0 ..  A = 3 3  α¯4 0 0 β4 0 .     α¯5 00 0 β5 .     ......     0 1  1 0 0   1000 ... G =  0100 ...    0 0010 ...    0001 ...    ...  ····  where α ∞, β ∞ satisfy the following conditions  { k}3 { k}3 α 2 < , β > β 0, if k (1) | k| ∞ k k+1 → →∞ X Exercise 10.4. Show that A is compact operator in ℓ2 if the condition (1) holds. Moreover, GA =(GA)∗, i.e. A is self-adjoint operator in P1 =(ℓ2, G). 94

Let us ﬁnd all eigen and associated vectors of A. Writing the equation Ax = λx for x =(x ,x ,...) ℓ we obtain 1 2 ∈ 2

∞ 0,x + α x , α¯ x + β x , α¯ x + β x ,... = λ x ,x ,x ,x ,... . (2) { 1 k k 3 1 3 3 4 1 4 4 } { 1 2 3 4 } Xk=3 If λ = 0 we have x1 = 0, λ = βk, xj = 0 for j = k (since βj = βk for j = k) 6 1 6 6 6 and x2 = αkβk− xk. This means that nonzero eigenvalues of A coincide with βk and the corresponding eigenvalues have the representation { } α yk = 0, k , 0,..., 0, 1, 0,... ,k =3, 4,..., { βk } where 1 occupies the k-th position. Now, suppose λ to be equal zero in (2). Certainly the vector y2 = 0, 1, 0, 0,... is the eigenvector corresponding to the eigenvalue 0. { } 2 Proposition 10.1. The root subspace L 0 of A consists of the only vector y (up to multiplication by constant) if and only if

1 α β− ℓ (3) { k k } 6∈ 2 1 Proof. Let the condition (3) holds. If (2) is fulﬁlled with λ0 then xk = α¯kβk− x1. Therefore, in the case x =0 (2) has the only solution x =0 and in the case− x =0 1 1 6 the solution x ℓ2. This means that L 0 does not contain another eigenvectors. A vector x 6∈is an associated with y2 if Ax =0 x + y2, i.e. · ∞ 0,x + α x , α¯ x + β x , α¯ x + β x ,... = 0, 1, 0, 0,... (4) { 1 k k 3 1 3 3 4 1 4 4 } { } Xk=3 It is easy to see agein that if (4) has a solution x then x ℓ . Hence, (3) implies 6∈ 2 dimL 0 =1. 1 1 α¯3 α¯4 On the other hand, suppose αkβ− ℓ2. Then y = 1, 0, , ,... is the k β3 β4 eigenvector of A if the condition{ }∈ {− }

∞ α 2 γ := 1 | k| =0 − βk Xk=3 1 1 α¯3 α¯4 2 holds and y = γ− 1, 0, , ,... is the associated with y if γ == 0. − {− β3 β4 } 6 95

Proposition 10.2. A system of root vectors of A is a Riesz basis in ℓ2 if and only if 1 α β− ℓ then the system of root vectors is not complete in ℓ and not minimal. { k k } 6∈ 2 2 1 k Proof. Let αkβk− ℓ2. Then the system y 1∞ is obviously complete and { } ∈ k { } minimal (prove this!). Moreover, y 2∞ is a Riesz basis in the subspace ℓ2 e1 , { } k ⊖{ } where e1 = 1, 0, 0,... (prove this!). Then y 1∞ is a Riesz basis in ℓ2. Now, suppose{ (3) to} be hold. Then the system{ } of root functions coincides with k k y 2∞. Obviously, it is not complete (the element e1 is orthogonal to y 2∞) and { } 2 k { } it is not minimal! We prove this showing y Span y 3∞. Suppose there exists a ∈ k { } 1 vector y = y , y ,... ℓ such that 0=(y, y )= y α β− y =0. { 1 2 }∈ 2 2 k k − k Now it follows: if y2 = 0 then y ℓ2 but if y2 = 0 then y = 0. Hence, the k 6 k 6∈ system y ∞ is complete in Span y . { }3 { }2∞ 1 2 We notice that in the case αkβk− ℓ2 the root subspace L 0 = y is degenerated since [y2, y2]=0. Because{ of} 6∈that the system of root functions{ } neither complete nor minimal.

10.3 Criteria for to be Pontrjagin subspace in Pκ. L Let us recall some well known facts on geometry of Pontrjagin space. The results which we present in this section are well known although the proofs sometimes are new. We always suppose to be closed saying to be a subspace. First, let us recall the following deﬁnitions.L L

Deﬁnition 10.6. A subspace in Pκ is said to be a Pontrjagin subspace if it is L Pontrjagin space with indeﬁnite metric inherited from Pκ.

Definition 10.7. A subspace is said to be a regular subspace in Pκ if the operator G = P G where P : H L is the opthoprothector and G is the restriction ofLG ontoL |Lis invertible.L The→ operator L G is called Gram operator.|L L L Definition 10.8. Let be a subspace in Pκ. An operator Q : H is said to be L 2 [ ] → L G-orthogonal projector onto if Q = Q and x Qx ⊥ . L − ∈ L Remark. Certainly G-orthogonal projector not always exists. But if it exists then it is bounded. Indeed, it is defined on the whole H and it is easy to prove that it is closed. Then by virtue of Closed graph theorem Q is bounded.

x x, y = Gx y. Qy = y y Q2y = Qy . n → n n → n n → ⇒ n n 96

Definition 10.9. A subspace is said to be projectively complete if L [ ] + ⊥ = H. (5) L L 0 [ ] Remark. If (5) holds then the sum is direct. Indeed, if = ⊥ then (5) implies 0 [ ] 0 L L∩L 0 [ ] + ⊥ = H, hence G( ) H. Since G is invertible we have = 0 . L ⊥ L L L ⊥ L { } Definition 10.10. A subspace in Pκ is said to be positive (uniformly positive) if [x,x] > 0 (> ε x 2 with some Lε > 0) for all 0 = x . Negative and uniformly negative subspaces|| || are defined in the same way. 6 ∈ L

Theorem 10.3. The following statements in Pontrjagin space Pκ are equivalent:

1. is positively complete; L 2. There exists a G-orthogonal projector onto ; L 3. is regular; L 4. is a Pontrjagin subspace; L 5. is non-degenerated; L [ ] 6. ⊥ is non-degenerated. L x Note. = ,x H is degenerated α = 1. L { αx ∈ 2} ⇔ ± Proof. Step 1. Let us prove 1) 2). If is projectively complete then⇔ according to Remark 3.2 the sum (5) is direct. This impliesL the existence of a uniquely defined on the whole H G-orthogonal projector onto . The implication 2) 1) follows from the definition. StepL 2. 2) 3). According⇒ to Remark 3.1 G-orthogonal projector Q onto is bounded. Now⇒ for x , x =1, we have L ∈ L || || G x QG > (G x, QGx) = (P Gx, QGx) = (Gx, QGx) = || L || || || | L | | L | | | 1 2 = (Q∗Gx, Gx) = (GQx, Gx) =(Gx, Gx) > G− . | | | | || || Here we used the fact [Qx,y]=[x,Qy] which is equivalent to (GQ)∗ = GQ and follows from the definition of Q. From the last inequality we obtain G x > ε x . Since G is self-adjoint we have it is inveritable. || L || || || L 97

+ + Step 2a. 3) 2). Suppose that G is inveritable. Then G = G G−, where G ⇒ L L L − L L and G− are uniformly positive on the subspaces ± = Im G±. This means that L + L L the norm in ( −) is equivalent to the norm 1 defined by the equality || · || 1/2 L L || · || x 1 = [x,x] . Then for any fixed y H a linear functional ϕy(x)=(x, Gy) || || | | + ∈ is continious in ( −) with respect to both norms and 1. By virtue of L L || · ||+ || · || the classical Riesz’ theorem there exists a vector x1 ( −) such that ϕy(x) = + ∈ L L ( ) + [x, y]=[x,x1− ]. Now, define Q±y = x1±. Then Q and Q− are G-orthogonal + + + projectors onto and −. Moreover, since G G− = G− G = 0 we have + +L L + L L L L Q Q− = Q− Q =0 . Hence Q = Q + Q− is G-orthogonal projector onto . Step 3. The equivalence 3) and 4) follows from the definitions. L Step 4. 4) 5). We have proved 4) 1) and according to Remark 3.2 1) 6). Step 5. 5) ⇒ 4). Let be non-degenerated.⇒ This is equivalent that Ker G =⇒ 0 . We have also⇒ L L { }

G = P (G+ G ) = P (G+ + G ) 2 P G = G 2 G− L L − − | L L − |L− L − | L | |L − L

. The operator G is uniformly positive in while G− is of ﬁnite rank. Now it follows from Fredholm| |L theorem that G is inveritableL sinceL Ker G = 0 . Step 6. The implications 5) 6) are obvious.L L { } ⇔ As a corollary we obtain the following important theorem.

Theorem 10.4. Any positive (negative) subspace in Pκ is uniformly positive (negative).

Proof. Let be a positive subspace. Then it follows G > 0, therefore is non- degenerated.L Now apply Theorem 3.1. L L Remark. First four statements in Theorem 3.1 are equivalent also in Krein space. Only on Step 5 we used the fact that G− is of ﬁnite rank.

10.4 Riesz basis theorem. First we present some simple lemmas. We omit their proofs because they can be found in any book concerning indeﬁnite metric. Let us denote λˆ := λ , λ¯ and k { k k}

ˆ = + ¯ . Lk Lλk Lλk 98

Lemma 10.1. Let λk 0∞ be a sequence of eigenvalues of A and λk be the corresponding root subspaces.{ } Then {L }

ˆ [ ] ˆ for all λˆ = λˆ Lk ⊥ Lj k 6 j and each subspace ˆ is non-degenerated with possible exeption ˆ = correspond- Lk L0 L0 ing to the eigenvalue λ0 =0.

Lemma 10.2. There are ﬁnitely many subspaces, say λ0 , λ1 ,..., λp containing associated vectors, moreover, the length of Jordan chain coLrrespondingL L to each eigenvector does not exceed 2κ +1.

Lemma 10.3. The dimension of each non-positive subspace in Pκ does not exceed κ. Now let us prove Theorem 1.2. Proof. Step 1. Let The root subspace 0 corresponding to 0 be degenerated and 0 be the isotropic subspace of . WeL denote L0 L0 ˆ = span k λk σ(A). L {L } ∈ 0 From Lemma 4.1 we obtain that 0 is also the isotropic subspace in . Then 0 L L D( 0) and = H. Thus we proved: the non-degeneracy of 0 is the neces- saryL condition⊥ L forL completeness 6 (and, certainly, for basisness) of theL system of root vectors. Step 2. Now, assume that is non-degenerated. We can represent it in the form L0 = ′ [+] ′′ L0 L0 L0 where 0′ Ker A while 0′′ is of finite dimension and both subspaces 0′ and 0′′ are invariantL ∈ under A (becauseL of Lemma 4.2 there are finitely manyL associatedL vectors, therefore we can choose 0′′ of finite dimensional). Certainly, 0′ and 0′′ are non-degenerated. L L L

In each subspace λk k>p and in 0′ we can choose G-orthogonal basis yk,s . {L } L ˆ p { } In the ﬁnite dimensional subspaces 0′′ and k 1 we can choose any basis yk,s L {L } ′ { ′′} consisting of root functions. By virtue of Theorem 3.1 Gram operators G 0 , G 0 , G are inveritable. Certainly, y for each ﬁxed k is minimal in L , L , ˆk 1∞ k,s 0′ 0′′ { L } { } L L ˆ ∞ respectively. This implies the existence of systems z such that {Lk}1 { k,h} 1 (yk,s, zk,h)=[yk,s, G− zk,h]= δsh. ˆk L 99

By virtue of Lemma 4.2 we have also

1 1 [yj,s G− zk,h]=(yj,s GG− zk,h)= δjkδsh. ˆk ˆk L L The last relation shows that the chosen system of root functions yk,s in is minimal. { } L Step 3. As before, let 0 is non-degenerated. Then is also non-degenerated. Indeed, if 0 is the isotropicL subspace of then 0 isL invariant under A (show this!). ByL virtue of Lemma 4.3 0 is ﬁniteL dimensional.L According to Lemma 4.1 A 0 has no non-zero eigenvalues.L Hence 0 . But 0 can not be the isotropic |L L ⊂ L0 L subspace of since there are no isotropic subspace in 0. L L [ ] Thus, we have proved that is non-degenerated. By virtue of Theorem 3.1 ⊥ is L [ ]L non-degenerated and, certainly, it is invariant under A. The operator A ⊥ =: A′ [ ] |L has no eigenvalues. According to Theorem 3.1 Pκ′ := ⊥ is the Pontrjagin subspace L with κ′ 6 κ. If κ′ > 0 then according to Theorem 1.1 A′ has κ′-dimensional invariant subspace and, consequently, has eigenvalues. This is a contradiction. If κ′ = 0 then Gram operator G [⊥] is positive and inveritable (Theorem 3.1) hence L [ ] G [⊥] is strictly positive, hence Pκ′ = ⊥ is the Hilbert space with the norm which L L is equivalent to the previous one. Since A is compact we have A′ is compact in Pκ′ . According to Hilbert theorem it has the eigenvalues. This is again the contradiction. Step 4. It was proved already that we can choose a system yk,s of root functions almost G-orthonormal. Let { }

+ + ′ = span +( ′ ) , ( ′ ) ′ . L {Lλk }k>p L0 L0 ⊂ L0 + Obviously, ′ is G-positive for p sufficiently large if ( ′ ) is positive in ′ . We L L0 L0 have also that ′ is of finite codimension. By virtue of Theorem 3.2 ′ is uniformly L L positive hence for yk,s, yj,h ′ we have ( G ′ >> 0 ) ∈ L L 1/2 1/2 [yk,s, yj,h]=(G ′ yk,s, yj,h)=(G ′ yk,s, G ′ yj,h)= δkjδsh. L L L 1/2 This means that the basis yk,s in ′ is equivalent to orthonormal basis G ′ yk,s { } L { L } in ′. Since ′ is of finite codimension in H we obtain the assertion of Theorem 1.2. L L

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11 Operator pencils arising in elasticity and hydrodynamics: The instability index formula Introduction The plan of the present section is the following. In subsection 1 we consider some concrete problems arising in elasticity and hydrodynamics. Further we prefer to work with abstract formulations of physical problems under consideration. For this purpose we provide general classes of operator pencils with unbounded operator coeﬃcients related to problems of origin. The main object of the paper is an operator pencil of the form A(λ)= λ2F +(D + iG)λ + T, where F and T are self-adjoint and boundedly invertible operators, while D > 0 and G are symmetric and T -bounded. The study of the pencil A(λ) is realized in Section 3. In particular, we introduce the concepts of the classical and the generalized spectra and investigate the relations between them. We associate the linear pencil

D +ßGT F 0 1 A(λ) := T λW := λ , J = T T − , − − J 0 − 0 J | | − with the quadratic pencil A(λ). It turns out that the operator T is dissipative in the 1/2 space H = HxH1, where H1 coincides with the domain of the operator T and 1/2 1/2 | | equipped with the norm ( , )1 = T , T . Generally, the spectrum σ(A) of the linearization A(λ) coincides· · with| | neither· | | the· classical nor the generalized spectrum of A(λ). However, we prove that σ(A) coincides with the generalized spectrum of A(λ) in the open right half plane if the operator W generates a Pontrjagin space metric. In this case σ(A) in the right half plane consists of ﬁnitely many eigenvalues, say κ(A), and the number κ(A) characterizes the index of instability of the equation

du d2u du A = F +(D +ßG) + Tu =0, u = u(t) dt dt2 dt The problem on stability for such kind of equations has a long background and apparently was originated by Kelvin and Tait [KT] (in the end of Section 3 we 107 present a short historical review related to this problem). The main result of the paper is the instability index formula κ(A)= ν(F )+ ν(T ) ε+(A) − where ν(F ) and ν(T ) are the numbers of the negative eigenvalues of the operators F and T respectively, while ε+(A) is expressed in terms of the lengths and the sign characteristics of Jordan chains corresponding to the pure imaginary eigenvalues of A(λ). In particular, if all the pure imaginary eigenvalues of A(λ) are of definite type then ε+(A) coincides with the number of the first type eigenvalues of A(λ) (see the definitions in Section 3).

The results of Section 2 on root subspaces of linear dissipative pencils seem at the ﬁrst sight to be isolated from the main subject of the paper. However, these results form a theoretical base to prove the index formula in Section 3. In our opinion, they have also an independent interest.

In Section 4 we return to the physical problems of origin and present the corollo- ries of our abstract results. Here we also demonstrate how the index formula can be applied to estimate the number of the nonreal eigenvalues of a self-adjoint operator pencil.

11.1 Classes of unbounded operator pencils Small oscillations of an elastic thin beam of unit length with external and internal damping (so called Kelvin-Voigt material) are described by the equation ∂4u ∂ ∂2 ∂2u ∂ ∂u ∂u ∂2u + α(x) + g(x) + β(x) + ρ(x) =0 (1) ∂x4 ∂t ∂x2 ∂x2 ∂x ∂x ∂t ∂t2 Here x [0, 1], t R+, and u(x, t) is the transverse displacement at position x and time∈t. The function∈ α(x) determines the internal damping and takes generally small values. The function β(x) > 0 determines the distribution of viscous damping, ρ(x) > 0 defines the mass distribution and g(x) is responsible for the forces of contraction or tension (see more details in [PI], for example). As the equation is considered on the finite interval, we have to submit solutions of (1.1) to some boundary conditions. For the sake of definitness we consider the case when both ends of the beam are clamped, i.e. ∂u(x, t) ∂u(x, t) u(0, t)= / = u(1, t)= / =0 (2) ∂x x=0 ∂x x=1 108

Separating variables u(x, t)= y(x)eλt, we obtain the following spectral problem

1 (4) ρ− (x) y (x) +(g(x)y′(x))′ (3) h 1 i 2 + λρ− (x) (α(x)y′′(x))′′ + β(x)y(x) + λ y(x)=0 h i y(0) = y′(0) = y(1) = y′(1) = 0. (4) Suppose that ρ(x), β(x) C[0, 1],g(x) C1[0, 1] and α(x) C1[0, 1]. According to the physical sense we∈ have ρ(x) > 0,∈ α(x) > 0, β(x) > 0∈and either g(x) > 0 or g(x) 6 0. Then quadratic eigenvalue problem (1.3), (1.4) is represented in the form 2 λ I + λ (Dα + Dβ)+ A + C y(x)=0, (5) where operators Dα, D β,A,C act in Hilbert space H = L2([0, 1], ρ(x)) with the scalar product 1 (y, z)= ρ(x)y(x)z(x)dx Z0 and are deﬁned by the equalities

1 (4) 1 (Ay)(x)= ρ− (x)y (x), (Dαy)(x)= ρ− (x)(α(x)y′′(x))′′ 1 1 (6) (Dβy)(x)= ρ− (x)β(x)y(x), (Cy)(x)= ρ− (x)(g(x)y′(x))′ on the domains

4 (A)= (D )= (C)= y y W [0, 1], y(0) = y′(0) = y(1) = y′(1) = 0 , D D α D | ∈ 2

(D )= L ([0, 1], ρ(x)) = H. D β 2 k N+ We denote by I the identity operator and by W2 [0, 1](k ) the Sobolev spaces. Naturally, it is more fruitful to study an abstract operator∈ pencil of the form (1.5) rather then problem (1.3), (1.4). We have only to extract the most essential properties of the operators (1.6). We observe that these operators satisfy the following conditions (the terminology of unbounded operator theory we borrow from the book [Ka]):

i) A = A∗ 0 (i.e. A is self-adjoint and uniformly positive), and T := A + C is self-adjoint≫ and bounded below; 109

ii) Dα and Dβ are nonnegative symmetric A-bounded operators. iii) the identity operator I and the operator C are A-compact (or T -compact) and hence T has ﬁnitely many negative eigenvalues.

Some results on spectrum of problem (1.3), (1.4) in the case α(x)= const were reported by Pivovarchik [PI]. The comprehensive study of abstract pencil (1.5) with Dα = αA, C = 0 was carried out by Lancaster and Shkalikov [LS]. Additional results in the case Dα = αA, C =0 were obtained in a recent paper by Shkalikov and Griniv [SG]. New problems appear6 in the case α = const, as pencil (1.5) in this situation has nontrivial essential spectrum. However6 , we leave an interesting problem on the spectrum localization of pencil (1.5) with Dα = αA for another occasion. We will deal with pencil (1.5) (and more general ones)6 mainly in view of the application of our index formula. A more interesting example for the application of the index formula comes from hydrodynamics. Namely, small transverse oscillations of ideal incompressible ﬂuid in a pipe of ﬁnite length are described by the equation which is obtained from (1.1) if we add in the left hand side of (1.1) the "gyroscopic" term

2sv∂2u/∂x∂t.

Here v is the velocity of the fluid and s depends on the mass of the pipe and the fluid (see [ZKM], for example). The physical meanings of the functions in (1.1) are subject to change in this situation. In particular, g(x)= v2. Assuming sv = const and repeating the previous arguments we come to the following quadratic spectral problem 2 λ I + λ (Dα + Dβ + iG)+ A + C y =0, (7) where Gy = 2sviy′, (G)= (A) − D D andDα, Dβ,A,C are defined as in (1.6). The last operators retain the properties i)-iii). The most essential properties of the operator G are the following:

iv) G is a symmetric T -bounded operator;

v) G is a T -compact operator.

It is also of interest to consider equation (1.1) on the semi-axis x R+ (see the papers of Pivovarchik [P2] and Griniv [Gr]). Assuming that the left∈ end of a 110 beam is clamped we deﬁne the operator coeﬃcients in (1.5) by equalities (1.6) on the domains

4 (A)= (D )= (D )= (C)= y y W [0, ], y(0) = y′(0) = 0 D D α D β D | ∈ 2 ∞ Obviously, this definition is correct if we assume in addition that all the functions 1 + ρ(x), ρ− (x), α(x), β(x),g(x) are bounded on R . In this case, the properties i)- ii) are retained, however, the property iii) is not true any more. This makes the problem much more complicated. Nevertheless, under some additional assumptions on the behavior of the function g(x) at (see [Gr]) the important property ∞ vi) T = A + C has finitely many negative eigenvalues remaims valid. Analogously, equation (1.1) with the additional "gyroscopic" term can be considered on the semi-axis R+ with respect to the variable x. In this case we obtain a pencil of the form (1.7) whose coefficients satisfy the properties i)-ii), iv) and also the property vi) under additional assumptions on the behavior of the function g(x) found in the paper [Gr].

11.2 Root subspaces of linear dissipative pencils and their properties In this section we deal with a linear dissipative operator pencil

A(λ)= T λW, − where W is a bounded self-adjoint operator, while T is a closed dissipative operator in Hilbert space H. This means that T is closed and

Im(Tx,x) > 0 for all x (T ) ∈ D and (T ) is the domain of T . Through all the section we also assume that there D + exists at least one point µ0 belonging to the open upper half plane C such that A (µ0) has a bounded inverse, i.e. µ0 ρ(A). If the operator W has a bounded∈ inverse then the spectrum σ(A) and the root 1 subspaces µ(A) of the pencil A(λ) coincide with those of the operator A = W − T . Hence, in thisL case spectral problems for the pencil A(λ) are equivalent to those for dissipative operators in Krein or Pontrjagin spaces (see [AI, ch.II, 2]). In the sequel § 111 we prefer to deal with the linear pencil A(λ). The motivation for this becomes clear when considering the corresponding operator differential equations. Moreover, at least formally, we obtain more general results, as we do not always assume that W generates a regular indefinite metric. The basic goal of this section is to prove formula (2.17). This formula is based on the well-known fundamental result on the existence of a maximal W -nonnegative A-invariant subspace in Pontrjagin space and on the explicit construction of maximal W -nonnegative subspaces corresponding to real normal eigenvalues of the pencil 1 A(λ) or of the operator A = W − T . In the paper [?] the author considered dissipative operator pencils of an arbitrary order n > 1 and constructed for such pencils regular canonical systems corresponding to real normal eigenvalues. This construction allows us to define the sign characteristics for Jordan chains and to realize the + construction of a maximal W -nonnegative subspace µ in the root subspace µ corresponding to a real eigenvalue µ. The additional detailsL for linear pencils wereL given in the unpublished manuscript [?]. We note also the papers of Kostyuchenko and Or azov [?] (devoted to the case of a self-adjoint operator T ) and Gomilko [G] related to this topic. However, our construction is new and, perhaps simpler, even for self-adjoint pencils. In addition we obtain the information on the connection of the middle elements of mutually adjoint canonical systems. This information is essentially used when considering half range completeness and minimality problems (see [Sh3]). Recently Ran and Temme [RT] investigated an analogous problem from another point of view. Here we present some results of [?] concerning this subject. Let µ be an eigenvalue of the pencil A(λ)= T λW and − 0 1 pj yj , yj ,...,yj , j =1,...,N, (8) be a canonical system of eigen and associated elements (or Jordan chains) corresponding to µ (see [Ke]). The linear span of all elements (2.1) is denoted (A) or Lµ simply µ and is called the root subspace corresponding to the eigenvalue µ. An eigenvalueL µ is said to be normal if A(λ) is invertible in a punctured neighborhood of µ and the number N = Ker(T µW ) as well as the lengths pj +1 of Jordan chains (2.1) are finite. It is known [Ke]− that the principal part of the Laurent expansion of 1 the function A− (λ) at the pole µ has the representation p N j ( ,xs)y0 + +( ,x0)ys · j j · · · · j j , (9) (λ µ)pj +1 s j=1 s=0 − X X − where the adjoint system

0 1 pj xj ,xj ,...,xj , j =1,...,N, (10) 112 is uniquely determined by the choice of system (2.1). It turns out that the adjoint system (2.3) is a canonical system of Jordan chains corresponding to the eigenvalue µ of the pencil A∗(λ)= T ∗ λW . Further the upper index− is always used for numeration of associated elements while the the lower one numerates eigenvalues and canonical chains simultaneously, i.e. each eigenvalue is counted as many times as its geometric multiplicity. The set of all eigenvalues of the pencil A(λ) is denoted σp(A). For the subset in σp(A) consisting of the normal eigenvalues we reserve the notation σd(A) (the discrete spectrum). Notice that canonical system of Jordan chains (2.1) is well defined for any µ σ (A) (possibly, consisting of infinitely many elements), however, adjoint ∈ p system (2.3) is well defined only for µ σd(A). As usually the indefinite scalar product (Wx,x) is denoted [x,x]. ∈ Although some of the subsequent propositions are essentially known, we present their proofs here for the reader’s convenience. New constructions axe started from Proposition 2.6. 0 Proposition 11.1. Let + be the minimal subspace containing the root subspaces corresponding to all µ LC+ σ (A). Then 0 is a W -nonnegative subspace. ∈ ∩ p L+ Proof. (Cf. [AI, Ch.2, Corollary 2.22]). We present here another, shorter proof. Suppose eigenvalues are numerated as many times as their geometric multiplicity. Let us consider the functions h h iµj t h it h 1 (it) 0 u (t)= e y + y − + + y , h =0, 1,...,p , j j 1! j · · · h! j j 0 pj C+ where yj ,...,yj are Jordan chains corresponding to the eigenvalues µ . It is h ∈ easily seen that the functions uj (t) satisfy the equation

iWu′(t)+ Tu(t)=0. h Any linear combination u(t)= cj,huj (t0) also satisﬁes this equation, therefore

[u(ξ),u(ξ)]′ =(Wu′(ξ))+(u(Pξ),Wu′(ξ)) =(iTu(ξ),u(ξ))+(u(ξ),iTu(ξ)) = 2Im(Tu(ξ),u(ξ)). − h As all the functions uj (t) vanish at , so does u(t). Integrating the last equality from t to we obtain ∞ ∞ ∞ [u(t),u(t)]=2 Im(Tu(ξ),u(ξ)) > 0. Zt 113

h In particular [u(0),u(0)] > 0 for all u(0) = cj,hyj . By the definition the set of these elements is dense in 0 , hence, 0 is a W -nonnegative subspace. L+ L+ P Proposition 11.2. Let (2.1) be a canonical system corresponding to a real eigenvalue µ. If [γ] is the integer part of a number γ then the elements p y0, y1,...,yαk , k =1,...,N, α = k , (11) k k k k 2 h i h h belong to (T ∗) and T ∗y = T y for all 1 6 k 6 N, 0 6 h 6 α . D k k k Proof. First we notice that T ∗ is well defined, as the operator T is closed by assumption (see [Ka, Ch.3, 5.5]). Now, let us prove the following: If x (T ) and § ∈ D Im(Tx,x)=0 then x (T ∗) and T ∗x = Tx. (Cf. [AI, Ch.2, Theorem 2.15]). To prove this fact, we introduce∈ D an indefinite product in the space H = H H as follows × x ,x , y , y = i(x , y ) i(x , y ). h{ 1 2} { 1 2}i 1 2 − 2 1 As T is dissipative, we have

x, x =2Im(Tx,x) > 0 for allx = x,Tx Γ(T ), h i { }∈ where Γ(T ) is the graph of T . If x (T ) and Im(Tx,x)=0 then by virtue of Cauchy–Schwarz–Bunyakovskii inequality∈ D we obtain

(x,Tz) (Tx,z) = x, z 6 x, x 1/2 z, z 1/2 =0 for allz = z,Tz Γ(T ). | − | |h i| h i h i { }∈ Hence, (Tz,x)=(z,Tx) for all z (T ). From the deﬁnition of the adjoint ∈ D operator we obtain x (T ∗) and T ∗x = Tx. Now let us prove∈ the D assertion of Proposition 2.2. As the elements of system (2.1) are Jordan chains, we have

h h 1 1 (T µW )y = W y − , o 6 h 6 p (y− := 0). (12) − k k k k In particular, Im((T µW )y0, y0)= Im(T y0, y0)=0. − k k k k 0 0 0 Therefore, y D(T ∗) and T y = T ∗y . Now we can end the proof by induction. k ∈ k k Suppose that for some h 6 αk we have proved that

s s s y D(T ∗)andT y = T ∗y fors =0, 1,...,h 1. k ∈ k k − 114

As 2h 6 pk, we ﬁnd

h 1 h h 1 h h 2 h+1 (W y − , y )=(y − , (T µW )y )=(W y − , y )= ... k k k − k k k =(y0, (T µW )y2h)=0. k − k h 1 h h h h h Hence, Im(W yk− + µW yk , yk ) = 0 and Im(T yk , yk ) = Im((T µW )yk h 1 h h h h − − W y − , y ) As before we deduce that y D(T ∗) and T y = T ∗y . k k k ∈ k k Proposition 11.3. Let (2.1) be a canonical system corresponding to a real eigenvalue µ . Then

[yh, ys]=0, j =1,...,N, h 6 [(p 1)/2], s 6 [(p 1)/2]. (13) k j k − k − If p = p then (2.6) hold for all s 6 [p /2], h 6 [p /2]. j 6 k j k Proof. Suppose pj 6 pk. Then it follows from our assumptions that h + s +1 6 pk. s s Taking into account (2.5) and the equalities T yj = T ∗yj (Proposition 2.2) we ﬁnd

h s h+1 s h+1 s 1 (y , W y )=((T µW )y , y )=(y , W y − )= ... k j − k j k j h+s+1 0 =(y , (T ∗ µW )y )=0. k − j and the equalities (2.6) follow.

Proposition 11.4. Let ν and µ be eigenvalues of the pencils A(λ) and A∗(λ) respectively. If ν = µ then the root subspaces ν(A) and µ(A∗) are W -orthogonal. In particular, truncated6 Jordan chains (2.4)L correspondinLg to a real eigenvalue µ of the pencil A(λ) are W -orthogonal to any root subspace ν(A) if ν = µ. L 6 0 p 0 q Proof. Let y ,...,y ν(A),x ,...,x µ (A∗) be Jordan chains and ν = µ. Using ∈ L ∈ L 6

s l s l s 1 l T y ,x = ν y ,x + y − ,x s l s l s l 1 1 1 (14) = y ,T ∗x = µ ¯ y ,x + y ,x − , y− := x− := 0 In particular, from these equalities we have [y0,x0]=0 . Now, the proof of the ﬁrst assertion is ended by induction with respect to the index s+l. The second assertion follows from Proposition 2.2. 115

Proposition 11.5. Let (2.1) and (2.3) be mutually adjoint canonical systems corresponding to normal eigenvalues µj which are enumerated according to their geometric multiplicity. Then the following biorthogonality relations hold:

h s yk ,xj = δk,jδh,pj s (15) − − where δm,n is the Kronecker symbol. Proof. (Cf.[Ke]). We have

h h h 1 h A(λ)y =[A (µ ) (λ µ ) W ] y = W y − (λ µ ) W y , 0 6 h 6 p , k k − − k k k − − k k k 1 where as before it is assumed that yk− := 0. Using the representation (2.2) we obtain

h 1 h yk = A− (λ)A(λ)yk s 0 0 s N2 pj ( ,xj )yj +...+( ,xj )yj h h 1 (16) · ·− = j=N s=0 pj +1 s + R(λ) (λ µk) W yk + W yk− 1 (λ µj ) − − − P P where R(λ) is a holomorphic operator function at the point µ = µ and N N +1 j 2 − 1 is the geometric multiplicity of the eigenvalue µ. We may assume that N1 = 1, N2 = N, p1 > p2 > ... > pN . Suppose that µk = µj. If we take h = 0 and compare the coefficients of the pj 1+6 s powers (λ µ )− − , 0 6 s 6 p , we find − j j eq : 29( y0,x0 y0 =0, (17) − k j j pj =p1 X eq : 210 y0,x1 y0 y0,x0 y1 y0,x0 y0 =0. (18) − k j j − k j j − k j j pj =p1 pj =p1 pj =p1 1 X X X− We do not write out the other coefficients corresponding to the indices s 2. We also notice that the third term in (2.10) should be omitted if there are no≥ Jordan chains of length (p1 + 1) 1. It follows from the definition of a canonical system N − that the elements y0 are linearly independent. Hence, from (2.9) we have j 1 0 0 eq : 211[yk,xj]=0, for all indices j such thatpj = p1. (19) Now, it follows from (2.10) and (2.11) that [y0,x0]=0, if p = p 1; [y0,x1]=0, if p = p . k j j 1 − k j j 1 116

Repeating the argument we ﬁnd [y0,xs] for all indices 0 s p . Using the last k j ≤ ≤ j equalities and taking h =1, 2,...,pk, we ﬁnd subsequently

1 s pk s yk,xj =0,..., yk ,xj =0 for all 0 6 s 6 pj.

The same arguments can be applied in the case µk = µj. Comparing the coeﬃcients of the powers (λ µ )ν in (2.10) it is found that, for h =0, 1,...,p , − j k h s yj ,xj = δh,pj s − − and relations (2.7) follow. Let a canonical system (2.1) correspond to a real normal eigenvalue µ. Denote 0 by Sµ the span of elements

y0, y1,...,yβk , k =1,...,N, β = [(p 1) /2] (20) k k k k k − 0 0 (if pk =0, we assume that βk = 1 and the element yk does not belong to µ. Let − 0 S us ﬁx an index k, 1 k N. If the number pk +1 is even we set Sµk := µ. If pk +1 ≤ ≤ S αj is odd we denote by the span of elements (2.12) combined with the elements y , Sµk j αj =[pj/2], where index j runs through all the values such that pj = pk. Similary, 0 by replacing chains (2.1) with adjoint chains (2.3) we construct subspaces ( )∗ Sµ and µ∗k . We emphasize that, according to our agreement about the enumeration of eigenvalues,S the subspaces are generally diﬀerent although µ = µ. Sµk k Proposition 11.6. For all nonzero real normal eigenvalues p the following equalities hold 0 0 = ∗ , = ∗ for all 1 6 k 6 N Sµ Sµ Sµk Sµk Proof. Suppose that yh andxh / . It follows from Proposition 2.2 that k ∈ Sµk k ∈ Sµk 0 1 αk xk,xk,...,xk , k =1,...,N, αk =[pk/2] , are chains of EAE of the pencil A(λ) as well as of A∗(λ). Since (2.1) is a canonical system, we have the representation

N h h s xk = cj,syj , if 0 6 h 6 αj =[pj/2] . (21) j=1 s=0 X X h We have assumed that xk / µk , therefore, at least one of the numbers Cj,s in (2.12) is not equal to zero for s∈ > S β = [(p 1) /2], p < p . In this case, however, j j − j k 117

pj s x − ∗ , i.e. p s 6 [p /2]. Applying Proposition 2.3 with respect to the j ∈ Sµj j − j pencilA∗(λ) we ﬁnd h pj s xk,xj − =0. On the other hand it follows fromh Propositioni 2.5 and representation (2.13) that

h pj s x ,x − = c k j − j,s h i h 0 0 ∗ Hence, the assumption xk / µk is not valid. The equality = is proved in a similar way. ∈ S S S Proposition 11.7. A canonical system (2.1) corresponding to a reed normal eigenvalue p of the pencil A( A) can be chosen in such a way that

α 0 if p +1 is even y j , yαl = ε δ , α =[p /2] , ε = j (22) j l j j,l j j j 1 if p +1 is odd ± j for all indices 1 j, l N. ≤ ≤ Proof. Fix an index k such that pk +1 is odd. Assume that there are q chains of the length pk +1, i.e. pj = pk for j = k,k +1,...,k + q 1. According to the 0 − deﬁnition of µk we have dim µk µ = q. Let Pk be the orthoprojector onto the subspace S . It follows fromS the⊖ S biorthogonality relations (2.7) that the self- Sµk adjoint operator PkW Pk has exactly q nonzero eigenvalues which correspond to an q orthogonal basis ϕs 1. We can replace, if necessary, chains (2.1) corresponding to indices l = k,k +1{ ,...,k} + q l,by their linear combinations ﬁnd obtain a new − q αk k+q 1 canonical system such that the system ϕl 1 coincides with ys k − . Then, after a proper norming, the relations (2.14) hold{ } for all indices l, j {= k,k} +1,...,k+q 1. We can repeat the same arguments for any other index r such that = . Taking− Sµr 6 Sµk into account that the subspaces µr and µk are W -orthogonal (Proposition 2.3), we obtain relations (2.14) for all indicesS suchS that 1 j, l N. ≤ ≤ Proposition 11.8. Let a canonical system (2.1) correspond to a real normal eigenvalue µ and satisfy relations (2.14). Then for all indices j such that pj = 2αj the αj elements xj of the adjoint system (2.3) have the representation

xαj = ε yαj + y, ε = 1, where y 0. (23) j − j j j ± ∈ Sµ In other words: there exists a canonical system (2.1) such that for Jordan chains of αj odd length the middle elements xj of its adjoint system have representation (2.15). 118

αj Proof. As x ∗ = , we have j ∈ Sµj Sµj xαj = c yαk + y, where y 0. j l l ∈ Sµ p =p Xi j

Now, if canonical system (2.1) satisfies relations (2.14) then cl = εjδj,l, and relation (2.15) follow. − A canonical system (2.1) which satisfies relations (2.14) or (2.15) is said to be regular. The numbers εj in (2.15) are said to be sign characteristics. We note that for linear self-adjoint pencils the sign characteristics are determined in a different way, namely, εj = 1 for Jordan chains of any length (see [GLR, Ch.3], and [KS, Lemma 2]). Simple± examples show that for dissipative pencils the definite sign characteristics can not be well defined for Jordan chains of even length. In this situation it is convenient to assume that the sign characteristics εj =0 for all chains of even length pj +1. It is supposed that this agreement holds through the rest of the paper. Let (2.1) be a regular canonical system corresponding to a normal reed eigenvalue + αj µ. Denote by − the span of elements (2.12) combined with y satisfying Lµ Lµ j relations (2.14) with εj = +1 (εj = 1). Then according to the definition of the sign characteristics we have −

N dim + = ε+ + [(p 1) /2] , where ε+ = max(0,ε ) . (24) Lµ k k − k k k=1 X Proposition 11.9. Let µ be a real normal eigenvalue of the pencil A(λ)= T λW . Then + is a maximal W -non-positive subspace in the root subspace . − Lµ Lµ + Proof. It follows from Propositions 2.2 and 2.7 that µ isa W -nonnegative subspace. + L Assume that µ ′ µ, where is also W -nonnegative subspace, and there L ⊂ L ⊂ L L + exists an element y ′ such that y µ . Obviously, y µ−, as the assumptions + ∈ L 6∈ L + 6∈ L y µ−, y µ imply [y, y] < 0. Therefore, y µ µ−. Now, using (2.7) we can ∈ L 6∈ L k 0 h 6∈ L ∪ L h ﬁnd an element yk µ such that [yk , y] = γ = 0. Denote z = ayk + γy. Then 2 ∈ S 6 [z, z]= γ (a +[y, y]) if a . On the other hand [z, z] 0, as z ′ | | → −∞ → −∞ ≥ ∈ L and ′ is by assumption W -nonnegative. This contradiction ends the proof. L Denote by the minimal subspace containing the root subspaces µ(A) cor- L + L responding to all the eigenvalues µ C and all the root subspaces µ(A) corresponding to normal real eigenvalues.∈ Analogously, let + the minimalL subspace L 119

C+ + containing µ for all µ σp(A) and all the subspaces µ corresponding to the normal realL eigenvalues.∈ For a∩ self-adjoint operator C we introduceL the (well-known) notations

π(C) = rank C+, where C+ =( C + C)/2, ν(C)= π( C) | | − Further, we use the following fundamental result. Theorem on a maximal nonnegative invariant subspace. Suppose W generates a Pontrjagin space, i.e. W is boundedly invertible and ν(W ) < ßnfty. 1 + If A = W − T and ρ(A) C = ∅ then there exists a maximal A-invariant W - nonnegative subspace H+∩ H,6 dim H+ = ν(W ), such that the spectrum of the +⊂ + restriction A/H+ lie in C , and in C coincides with the spectrum of A.

Proof. In the case T = T ∗ this is a well-known Pontrjagin theorem [P]. For a maximal W -dissipative operator A in Pontrjagin space the theorem was proved by Krein and Langer [KL], and by Azizov [A] (see [AI] and references therein).

Theorem 11.1. The subspace + deﬁned above is a maximal W -nonnegative subspace in . If W generates a PontrjaginL space and all the real eigenvalues of the pencil A(Lλ) are normal then + is a maximal W -nonnegative subspace in the whole space H. L

Proof. It follows from Proposition 2.4 and the deﬁnition that + is a W -nonnegative + L R subspace. As µ is a maximal W -nonnegative subspace in µ for any µ σd(A) (Proposition 2.9),L we have that + possesses the same propertyL in . ∈ ∩ Now, let W generate a PontrjaginL space and all the real eigenvaluesL of the pencil A(λ) are normal. According to the generalized Pontrjagin theorem there exists a maximal W -nonnegative subspace H+ in H, dim H+ = ν(W ) , such that 0 + 0 + + H , where + is deﬁned in Proposition 2.1. As the subspace H µ is L ⊂ ⊂ L L + ∩ L W -nonnegative in µ and µ is a maximal nonnegative subspace in µ (Proposition L + L + L 2.9), we have: dim(H µ) 6 dim µ (see, for example, [AI, Ch.I, §4] ). Then it follows that ∩ L L

dim H+ = dim 0 + dim H+ 6 dim +, σ := σ (A). L+ ∩ Lµ L d d µ R σd ∈X∩ On the other hand, it is known ([AI, Ch.I, §4]) that dim + 6 ν(W ) = dim H+. Hence, dim + = dim H+ and from this it follows thatL + is a maximal W - nonnegativeL subspace in the whole H. L 120

Corollary 11.1. Let W be boundedly invertible, ν(W ) < , and a11 the real eigenvalues of A(λ) be normal. Then the following formula is∞ valid

κ(A)+ ε+ + [(p 1) /2] = ν(W ), ε+ = max(0,ε ) (25) k k − k k µk R σd X∈ ∩ + Here κ(A) is the total algebraic multiplicity of all eigenvalues in C and εk(pk + 1) are the sign characteristics (the lengths) of Jordan chains of regular canonical systems corresponding to real normal eigenvalues µk. Proof. It follows from formula (2.16) and Theorem 2.10. Remark. Formula (2.17) is not applicable if the pencil A(λ) has real eigenvalues which are embedded into the essential spectrum. In this case we do not know how to determine the sign characteristics and how to realize the explicit construction of a maximal W -nonnegative subspace in the the root subspace µ. However, the following inequality is always valid (cf. [AI, Ch.2. Theorem 2.26])L

κ(A)+ [(p 1) /2] 6 ν(W ), σ := σ (A) (26) k − p p µk R σp X∈ ∩ This inequality is much more simple and follows directly from Propositions 2.1, 2.3 and 2.4. It expresses the fact that the linear span of all root subspaces µ corresponding to µ σ (A) C+ and all the truncated root subspaces 0 correspondingL ∈ p ∩ Sµ to µ σp(A) R forms a W -nonnegative subspace (not necessarily a maximal one). Indeed,∈ using∩ (2.17) we can improve (2.18) and write the following inequality

κ(A)+ ε+ + [(p 1) /2] 6 ν(W ), σ := σ (A) (27) k k − p p µk R σp X∈ ∩ where ε+ = max(0,ε ) if µ σ and ε+ =0 if µ σ σ . k k k ∈ d k k ∈ p\ d 11.3 Quadratic dissipative pencils and the instability index formula In this section we study a quadratic operator pencil of the form (3.1) A(λ)= λ2F +(D + iG)λ + T (28) Further it is always assumed that the coeﬃcients in (3.1) are operators in Hilbert space H satisfying the following conditions: 121

i) F is a self-adjoint bounded and boundedly invertible operator;

ii) T is deﬁned on the domain V (T ), T = T ∗ and T is boundedly invertible; iii) D and G are symmetric T -bounded operators (i.e. D and G are symmetric, (D) (T ) and (G) (T ). Moreover, D 0. D ⊂ D D ⊂ D ≥ These assumptions imply that A(λ) is a quadratic dissipative pencil with respect to the imaginary axis in the following sense (see [Sh3])

Im(ζA(iζ)x,x)= ζ2(Dx,x) > 0 for all x (T ) and ζ R ∈ D ∈ One may expect that the quadratic dissipative pencil (3.1) can be transformed into a linear dissipative pencil. Indeed, such a linearization will be realized below. How- ever, working with unbounded pencils we come to some new problems which do not arise when considering pencils with bounded coefficients. In particular, the spectrum of a linearization may not coincide with the spectrum of the original pencil. According to our assumptions A(λ) is well defined for each λ C on the domain (T ). Hence, the first natural definition of the resolvent set ρ(A∈) is the following: ζD ρ(A) if A(ζ) with the domain (T ) has a bounded inverse. To give another ∈ D definition, we consider the scale of Hilbert spaces Hθ,θ R (H0 = H) generated 1/2 ∈ by the self-adjoint operator S2 := T := T 2 . Namely, if θ > 0 we set H = | | θ x x Sθ with the norm x = Sθx . If θ< 0, the space H is defined as | ∈ D k kθ θ the closure of H with respect to the norm x = Sθx . θ Let us associate the pencil k k

Aˆ(λ)= λ2Fˆ + λ(Dˆ + iGˆ)+ J with the pencil A(λ). Here

1 1 1 1 1 1 1 Fˆ = S− FS− , Dˆ = S− DS− , Gˆ = S− GS− ,J = T − T | | Obviously Fˆ and J are bounded. From the next Proposition it follows that Dˆ and Gˆ are also bounded in H.

Proposition 11.10. Let S be an uniformly positive self-adjoint operator and B 2 θ 2 θ be a symmetric operator such that (B) (S ). Then the operator S − BS− θ 2 D ⊃ D deﬁned on the domain (S − ) is bounded in H for all 0 θ 2. Equivalently, B D ≤ ≤ is bounded as an operator acting from Hθ into Hθ 2. − 122

2 Proof. As B is closable, the assumption (B) (S ) implies that B : H2 H is a bounded operator (this follows immediatelyD ⊃ fromD the closed graph theorem).→ Hence, the adjoint operator B∗ : H H 2 is also bounded. As B∗ B, we have → − ⊃ that B : H H 2 is bounded. Now, applying the interpolation theorem (see [LM, → − Ch.l], for example) we find that BB : Hθ Hθ 2 is bounded for all 0 θ 2. → − ≤ ≤ Let σ(Aˆ) be the spectrum of the pencil Aˆ(λ) with bounded operator coefficients in the space H. It is easily seen that σ(Aˆ) coincides with the spectrum of A(λ) considered as the operator function in the space H 1 on the domain (A) = H1. Both our definitions of the spectra are better understood− (especially forD the special- ists working with partial differential operators) if we say the following: σ(A) is the spectrum of the pencil A(λ) considered in the “classical” space H while σ(Aˆ) is its spectrum in the generalized space H l. Generally, σ(A) = σ(Aˆ). What is− the connection between the classical and the generalized spectra?6 Some light is cast on this problem by the next propositions. It will be convenient to define in the complex plane the open set ρ (A) := ρ(A) σ (A). m ∪ d The set ρm(Aˆ) is defined analogously. In the other words ρm(A) and ρm(Aˆ) are the 1 domains where the operator functions A− (λ) is finite meromorphic in the spaces H and H 1, respectively. − Proposition 11.11. In the domain pm(A) fl pm(.4) all the eigenvalues and Jordan chains of .4(A) in the spaces H and H- coincide.

11.4 Applications In this section we shall apply the obtained abstract results to concrete problems considered in Section 1.

Theorem 11.2. Formula (3.11) or its simplifications (3.12) or (3.13) are valid for operator pencil (1.7) associated with the problem of small oscillations of ideal incompressible fluid in a pipe of finite length if the condition KerT = 0 is fulfilled (T := A + C). For a pipe of infinite length the assertion of Theorem{ 3.7} is valid if g(x) is such a function that KerT = 0 and ν(T ) < . { } ∞ Proof. The conditions i)-ii) and iv) of Section 1 imply conditions i)-iii) of Section 3 if it is assumed in addition that KerT = 0 . Moreover, for a pipe of finite length the assumptions of Corollary 3.8 are fulfilled.{ } For a pipe of infinite length the operators G and I are not T -compact and we must use Theorem 3.7. In the last 123 case we can not guarantee the absence of pure imaginary eigenvalues belonging to the non-discrete spectrum. If KerT = 0 then λ = 0 is an eigenvalue of pencil (3.1). In this case the analogue of formula6 { } (3.11) can also be obtained. For this purpose one has to modify the results of Section 2 for the case KerW = 0 . Technically this is not a trivial 6 { } work. However, the estimates for the number κ(Aˆ) can be obtained easily if KerT = 0 . 6 { } Theorem 11.3. Suppose that a pencil A(λ) is defined by (3.1) and its operator coefficients satisfy the assumptions i)-iii) of Section 3 with the possible exception that the operators F and T are not necessarily boundedly invertible. Suppose that there exists a point µ, Reµ > 0 such that Aˆ(µ) is boundedly invertible. Then

κ(Aˆ) ν(F )+ ν(T ). (29) ≤ Proof. Let us consider the pencil

2 Aτ (λ)= λ (F + τI)+(D + iG)λ + T + τI, τ> 0.

Obviously, ν(T + τI)= ν(T ), ν(F + τI)= ν(F ), if τ (0,τ0) and τ0 is suﬃciently small. By virtue of Theorem 3.7 we have ∈

κ(Aˆ ) ν(F )+ ν(T ) for all 0 <τ<τ . (30) τ ≤ 0 Repeating the arguments from the proof of the Theorem 3.7 and taking into account that µ ρ(Aˆ) for some µ with Reµ > 0 we obtain that the spectrum of ∈ Aˆ(λ) := Aˆ0(λ) in the open right half plane consists only of normal eigenvalues. These eigenvalues continuously depend on τ (see [Ka], Ch. 7). Then (4.2) implies (4.1). The results of Sections 2 and 3 can also be applied to self-adjoint pencils. Lan- caster and Shkalikov [LS] considered an operator pencil L(λ) deﬁned by (1.5) with C =0, Dα = αA and obtained the following estimate η/2 min π(L(k)), π(L) := ν( L), (31) ≤ k R − ∈ where η is the number of non-real eigenvalues of the pencil L(λ) counting with algebraic multiplicities. Using an analytic approach Shkalikov and Griniv proved a sharper estimate for the case C = 0 and reproved (4.3) for C = 0 (if C is an A-compact operator). Here we reﬁne the corresponding results from6 [LS] and [SG]. 124

Theorem 11.4. Let L(λ)= λ2F + λD + T, 2 where T = T ∗ and F ,D are symmetric and T -bounded operators. Let S = T + I | | and the scale of Hilbert spaces Hθ be generated by the operator S 0. Suppose that there exist real points a and b belonging to ρ(Lˆ) such that ≫ π(L(a)) < , ν(L(b)) < . ∞ ∞ Then the non-real spectrum of L(λ) in the space H 1 consists of ﬁnitely many, say η, non-real eigenvalues, and the following estimate− is valid η/2 π(L(a))+ ν(L(b)) δ+(L), (32) ≤ − + where δ (L) is the number of real eigenvalues µk of L(λ) counting with multiplicities such that

µk a (b a) − (L′(µ )y, y) > 0 for all y KerL(µ ). − b µ k ∈ k − k Proof. We use the same idea as in [LS] where estimate (4.4) was obtained in a slightly different situation not taking into account the number δ+(L). It was shown in Section 3 that the Spectrum of L(λ) in the space H 1 coincides with the spectrum 1 1 − of Lˆ(λ)= S− L(λ)S− in the space H. The pencil Lˆ(λ) has the bounded operator 1 coefficients Fˆ, Dˆ, Tˆ. After the substitution λ = (bξ + a)(ξ + 1)− we obtain the quadratic pencil L˜(ξ):=(ξ + 1)2Lˆ(λ(ξ)) = ξ2F˜ + ξD˜ + T,˜ F˜ = Lˆ(b), T˜ = Lˆ(a). Let us consider the linearization of L˜(ξ) D˜ T˜ F˜ 0 L(ξ)= ξ . − T˜ 0 − 0 T˜ − Suppose that ξk is a simple (or semi-simple) real eigenvalue of L˜(ξ) with a corresponding eigenvector yk. then the sign characteristic εk (see Section 2) is defined as follows F˜ 0 ξ y ξ y ε = k k , k k = ξ(L˜ (ξ )y , y ) k ˜ y y ′ k k k 0 T k k H H − × 1 = ξ λ′(ξ )(Lˆ′(λ )y , y )=(b a)(µ a)(b µ )− (Lˆ′(λ )y , y ). k k k k k − k − − k k k k Now apply Corollary 2.11. We note that the estimate (4.4) is also new for matrix pencils. 125

Bibliography for Section 11

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[ZKM] V.N. Zefirov, V.V. Kolesov and A.I. Miloslavskii, On eigenfrequences of a strightline pipe, Izv. Acad. Nauk SSSR, Ser. Mech. Tverdogo Tela (1985), no. 1, 179-188 (Russian); English transl. in Math. USSR Izv. Ser. Mech (1985). 128 12 Factorization of elliptic pencils and the Mandel- stam hypothesis Introduction This section is a modified and extended version of section 6, where the main attention was paid to the finite dimensional case. Here we deal with pencils which present the abstract models of concrete essentially infinite dimensional problems. Some problems of mathematical physics (one of them will be discussed below) can be written abstractly in the form

d2u du (u)= F + iG +(H ω2R)u =0. (1) A − dy dy − Here F,G,H, and R are symmetric operators on a suitable Hilbert space satisfying certain additional conditions which ensure the elliptic nature of this equation,H and ω is a physical parameter (frequency) which appears after the separation of the time variable. Physical meaning have solutions of equation (1) which are bounded as y and satisfy the so-called radiation principle. Diﬀerent approaches to formulate→ ∞ the radiation principle have been widely discussed in physical and mathematical literature (see, for example, Sveshnikov [Sv], the books of Zilbergleit and Kopilevich [ZK], Vorovich and Babeshko [VB]). The formulation of the radiation principles is based on the preliminary spectral analysis of the pencil T (λ)= λ2F + λG + H ω2R. (2) ω − We say λk,vk with vk = 0 is an eigenpair of the pencil Tω(λ) if Tω(λk)vk = 0. Any eigenpair{ }λ ,v generates6 the solution { k k} iλky uk = e− vk (3) of equation (1). Those solutions which correspond to the real eigenvalues λk are of particular interest, they are called propagating waves. Among propagating waves there are the outgoing and incoming ones. It was understood after the author’s discussions with physicists, that the Mandelstam hypothesis can be formulated as follows (see [BS], [ZK], although the problem is not clearly formulated there): given an element x there is a unique solution u(y) of equation (1) such that u(0) = x, and as y ∈Hthe solution u(y) asymptotically coincides with a linear combination of outgoing→∞ waves. 129

This problem is also related to those settled by Reyleigh on the wave diffraction on a periodic surface. Some of them are treated in the book of Wilcox [W]. This connection, however, is not easily seen, and its demonstration is left for a future occasion. Our first aim is to define an abstract model of strongly elliptic equations in wave- guide domains whose symbols are quadratic selfajoint pencils. The main goal is to prove the factorization theorems for these pencils and investigate the properties of a right divisor. The results obtained enable us, in particular, to approve the Mandelstam hypothesis. Our starting point was a celebrated paper of Krein and Langer [KL] which deals with pencils of the form

L(λ)= I + λB + λ2C.

Here I is the identity operator, B is bounded and selfadjoint, while C is positive and compact. The fundamental theorem of [KL] yields the factorization

L(λ)=(I λZ )(I λZ). − 1 − Among possible divisors there is an operator Z whose spectrum σ(Z) lies in the closed upper (or lower) half plane and coincides with the spectrum of L(λ) in the open half plane. A further analysis of an operator Z occuring in this factorization was given in the papers of Kostyuchenko and Orazov [KO1] and Kostyuchenko and Shkalikov [KS]. However, while attempting to apply the method of Krein and Langer to attack the factorization problem for qudratic pencils with unbounded coeﬃcients, one faces new serious obstacles. Moreover, a further analysis of divisors has to be carried out after the factorization is already proved. In particular, to prove the Mandelstam hypothesis we have to show that among possible factorizations

T (λ)=(λ Z )F (λ Z), ω − 1 − there is the only operator Z which generates a C0 (or holomorphic) semigroup in an appropriate Hilbert space. The plan of this paper is the following. In Section 1 we deﬁne strongly elliptic pencils as relatively compact pertubations of uniformly positive ones. For pencils with discrete spectrum our deﬁnition is equivalent to the asymptotic inequality

T (λ) ε(λ2 + H), for λ R, λ >r , ω ≥ ∈ | | 0 130 provided r is sufficiently large. This assumption can be easily checked for concrete 0 o elliptic systems, since it is equivalent to the Gording inequality (this is shown in Section 3). Following the paper [S1] we define the "classical" and the "generalized" spectra of Tω(λ). We show that the classical and generalized spectra of a strongly elliptic pencil coincide in the union of a ball centered in the origin and a sufficiently small double sector containing the real axis. Moreover, in this domain the spectrum consists of finitely many normal eigenvalues. For large values of λ inside a double sector we prove the resolvent estimates which play an important| role| in the sequel. They look similar to the classical a priori estimates for regular elliptic boundary value problems obtained by Agmon, Douglas and Nirenberg [ADN], [AN] and Agranovich and Vishik [AV]. Nevertheless, estimates obtained in Section 1 are of different nature, in particular, they can be used for elliptic systems on non-smooth domains. One can feel the difference while considering the example in Section 6. In Section 2 we give more details about the real spectrum of Tω(λ). In particular, we show that the outgoing waves correspond to those eigenpairs which have the positive sign characteristics

εk =(Tω′ (λk)vk,vk). In Section 3 we prove the factorization theorem for positive strongly elliptic pencils. We could obtain this theorem (although is not easy) using classical results on the factorization of non-negative operator functions on the real line (see the exposition of this theory in the books of Foias and Nagy [FN] and of Rosenblum and Rovnjak [RR]). However, we preferred to give a new approach based on the semigroup theory, as it seems more natural for the problem in question. Moreover, we believe that this method can be modified to fit arbitrary strongly elliptic pencils not positive ones only. In Section 4 we prove the factorization theorem for strongly elliptic pencils (not necessarily positive) under an additional assumption (the so-called Keldysh-Agmon condition). The proof is based on the preliminary analysis of the half-range completeness and minimality problem for the pencil Tω(λ). To solve these problems we borrow the ideas from the papers [KS] and [SS]. In this exposition, however, we get rid of some superfluous assumptions and presented the material in a different and shorter way. In particular, in contrast to the cited papers, now we can apply our results in the case when the operator H is generated by an elliptic operator (or system) on a non-smooth domain. The results of Section 3 and 4 are used in Section 5 to approve the Mandelstam hypothesis. Finally, in Section 6, we demonstrate how the obtained results can be applied to the elliptic system of differential equations of elasticity theory. 131

The second part of the paper (Sections 4–6) is a revised version of results on elliptic pencils presented by the author in the unpublished manuscripts [S2], [S3].

12.1 Elliptic pencils and their spectrum Definition of regular elliptic and strongly elliptic pencils. In what follows we always assume that the coefficients of equation (1) or a quadratic pencil Tω(λ) of the form (2) are operators on a separable Hilbert space having the following properties (we borrow the terminology from the book of KatoH [Ka]): F is a bounded and uniformly positive operator (0 F ); H is a selfadjoint uniformly positive operator with≪ domain≪∞ (H) (H = D ⊂ H H∗ 0); ≫ 1/2 G is a symmetric operator (G G∗) with domain (G) (H ); ⊂ D ⊃ D 1 R is an H-compact positive operator (i.e. R> 0, (R) (H) and RH− is compact on ), and the closure of the operator MRD M has⊃ D trivial kernel. It is worthH noting that for any symmetric H-boundedH H operator R the closure of MR M exists and is a bounded operator on (see the remark explaining the boundednessH H of the operator C defined in (7)). H The parameter ω plays a role in the sequel only in cases when we appeal to physical considerations. For fixed ω it will be convenient to denote S = H ω2R and consider the pencil − T (λ)= λ2F + λG + S (1) implying that

1/2 0 F , G G∗, (G) (H ), ≪ ≪∞ ⊂ D ⊃ D (2) S = S∗ is a relatively compact perturbation of H = H∗ 0. ≫ We use the scale of Hilbert spaces θ generated by the "main" operator H θ/2 H. Namely, for θ 0 the space θ coincides with (H ) endowed with the θ/2≥ H D norm x θ = H x , while θ is the dual space to θ with respect to . The followingk k factk will bek used in theH− sequel: If S 0 thenH the scale of HilbertH spaces ≫ generated by S coincides with θ for 0 θ 2. This fact follows from the assumption (S)= (H) and the interpolationH ≤ theorem≤ (see, e.g., [LM], Ch 1). D D Further, by writing T (λ) instead of Tω(λ) we always assume that T (λ) is of the form (1) with coefficients satisfying conditions (2). The definition of a regular elliptic boundary value problem (see [AN], [AV], [LM]) is expressed algebraically in terms of principle symbols of a differential equation and boundary operators (the so-called ellipticity condition for the equation and the 132 complementing Lopatinskii condition for boundary operators). Suppose that we consider a regular elliptic problem in a wave-guide domain Ω R (Ω is a smooth bounded domain in Rn) and write it in abstract form (1) (homogeneous× boundary conditions are included in the domain of the main operator H). It follows from the results of [AN] and [AV]: a problem is regular elliptic if and only if T (λ) is invertible for λ R and λ >r , with r large enough and for these values of λ ∈ | | 0 0 1 1/2 1 2 1 HT − (λ) + λ H T − (λ) + λ T − (λ) const. (3) k k | | k k | | k k≤ These arguments lead to the following definition (as we agreed the parameter ω is omitted).

d Deﬁnition 12.1. A pencil T (λ) or equation T idy u(y)=0 is said to be regular elliptic if estimate (3) holds for λ R, λ >r0. ∈ | | In this paper, however, we deal mostly with equations which are abstract gener- alizations of strongly elliptic equations (see, e.g., the book of Fichera [Fi]). Deﬁnition 12.2. A pencil T (λ) is said to be uniformly positive if there exists a number ε> 0 such that

T (λ) ε(λ2 + H) for all λ R. (4) ≥ ∈ It follows from the definition that S 0 if T (λ) is uniformly positive. As 1 ≫ 1 (S) = (H), both the operators SH− and HS− are defined on the whole , andD it followsD from the definition that they are closed. Hence, by the closed graphH theorem these operators are bounded and then there exist positive constants c0, c1 such that c Hx Sx c Hx , x (H). 0k k≤k k≤ 1k k ∈ D By virtue of the Heinz inequality (see [Ka], Ch.5.4 ) we have c0H S c1H. Therefore, (4) implies also ≤ ≤

T (λ) ε (λ2 + S), λ R, ≥ 1 ∈ with ε1 = ε/c1. Actually, we have just showed that the operator H in Deﬁnition 1.2 can be replaced by any operator S = S∗ 0 such that (S)= (H). ≫ D D Deﬁnition 12.3. A pencil T (λ) of the form (1) is said to be strongly elliptic if there exists an H-compact positive operator V such that T (λ)+ V is uniformly positive. 133

Proposition 12.1. Let T (λ) be a strongly elliptic pencil. Then there exist numbers ε> 0 and r0 > 0 such that T (λ) ε(λ2 + H) for all λ R and λ >r . (5) ≥ ∈ | | 0 Proof. By the deﬁnition we have T (λ) ε(λ2 + H) V, (6) ≥ − 1 where V is an H-compact positive operator. Obviously, if VH− is compact in then V is H-bounded with zero H-bound, i.e. for any ε > 0 there exist c = c(Hε) such that Vx ε Hx + c x , c = c(ε), x (H). || || ≤ || || || || ∈ D By virtue of the Heinz inequality we have V εH + cI, ≤ where I is the identity operator. Taking in the last inequality ε/2 instead of ε we obtain (5) from (6). The inverse assertion of Proposition 1.4, generally, is not true. Examples can be easily given by considering bounded operators G and S on . However, we can invert the statement of Proposition 1.4 assuming that H has discreteH spectrum or, equivalently, the identity operator is H-compact.

1 Proposition 12.2. Let H− be compact in .Then condition 5 implies that T (λ) is strongly elliptic. H Proof. According to (2) G is P -bounded operator and S H is H-compact. Hence, G is H-compact and for anyHλ R, we have − ∈ T (λ)= λ2F + H + K(λ),

1 where K(λ)H− is compact. Given ε> 0 there exists c = c(λ) > 0 such that (K(λ)x,x) ε(Hx,x)+ c(x,x). | |≤ If ε =1/2 and c is the maximum of c(λ) on the interval ( r ,r ) then 0 − 0 0 1 T (λ)+ c I (λ2F + H), for λ ( r ,r ), 0 ≥ 2 ∈ − 0 0 and together with (5) this implies that T (λ)+ c0I is uniformly positive. 134

Location of the spectrum and the resolvent estimates. In [S1] three different approaches are proposed to define the spectrum of a pencil with unbounded coefficients. In particular, the "classical" and the "generalized" spectrum of T (λ) are defined as follows. We say µ belongs to the classical spectrum of the pencil T (λ) if T (µ) is not boundedly invertible in . This concept is natural but not always convenient (see [S1]). To define the generalizedH spectrum, let us consider F , G and S as the operators acting on the space 1 with domain = 1 (recall that = (Hθ/2) is the scale of Hilbert spaces generatedH− by the operatorD HH). Since all Hθ D these operators are H-bounded and symmetric, they are well defined in 1 with domain (see details in [S1]). Now we can consider T (λ) as an operatorH function− H1 in the space 1 defined on the domain (T )= 1. We say that µ belongs to the H− D H generalized spectrum of the pencil T (λ) if T (µ) is not boundedly invertible in 1. The complement of the generalized spectrum is said to be the generalized resolventH− set of T (λ). It can be easily checked (see [S1]) that µ belongs to the generalized spectrum of T (λ) if and only if µ belongs to the spectrum of the pencil

L(λ)= λ2A + λB + C, with bounded in coeﬃcients H A = MF M, B = MG M, C = MS M. (7) H H H H H H 1 We have to explain why C is bounded. The operator SH− is deﬁned on the 1 1 whole and is closed. Then SH− and its adjoint H− S are bounded, and according to theH interpolation theorem the operator C is bounded, too. Generally, we can not claim that the generalized and the classical spectra of T (λ) coincide. In the subsequent theorems we clarify the relationship between these concepts.

Theorem 12.1. Let ρcl(T ) and ρgen(T ) be the classical and the generalized resolvent sets of a strongly elliptic pencil T (λ). Then

ρ (T ) ρ (T ). cl ⊂ gen

The real line belongs to ρcl(T ) ρgen(T ) with the possible exception of ﬁnitely many ∩ 1 normal eigenvalues whose algebraic multiplicity coincide in both sences. If H− is compact then the classical and the generalized spectra coincide in the whole and consist of normal eigenvalues. C 135

Proof. The last assertion of the theorem and the coincidence of the algebraic multiplicities of the normal eigenvalues in both sences are proved in [S1], §3. Let λ ρ (T ). Then ∈ cl T (λ): and T (λ): H2 →H H2 →H are isomorphisms, hence, so are the operators

T ∗(λ)= T (λ): 2 and T ∗(λ)= T (λ): 2. H→H− H→H− From the interpolation theorem (see [LM], Ch1) we obtain that T (λ): 1 1 H → H− is an isomorphism, i.e. λ ρgen(T ). Let us prove the second∈ statement. The assumption (G) (H1/2) implies 1/2 D ⊃ D that GH− is defined on the whole and it follows from the definition that it is 1/2 H closed. Hence, GH− is bounded and its norm c. Then for any ε> 0 we have ≤ 2 2 1/2 2 2 2 1 2 Gx c H x c (Hx,x) 2c (ε Hx + ε− x ), x ( ). k k ≤ k k≤ ≤ k k k k ∈ D H This means that G is H-bounded with zero H-bound, and so is the operator K(λ)= λ2F + λG + S H + V − for any H-compact operator V and λ . We can choose a positive operator V such that ∈ C T (λ)+ V ε2(λ2 + H), λ R. ≥ ∈ It follows from the stability Theorem V.4.11 of [Ka] that T (λ)+V = H +K(λ) 0 is self-adjoint for any fixed λ R. Therefore, T (λ)+ V is boundedly invertible≫ in for all λ R (and, hence, in∈ a neighborhood of any point λ R). We have the Hrepresentation∈ ∈ 1 T (λ)= I V (T (λ)+ V )− (T (λ)+ V ), − 1 R where V (T (λ)+ V )− is a holomorphic operator function in a neighborhood of whose values are compact operators. It follows from the theorem on holomorphic operator function (see [GGK], Ch. XI) that the spectrum of T (λ) in a neighborhood of R consists of finitely many isolated eigenvalues of finite algebraic multiplicity. According to Proposition 1.4 all the real eigenvalues are located in a finite interval [ r ,r ]. This ends the proof. − 0 0 For δ > 0 and 0 <ϕ π/2 we denote ≤ B = λ : λ δ , Λ− = λ : argλ <ϕ , Λ− = λ : π argλ <ϕ δ { | |≤ } ϕ { | | } ϕ { | − | } + and Λ = Λ Λ−. ϕ ϕ ∪ ϕ 136

Theorem 12.2. Let T (λ) be strongly elliptic. Then there exist positive numbers ϕ and δ such that the union Λ B with the possible exeption of ﬁnitely many normal ϕ ∪ δ eigenvalues belongs to the classical resolvent set of T (λ) (and hence to ρgen(T )). Moreover, the estimate

2 1 1 1 λ T − (λ) + λ PT − (λ) + PT − (λ) P const (8) | | k k | |kH k kH H k≤ holds for all λ Λ , λ >r if r is large enough. ∈ ϕ | | 0 0 + Proof. Let us prove (8) for λ Λ , the same arguments can be applied for λ Λ−. ∈ ϕ ∈ ϕ If λ = reiθ, then

T (λ)= T (r)+ r2(e2iθ 1)F + r(eiθ 1)G. (9) − − This equality and Proposition 1.4 yield the estimate Re (T (λ)x,x) ε(r2(x,x)+(Hx,x)), x , λ Λ , λ >r , (10) ≥ ∈H2 ∈ ϕ | | 0 for suﬃciently small ϕ and large r0. We noticed already that the coeﬃcients of the pencil L(λ)= MT (λ) M are bounded operators. From (9) we have H H L(λ) y Re(L(λ)y, y) εr2(y, y), y , λ Λ , λ >r . k kk k≥ ≥ ∈H1 ∈ ϕ | | 0 By continuity this inequality holds for all y and implies that zero does not belong to the numerical range of L(λ). Then L∈(λ H) is invertible, and

1 1 L− (λ) ε− . (11) || || ≤ From this we have that

1 T (λ)= PL− (λ) P : 1 1 H H H →H− 1 1 1 is an isomorphism, and T − (λ) exists in 1. Now, L− (λ) = PT − (λ) P and from (11) we obtain the estimate of the thirdH− term in (7). H H It follows from Proposition 1.5 that T 1/2(r)x ε (r2 + H)1/2x , r>r , x (H). k k≥ k k 0 ∈ D By virtue of Theorem 1.6 T (r) is invertible for r>r0, hence,

1/2 1 2 1/2 1 T − (r) ε− (r + H)− ε− r, k k≤ k k≤ (12) 1/2 1/2 1/2 1 GT − (r) c H T − (r) cε− . k k≤ k k≤ 137

We have

1/2 1/2 1/2 1/2 T (λ)= T (r)(I + G(λ))T (r), G(λ)= T − (r)(T (λ) T (r))T − (r). − + It follows from representation (9) and estimates (12) that G(λ) 1/2 if λ Λϕ and ϕ is suﬃciently small. Hence, T (λ) is invertible in fork λ k≤Λ+, and ∈ H ∈ ϕ 1 1/2 2 2 2 T − (λ) 2 T − (r) 2ε− r , k k≤ k k ≤ 1/2 1 1/2 1/2 1/2 2 1 H T − (λ) 2 H T − (r) T − (r) 2ε− r− . k k≤ k k k k≤ This completes the proof. Remark. We say T (λ) is positive if T (λ) > 0 for all λ R. We claim: A positive strongly elliptic pencil is uniformly positive. Indeed,∈ if T (λ) > 0 for all λ R then λ is not an eigenvalue of T (λ), and according to Theorem 1.6 T (λ) ∈ is boundedly invertible in as well as in 1. Therefore T (λ): 1 1 is a H H− 1 H → H− continuous bijection for λ [ r0,r0], hence, so is T − (λ): 1 1. This yields 1 ∈ − H− →H the estimate PT − (λ) P const, which implies T (λ) εH. Bearing in mind Proposition 1.4,kH we ﬁnd thatH k≤T (λ) is uniformly positive. ≥

12.2 The real spectrum of a strongly elliptic pencil We noticed in the Introduction, that the real eigenvalues of a pencil T (λ) play a significant role in physical considerations, as they correspond to waves propagating the energy at the infinity (or from the infinity). We already proved that strongly elliptic pencils may have only finitely many real eigenvalues. In this section we obtain additional valuable information. First, recall that a point µ is said to be a normal eigenvalue of T (λ) if it is an isolated point of the spectrum∈ C of T (λ) and the principal part of the Laurent 1 expansion of the resolvent T − (λ) in a neighborhood of µ admits a representation of the form N pk pk s s ( , z − ) x · k k . (1) (λ µ)pk+1 s s=0 − Xk=1 X − Here

0 pk xk,...,xk , k =1,...,N, (2) is a canonical system of eigen and associated elements of T (λ) and

0 pk zk,...,zk , k =1,...,N, (3) 138 is the adjoint canonical system which is uniquely deﬁned by the choice of system (2). Let µ R. Since the classical and the generalized spectra of T (λ) coincide in ∈ a neighborhood of R, the elements of systems (2) and (3) belong to 2. It follows from [KS], Lemma 2.1 that there exists a canonical system (2) such thatH

s s xk = εkzk, k =1,...,N, s =0,...,pk, where εk = 1. Such a canonical system is called normal and the numbers εk are called the sign± characteristics of the corresponding Jordan chains. A real eigenvalue µ is said to be of positive (negative) type if

(T ′(µ)y, y) > 0 (< 0) for all y KerT (µ). ∈ Proposition 12.3. If µ is a semi-simple real eigenvalue of T (λ) and 0 0 1 V = ε( , y ) y is the residue operator of T − (λ) at the pole µ then · k k

P (L′(µ)Vy,Vy)=(y,Vy) for all y KerT (µ). (4) ∈ In particular, µ is of positive (negative) type if and only if all the sign characteristics are positive (negative).

Proof. For y KerT (µ) we have V y KerT (µ) and ∈ ∈ 1 1 y = T (λ)T − (λ)y =(T (µ)+(λ µ)T ′(µ)+ ... )(V (λ µ)− + R(µ)+ ... )y = − −

= T ′(µ)V y + T (µ)R(µ)y + o(1), where o(1) 0 as λ µ and R(µ) is a bounded operator on . Taking the scalar product with→V y and→ letting λ µ, we obtain (4). H → Theorem 12.1. Let a pencil Tω(λ) of the form (2) be strongly elliptic. Then for all ω > 0 with possible exception of some values ω (the so-called resonant k → ∞ frequences) there is an even number, say 2κ, of real eigenvalues of Tω(λ) counting geometric multiplicities. They all are of deﬁnite type and exactly κ of them are of positive (negative) type.

Proof. Consider the pencil

L (λ)= MT (λ) M = L(λ) ω2R , R = MR M. ω H ω H − 0 0 H H 139

The assumptions on the operators (see Section 1) ensure us that the coefficients of L (λ) are bounded operators on , moreover, R > 0. By virtue of ω H 0 Theorem 1.6 there are finitely many normal eigenvalues of Lω(λ) on the real axis. To prove that they are of definite type with possible exception of isolated values ωk we apply the known results of pertubation operator theory which are based→ on ∞ theorems due to Rellich and Nagy (see Ch 9 of [RN]), Krein and Lyubarskii [KL], Kostyuchenko and Orazov [KO2]. A concentrated exposition of this material can be found in the paper of Shkalikov and Hriniv [SH], Propositions 1.6-1.9. The only reservation: the condition R 0 assumed in [SH] 0 ≫ can be replaced by R0 > 0 provided the coefficients of the pencil Lω(λ) are bounded. The main idea of proving this result is the following. Let µ be a real eigenvalue of the 2 pencil Lω(λ) with fixed ω = ω0, and let θ0 = ω0. We notice that Lω(λ) is a linear 2 selfadjoint pencil with respect to the parameter θ = ω and its eigenvalues θj(λ) according to the Rellich-Nagy theorem depend analytically on λ in a neighborhood of an eigenvalue λ = µ, namely, θ (λ)= θ + a (λ µ)pj + ... j 0 j − with some 0 = aj R and integer pj > 0. Then λj(θ) represent the branches of the inverse algebraic6 functions∈

1 1/pj λ (θ)= µ +(a− (θ θ )) + ..., k =0,...,p 1, j,k j − 0 j − and pj coincide with the lengths of the corresponding Jordan chains. Hence, λj(θ) move locally either in the complex plane or leave on the real axis depending monotonically on θ, moreover, the condition R0 > 0 implies that the real branches λj(θ) are strictly monotone functions. Thus, all the real eigenvalues in a small punctured neighborhood of µ are semi-simple. Further, it turns out (see Proposition 2.3 below) that the sign characteristics of the real eigenvalues λj(θ) coincide with sign λj′ (θ). Taking into account Proposition 2.1, we obtain that all the real eigenvalues of Tω(λ) in a small right (left) neighborhood of µ are of positive (negative) type. Hence, the resonant frequences are isolated points. Let us prove the other statements. Fix a non-resonant frequency ω, and ﬁx a positive H-compact operator V such that Tω(λ)+V is uniformly positive. Consider the pencil T (λ)+ ρ(V + I), 0 ρ 1. (5) ω ≤ ≤ Obviously, the closure of M(V + I) M is positive in . Now apply Proposition 1.9 from [SH] which says:H H H + E (ρ)+ E−(ρ)= const, 140

+ where E (ρ) and E−(ρ) are the number of real eigenvalues of positive and negative type, respectively. This equality holds also for the resonant values of θ if the numbers E±(ρ) are deﬁned as in [SH]. Since for ρ = 1 the pencil (5) is uniformly positive, we have + + E (1) E−(1) = 0, hence E (0) = E−(0). − This ends the proof. Let µ, f be a normal eigen-pair corresponding to a simple or semi-simple { 0} eigenvalue of Tω(λ) with a ﬁxed ω = ω0 > 0. As we mentioned above the eigenvalue 2 2 θ0 = ω0 admits an analytic continuation θj(λ)= ω when λ runs in a neighborhood of µ. The value θ′(λ) λ=µ is called the group velocity (see, for example, [ZK] or [VB]) of the wave solution iµy u(y)= e− f0.

Proposition 12.4. If µ is a deﬁnite type eigenvalue of a pencil Tω(λ) and µ, f0 is a corresponding normal eigen-pair then { }

2 (Tλ′ (µ)f0,f0)= θλ′ (µ)(Rf0,f0), θ = ω , (6) i.e. the sign characteristic of an eigen-pair coincides with the sign of its group velocity.

Proof. Let f(λ)= f0 +(λ µ)f1 + ... be the eigen-element of Tω(λ) corresponding − 2 to λ = λ(θ). Denoting θ0 = ω0 we obtain

[T (µ) θ + T ′(µ)(λ µ) (θ θ )R + ... ][f +(λ µ)f + ... ]=0, − 0 − − − 0 0 − 1 therefore θ θ0 (T ′(µ)f ,f )+ o(1) = − (Rf ,f ). 0 0 λ µ 0 0 − Letting λ µ, we get (6). → Corollary 12.1. For any non-resonant frequency ω equation (1) with strongly elliptic symbol (2) possesses ﬁnitely many, say 2κ 0, propagating waves and exactly κ of them are outgoing (incoming), i.e. have positive≥ (negative) group velocity or the sign characteristics. 141

12.3 Factorization of positive strongly elliptic pencils

In this section we use abstract Sobolev spaces. Namely, by WmRH we denote the + (j) space consisting of m-valued functions u(y) deﬁned on R , such that u (y) exist in the generalized senseH for j m as -valued functions and the integral ≤ H ∞ (m) 2 2 2 u (y) 0 + u(y) m dy =: u Wm 0 k k k k k k Z converges. The detailed information on abstract Sobolev spaces can be found in the book of Lions and Magenes [LM]. We recall here some facts which we need below. According to the theorem on intermediate derivatives we have

(j) + u (y) L2(R , m j), j =1,...,m, if u(y) WmRH. ∈ H − ∈ An important role in the sequel plays the trace theorem which we formulate (as it needed) in the case m =1. Trace Theorem. A function u(y) W1RH is continuous and uniformly bounded on R+ an -valued function and∈ the trace operator H1/2 : W RH , u = u(r), (1) Tr 1 →H1/2 Tr + is bounded for any ﬁxed r R , moreover, r c with a constant c not depending on r R+. ∈ kT k≤ ∈ + + If u(y) W2(R , ) then u′(y) W2(R , 1). As (G) 1, we have ∈ + H ∈ H D ⊃ H Gu′(y) L (R , ). Therefore ∈ 2 H d u = T i u = Fu′′(y)+ iGu′(y)+ Su(y) A dy − is well deﬁned in the space L2RH with domain ( ) = W2RH. Let 0 be the restriction of on the domain D A A A o ( )= W RH := y y W RH, y(0) = 0 . D A0 1 { | ∈ 1 } Lemma 12.1. Let T (λ) be strongly elliptic. Then there exist a number ε> 0 and an H-compact self-adjoint operator V 0, such that ≥ 1/2 1 ε u V u ( u, u) ε− u , u ( ). (2) k kW1 −k kL2 ≤ A0 ≤ k kW1 ∈ D A0 If in addition T (λ) is positive then the left hand side estimate holds with V =0. 142

Proof. Denote ∞ uˆ(λ)= u(y)eiλy dy, λ R. ∈ Z0 It follows from the Plancherel theorem that

fˆ(λ), gˆ(λ) = f(y),g(y) , for f,g L RH, ∈ 2 where the scalar product , is taken in L2RH. As T (λ) is strongly elliptic, there is an H-compact operatorVsuch that T (λ)+ V ε(λ2 + H), λ R. ≥ ∈

We can suppose that V = V ∗, otherwise the Fridrichs extension of V should be o considered. Bearing in mind that for all functions u(y) W RH = ( ) ∈ 1 D A0

∞ iλy iλuˆ(λ)= u′(y) e dy, − Z0 we ﬁnd that

( + V )u, u = Fu′, u′ i Gu, u′ + (S + V )u, u A0 − = (T (λ)+ V )ˆu(λ), uˆ(λ) ε (λ2 + H)ˆu(λ), uˆ(λ) ≥ 2 = ε u′, u′ + Pu, Pu = ε u , u ( 0). H H k kW1 ∈ D A h i This implies the left hand side estimate of (2). The right one is trivial and follows from the inequality

2 Gu, u′ c Pu, Pu u′, u′ c u . | |≤ H H ≤ k kW1 To get the last statement of Lemma, recall Remark 1.8. o o Let W 1 RH be the dual space to W1 RH with respect to L2RH. For any o − v W RH and u W (R+, ) ∈ 1 ∈ 2 H

( u, v)=(Fu′, v′) i(Gu,v′)+(K Pu, Pv), (3) A − H H 143

where K = MS M = K∗ is a bounded operator on . For any fixed u W1RH H H H o ∈ the right hand side of (3) represents a continuous linear functional on W1 RH. Ac- cording to the definition of a dual space, any such a functional admits a representa- o tion (f,v), with f W 1 RH. Hence admits the extension ∈ − A o : W1RH W 1 RH. (4) A → − A function u(y) W RH is called a generalized solution of the equation ∈ 1 d T i u(y)=0 (5) dy if u(y) belongs to the kernel of operator (4). Lemma 12.2. Let T (λ) be a positive strongly elliptic pencil. Then for any x 1/2 there is a unique generalized solution u(y) of equation (5) such that u(0)∈ = Hx. Proof. This statement is familiar from PDO theory; its abstract version is proved in the same way, one should use only the Friedrichs theorem instead of the Lax-Milgram lemma. Namely, taking into account Lemma 3.1 and the Friedrichs theorem (see [RN], Ch8), we obtain that 0 admits the only self-adjoint extension F 0 such o A A ≫ that ( 1/2)= W RH. Hence, D AF 1 o o F : W1 RH W 1 RH (6) A → − is an isomorphism. Since the trace operator defined in (1) is surjec- T0 tive, for any x 1/2 there is a function v1(y) W1RH such that ∈ H o ∈ v1(0) = x. Then v1(y) W 1 RH and taking into account that map- A ∈ − o ping (6) is an isomorphism, we find a function v2(y) W1 RH such that v (y) = v (y). Hence, the function u(y) = v (y) ∈v (y) is a generalized AF 2 A 1 1 − 2 solution of the equation u(y)=0 and u(0) = v1(0) = x. The uniqueness follows from the condition Ker A =0. AF Lemma 12.3. Let u(y) be a solution of equation (5) on the semiaxis R+ in the following sense: u(y),u′(y) CRHP, u′′(y) CRH, ∈ ∈ and equation (5) holds as an equality in . If u(y) W RH, then H ∈ 1 1/2 1/2 S u(y) = F u′(y) , y 0. (7) k k k k ≥ 144

Proof. Consider in 2 = the operator H H×H 1/2 1/2 1/2 1/2 F − GF − F − S T = − − , (8) S1/2F 1/2 0 − acting in 2 = (the linearization of T (λ)). Obviously, T is symmetric (and even selfadjoint)H H×H in the Krein space = 2, J with the fundamental symmetry I 0 K {H } J = . It is easy to see that equation (5) is equivalent to the following 0 I one − F 1/2u (y) Tu(y)= iu (y), u(y)= ′ . ′ iS1/2u(y) − Using this equation we ﬁnd (diﬀerentiation is allowed by our assumptions)

(Ju(y), u(y))′ =(Ju′(y), u(y))+(Ju(y), u′(y)) =

= i(JTu(y), u(y))+ i(u(y), JTu(y))=0. − Therefore, (Ju(y), u(y)) = const. The condition u(y) W1RH, obviously, implies (Ju(y), u(y))=0 and (7) follows. ∈

Theorem 12.1. Let T (λ) be a strongly elliptic positive pencil. Then there exists a closed operator Z in the space with domain (Z) , such that H D ⊂H1 T (λ)x =(F λ Z )(λ Z)x for all x (Z), (9) − 1 − ∈ D where Z1 = (G + F Z) and the equality is understood in 1. Moreover, (a) Z has− a representation Z = KS1/2 where K is a partialH− isometry in whose image Re(K)= ; H H (b) -iZ generates a holomorphic semigroup in the spaces θ, 0 θ 1/2; (c) the generalized solutions of equation (5) satisfy the eqHuation≤ ≤

u′(y)= iZu(y). − Factorization (9) with these properties is unique.

Proof. Let x . By virtue of Lemma 3.2 there is a generalized solution of ∈ H1/2 equation (5) such that ux(0) = x. Deﬁne the operator function U(t) on R as follows

U(t)x = u (t), t 0.r x ≥ 145

Note that according to Lemma 3.2 the restriction of the trace operator 0 to Ker W (R+, ) is a bounded isomorphism onto . Hence the inverse operatorT A∈ 1 H H1/2 1 1 − : Ker , − x = u (t) T0 H1/2 → A T0 x 1 is bounded, as well as the operator U(t) = t 0− acting in 1/2 (for any t 0). It follows from the definition of the operatorTUT(t) and from theH trace theorem≥ (see the formulation at the beginning of this section) that U(t + s)= U(t)U(s),U(0) = I, U(t) const, k k≤ s-lim U(t)= U(s), 0 s t, t s → ≤ ≤ where the strong limit is understood in 1/2. This means that U(t) is a uniformly H iZt bounded C0-semigroup in the space 1/2 (see, e.g., [Yo]). If U(t) = e− where iZ is the generator of U(t), then propertyH (c) of Theorem 3.4 is satisfied, and by Lemma− 3.2 it defines Z uniquely. It is known from semigroup theory that Z and Z2 (as well as the other powers) are closed operators in whose domains (Z) and (Z2) are densely defined in H1/2 D D 1/2. H Let x (Z2) and u (t) is the corresponding generalized solution of ∈ D ⊂ H1/2 x (5). In view of the semigroup properties the functions ux′ (t),ux′′(t) are continuous in + 1/2 on R and H 2 u′ (t)= iZu (t) u′′(t)= Z u (t). x x x − x The operator G : 1 is bounded, therefore, Gux′ (t) is continuous in 1. H→H− H− Since ux(t) is a generalized solution, we have the equality

Fu′′(t)+ iGu′ (t)= Su (t) (10) − x x − x o which is understood as an equality in W 1 RH. The left hand side is a continuous − function in 1, hence, so is the function Sux(t). Equivalently, ux(t) is continuous in . In particular,H− x = u (0) and (10) gives H1 x ∈H1 (F Z2 + GZ + S)x =0, for x (Z2), (11) ∈ D where the equality is understood in 1. Our further aim is to extend (11)H to− a larger domain. Notice, if x (Z2) then the conditions of Lemma 3.3 are fulﬁlled and we have ∈ D

1/2 1/2 F u′ (t) = S u (t) , t 0. k x k k x k ≥ 146

In particular, we have the equality F 1/2Zx = S1/2x which gives (for x (Z2)) the representation Z = KSk 1/2, wherek K isk a partialk isometry in . ∈ D 2 H Since (Z ) is dense in 1/2 and 1/2 is dense in , we have Re(K)= . Hence, D H H 1 1/2 H H Z is boundedly invertible in and Z− = S− K∗. This enables us to extend Z H 1/2 from 1/2 onto with domain D (Z) = Re (S− K∗). Further (and in (9)) we omit theH index Hand imply that ZH acts in and its domain (Z) is understood as described. Certainly,H (Z) and itH coincides with Dif and only if K is D ⊂ H1 H1 a unitary operator. Now, both terms Gx and Sx are in 1 for x (Z), so equality (11) can be extended to all x (Z). This is equivalentH− to∈ the D factor- ∈ D 1/2 ization (9), moreover, for Z1 we have the representation Z1 = S K∗ as well as Z = (G + F Z). Then we obtain 1 − 1 1 1/2 (λ Z)− = T − (λ)(F λ S K∗) − − where the both sides are understood as operators in . Applying Theorem 1.7 to the right hand side of the last identity we obtain theH right hand side of the last identity we obtain 1 c x (λ Z)− x k k (12) k − k≤ 1+ λ | | in a double sector Λϕ containing the real axis. Let us prove that (12) holds also for + all λ from the upper half plane . Since iZ is a generator of a C0-semigroup in we have (see [Yo, Ch. 9]) C − H1/2 1 1 cx (λ Z)− x (λ Z)− x for all x and λ − Λ , k − k≤k − k1/2 ≤ 1+ λ ∈H1/2 ∈C \ ϕ | | with a constant cx depending on x, and the estimate holds in the whole upper + half plane outside an arbitrary small double sector Λϕ containing the real axis. Applying theC Phragmen-Lindelöf theorem (see [Bo], for example) we obtain estimate (12) for all λ − and x 1/2 with the same constant c as it was in (12). By continuity (12)∈ can C be extended∈ H for all x R. This implies that iZ generates a holomorphic semigroup in . ∈ − H Actually, iZ generates a holomorphic semigroup in the space 1/2, too. To prove this, we− consider the pencil H d T (λ)= T (eiϕλ) and ϕu = T i . ϕ A ϕ dy For sufficiently small ϕ we can reprove Lemma 3.1 changing ( 0u, u) in (2) by Re( ϕu, u). This is possible,| | since the Friedrichs extension existsA for the sectorial A0 147 operators (see [Ka], Ch. 6). Repeating the arguments we find that there exists an operator iZ which generates a C -semigroup in , Z possesses property − ϕ 0 H1/2 ϕ (c) and realizes a factorization of the form (9) for the pencil Tϕ(λ). From this iϕ we obtain Zϕ = e Z. Then the minimal resolvent growth estimate of the form (12) holds for Z in a small double sector containing the real axis. Hence, the C0- semigroup generated by iZ is, actually, a holomorphic semigroup. Now, applying the interpolation theorem− we get assertion (b). This ends the proof.

12.4 Elliptic pencils satisfying the Keldysh-Agmon condition 12.5 The resolvent growth condition In this section we will use the condition which in general form can be formulated as follows. 1 The resolvent growth condition. Assuming that T − (λ) Px(λ) is holo- + H morphic in the upper (lower) half plane ( −) where C C x(λ)= x + λx + + λnx 0 1 · · · n is an -valued polynomial, we have H 1 m + PT − (λ) Px(λ) C λ for all λ ( −), λ >r , (1) kH H k≤ | | ∈C C | | 0 with some constants c and m. This condition is by no means obvious to verify and we formulate the other one which can be checked out more easily. Keldysh-Agmon condition. T (λ) is of the form (1) and 1 (a) the operator H has discrete spectrum (i.e. H− is compact and its eigenvalues are subject to the estimates λ (H) cjp, j =1, 2,..., (2) j ≥ with some constants c and p; (b) either p 2 or p< 2 but there are rays γj = λ arg λ = θj , j =1,...,N, ≥ + { } in the upper (lower) half plane ( −) such that C C

0 <θ <θ < 2π/p, j =1,...,N 1; max(θ ,θ θ , π θ ) < 2π/p, j j+1 − 1 j+1 − j − N and 1 m PT − (λ) P c(1 + λ ), for λ γ , kH H k≤ | | ∈ j with some constants c and m. 148

Proposition 12.5. If T (λ) is a strongly elliptic pencil then the Keldysh-Agmon condition implies the resolvent growth condition, moreover, one can take in (1) m = n.

Proof. First, notice that (2) implies that the generalized and the classical spectra of T (λ) coincide (Theorem 1.6). The essense of the matter is that condition (a) 1 together with (G) ( P ) imply that PT − (λ) P is an -valued meromorphic operatorD function⊃ D of orderH 2/p. The proofH is basedH on the resultsH of Keldysh [Ke], Agmon [Ag], Matsaev [Mat] et. al. (see historical remarks and details in [S3], §2). Now, if 1 F (λ)= PT − (λ) Px(λ) H H is holomorphic in + and x(λ) is a polynomial then condition (b) and the Phragmen- Lindel¨of theoremC imply that F (λ) has a polynonial growth in +. According to Theorem 1.7 C

F (λ) < c(1 + λ n), λ R, λ >r , n = deg x(λ). (3) | | | | ∈ | | 0 Since F (λ) is of order zero in +, by virtue of the Phragmen-Lindelöf theorem the estimate (3) holds asymptoticallyC for all λ +. ∈C 12.6 Half-range completeness and minimality In what follows we consider for simplicity a generic situation when T (λ) has only semi-simple real eigenvalues of definite type. For a pencil of the form (2) this is true according to Theorem 2.2 for all values of ω with the possible exception of isolated resonant frequences ωk . Let T (λ) have discrete→∞ spectrum and let the eigenvalues of T (λ) be numerated according to their geometric multiplicity (i.e. every eigenvalue λk is repeated n = nul T (λk) times). In this case we have a one-to-one correspondence between the eigenvalues λk and canonical Jordan chains of the form (2). As we agreed, all the real eigenvalues are supposed to be semi-simple. The eigen-elements corresponding to every real eigenvalue are assumed to form a normal canonical system (see Section 2). Take all the chosen Jordan chains of T (λ) corresponding to the eigenvalues from the open upper (lower) half-plane and all the eigen-elements corresponding to the real eigenvalues of positive (negative) type. Denote the system consisting of all these + elements by E (E−) and call it the first (second) half of the root elements of T (λ). Let us recall the well-known definitions. A system e ∞ is said to be minimal { k}1 in Hilbert space if there exists an adjoint system e∗ ∞ such that (e , e∗)= δ , H { k}1 k j kj 149 where δ is the Kronecker symbol. Equivalently, e ∞ is minimal if any element jk { k}1 ek is not contained in the closed linear span of the other ones. A system ek 1∞ is said to be complete in if there is no non-zero element in which is orthogonal{ } to all the elements of theH system. H Theorem 12.1. The first and the second half of the root elements of a pencil T (λ) form minimal systems in provided T (λ) is strongly elliptic and has discrete spectrum. H Proof. Let us work with the system E+, for example. By virtue of Propostition 1.4 there is a number r > 0 such that the pencil T (λ)= T (λ r ) has only positive 0 1 − 0 eigenvalues on the real axis (to prove the minimality of E− one should consider the 0 p pencil T1(λ) = T (λ + r0)). The Jordan chains xk,...,xk of the pencil T (λ) are changed after this transformation in the following way

0 0 1 1 1 0 p p 1 p 1 p 0 ξ = x , ξ = x r− x , ... ,ξ = x r− x − r− x , k k k k − 0 k k k − 0 k −···− 0 k 0 0 while the sign characteristics of the pairs λk, yk and λk + r0, yk are the same. Hence, it suﬃces to prove the minimality for{ the} case when{ T (λ) has} only positive real eigenvalues and S > 0. Let us consider the system

F 1/2(λ xs + xs 1) xs = k k k− , xs E+. (4) k S1/2xs k ∈ k s It is an easy exercise to show that the xk are the root elements of the operator T deﬁned by (8) (T is the linearization of T (λ)). As we mentioned T is a symmetric operator in the Krein space = 2, J with the fundamental symmetry J = I 0 K {H } . From this we have the biorthogonality relationships (see, e.g., [AI], 0 I Ch.1) − s h (Jxk, xj )=0 for all j, k, s, h except for the case k = j and λ R. For λ R we have k ∈ k ∈ (Jx , x )= λ2(Fx ,x ) (Sx ,x ) k k k k k − k k 2 =2λk(Fxk,xk)+ λk(Gxk,xk)= λk(T ′(λk)xk,xk)= λkεk, where ε is the sign characteristic of the pair λ , y . Hence, k { k k} (Jxs , xh)= δ λ ε , xs,xh E+, (5) k j kj k k k k ∈ 150

+ where εk =0 for the nonreal λk and λkεk > 0 for λk R and xk E . Let x be a ﬁnite linear combination of elements (4) ∈ ∈

1/2 s s 1 v1 s F (λkxk + xk− ) s s v := = ck 1/2 s =: ckxk. (6) v2 S xk X X From (5) we obtain v1 v2 , therefore v 2 v1 . Recall that the system s k k≥k k k k ≤ k k xk is minimal as the system of the root elements of the operator T with discrete spectrum. Then the inequality v1 v 2 v1 implies (if we use the second deﬁnition of minimality) that k k≤k k ≤ k k

1/2 s s 1 s + F (λ x x − ) , x E , { k k − k } k ∈ is a minimal system in . Hence E+ is minimal in , too. H H Theorem 12.2. The ﬁrst and the second half of the root elements of a pencil T (λ) form complete systems in 1 provided T (λ) is strongly elliptic and the Keldysh- Agmon condition holds. H

Proof. As before, we deal with the system E+. Suppose that there is an element f such that ∈H1 (f,xs) =( Pf, Pxs )=0 for all xs E+. (7) k 1 H H k k ∈

Choose a number r0 such that T (λ) > 0 for λ>r0 and consider the function

1 1 F (λ)= ( PT − (λ) Pg,g), g = Pf . (8) λ r H H H ∈H − 0 The principal part of F (λ) in a neighborhood of a real pole λk has the representation ε (g, Px )( Px ,g) k H k H k (λ r0)(λ λk) X − − where ε are the sign characteristics corresponding to the eigen-pair λ ,x . Due k { k k} to (7) all the terms with εk > 0 in the last expression are equal to zero. Since λ r < 0, all the residues of F (λ) at the poles λ R are non-negative. The k − 0 k ∈ residue at the additional pole λ = r0 is non-negative, too. Taking into account the 1 + representation (1) of T − (λ) in a neighbourhood of a non-real pole λk and assumption (7), we ﬁnd that F (λ) is holomorphic in +. By the Schwarz∈ symmetry C C 151

1 principle it is holomorphic in −. Proposition 4.1 gives us F (λ) = O(λ− ) when λ uniformly in . C →∞Let us show that theC residue of F (λ) at equals zero. Given ε> 0 we can ﬁnd ∞ g0 1 such that g g0 1 <ε. If we put g0 in (8) instead of g then by virtue of ∈H k − k 3 Theorem 1.7 the corresponding function vanishes at as O(λ− ) when λ 1 ∞ → ±∞ uniformly in . Therefore, F (λ)= o(λ− ) as λ uniformly in , i.e. the residue at is equalC to zero. Now, recall that all the→∞ residues of F (λ) atC the ﬁnite poles are∞ non-negative. This is possible only if all they are equal to zero, in particular,

( PT (r ) Pg, g)=0. H 0 H This implies g =0. Corollary 12.2. The ﬁrst and the second half of the root elements of a strongly elliptic pencil T (λ) satisfying the Keldysh-Agmon condition form complete and minimal systems in spaces for all 0 θ 1. Hθ ≤ ≤ Proof. It follows from the deﬁnitions: if a system is minimal (complete) in ( ) H H1 then it has the same property in θ for θ > 0 (θ < 1). Now apply Theorems 4.2 and 4.3. H

12.7 Factorization The obtained results enable us to construct a divisor of an elliptic pencil. Theorem 12.3. Let T (λ) be a strongly elliptic pencil satisfying the Keldysh-Agmon condition. Then T (λ)x =(λ Z )F (λ Z)x (9) − 1 − where (a) Z and Z1 admit a representation

r Z = K P, r Z = PK − 0H − 1 H 1 with bounded and boundedly invertible in operators K0 and K1, provided r R is not an eigenvalue of T (λ). In particular,HZ is a closed operator on with domain∈ H (Z)= 1 ; D (b) theH spectra of T (λ) and λ Z coincide in the upper half-plane, while on the real axis λ Z inherits only the− positive type eigen-pairs of T (λ), i.e. the system of the root functions− of Z coincides with the ﬁrst half of the root functions of T (λ); 152

(c) iZ generates a holomorphic semigroup in all spaces θ, 0 θ 1. Equality (9) holds for all x and is understood in senseH of≤ operators≤ acting ∈H1 from 1 to 1. Factorization (9) with property (b) is unique. H H− Proof. As in Theorem 4.2 we may assume that T (λ) has only positive eigenvalues, otherwise we have to work with T (λ r0), r0 1. Let us consider the set of all ﬁnite− linear≫ combinations of elements (4). The + elements of this set have representation (6). If the system E is complete in 1 then the system P (E+) is complete in . Therefore, Theorem 4.3 implies thatH H H the linear span of the elements v2 in (6) form a dense subset in as well as the elements v . Deﬁne the operator{ }K by H { 1}

Kv1 = v2. (10)

It was shown in Theorem 4.2 that v2 v1 . Hence, K is densely deﬁned on and can be extended as a contractionk k≤k on the wholek . The image of K is denseH in . H H Denote by E0 the subsystem of E+ consisting of all elements xs E+ corre- k ∈ sponding to the non-real eigenvalues. Let 0 be the closure in of the linear span 0 H H generated by E . Denote κ = codim 0 (κ coincides with the number of positive type eigenvalues counting with geometricH multiplicity). It is clear from (5) that Kv1 = v1 for v1 0, hence, K( 0) is a closed subspace in . By virtue kof Corollaryk k 4.4k the system∈ H P (E+) isH minimal and complete in .H This implies that codim K( ) = κ. Hence,H there is a unitary operator U in H such that the H0 H restriction of U onto 0 coincides with K, i.e. U K is of ﬁnite rank. We noticed already that the imageH of K is dense in . Now,− it follows from the Fredholm theorem that K is boundedly invertible onH . 1/2 1 1/2 H Denote Z = F − K− S , where S = T (0) > 0. From (6) and (10) we have

s s s 1 s + Zx = λ x + x − , x E . (11) k k k k k ∈ s Since xk are the root elements of T (λ), we have (F Z2 + GZ + S)xs =0 for all xs E+. k k ∈ The linear span of E+ is dense in , hence, H1 (F Z + G)Z = S, (12) − 153

where the equality is understood in the sense of operators acting from 1 to 1. 1 H H− Denoting Z = (F Z + G)F − we obtain from (12) the factorization 1 − T (λ)=(λ Z )F (λ Z). − 1 − 1/2 As (S)= (H) we have S = K2 P with a bounded and boundedly invertible D D H 1/2 1 operator K . Hence, Z = K P with K = F − K− K . We have also 2 0H 0 2 1/2 1 1 1 1 1/2 Z = SH− K− F − = PK∗K K− F − =: H K . 1 0 H 2 2 0 1 Thus (a) is proved. The assertion (b) follows from (11). The uniqueness of a factorization with property (b) follows from the completeness of the system E+. It remains to prove (c). To this end we obtain from (9)

1 1 (λ Z)− = T − (λ)(λ PK )F. (13) − −H 1 Applying Theorem 1.7 we obtain

1 1 (λ Z)− C λ − , λ Λ , λ >r . (14) k − k≤ | | ∈ ϕ | | 0 1 Moreover, (λ Z)− is holomorphic in −. By virtue of (13) and Proposition 1− C 4.1 (λ Z)− has a polynomial growth in C−. Consequently, (14) holds for all − λ Λ −, λ > r . Thus, iZ generates a holomorphic semigroup in . Since ∈ ϕ ∪C | | 0 H Z : 1 is an isomorphism, iZ possesses the same property in 1. Applying the interpolationH → H theorem we obtain assertion (c). This ends the proof.H

12.8 The Mandelstam hypothethis In this section we solve the problem

d T i u(y)=0 (1) dy u(0) = f (2) u(y)= u (y)+ u (y), u (y) 0 as y , (3) + 0 0 → →∞ where u+(y) is a linear combination of outgoing waves (3). Below we clarify the understanding of this problem and prove the solvability in the classical sense and the uniqueness in the generalized sense. We may say that 154

(1)-(3) is the half-range Cauchy problem because instead of two initial conditions at y =0 we set only one, but force a solution to behave at in a special way. Further we denote by C (a, b; ) the space of continuous∞ on (a, b) -valued 2 H H2 functions whose derivatives v′(y) and v′′(y) exist in and -norm and belong H1− H to C(a, b; 1) and C(a, b; ), respectively (the continuity at the ends of (a, b) is not assumed!)H H

Theorem 12.1. Let T (λ) be strongly elliptic and assume that the Keldysh-Agmon condition holds. Then for any θ [0, 1] and any f there exists a function ∈ ∈ Hθ u(y) C2(0, ; ) satisfying equation (1), having representation (3) with exponentially∈ decaying∞ uH(y) and satisfying initial condition (2) in the following sense k 0 k2

lim u(y) f θ =0. (4) y +0 → k − k Proof. We ﬁnd a solution of the problem in question by means of the operator Z which was constructed in Theorem 4.5. Namely, denote

1 iλy 1 u(y)= + e (λ Z)− f dλ, (5) 2πi   − Zγ ZΓ   where γ surrounds only real eigenvalues of Z, while Γ lies in the upper half-plane and is asymptotically directed along the rays arg λ = δ and arg λ = π δ with sufficiently small δ > 0. By virtue of Theorem 4.5 iZ generates a holomorphic− semigroup in θ, hence integral (5) is well defined and (4) holds (see [Yo, Ch. 9]). Moreover, theH functions Zku(j)(y) are well defined for y > 0,k,j 0 and are ≥ continuous in θ . Since Z : 1 is an isomorphism, we obtain that u(j)(y) are continuousH ⊂ H for y > 0 in H. The→ equality H H1 (G + F Z)Zx = Sx − holds for all x , in particular, for x . As u′(y) = iZu(y) we obtain ∈ H1 ∈ H2 that iGu′(y) Fu′′(y) C(0, ; ), equation (1) is satisfied in and u(y) − ∈ ∞ H H ∈ C2(0, ; ). Representation (3) with an exponentially decaying function u0(y) follows∞ fromH (5).

Theorem 12.2. A generalized solution u(y) of problem (1)–(3), such that u(y) L (0,ε; ) with some ε> 0, is unique. ∈ 1 H1 155

Proof. In Section 3 we assumed that generalized solutions u(y) belong to 1 W2 (0, ; ). Here our assumptions are weaker: we assume only u(y) 1 ∞ H ∈ W2 (ε, ; ) and u(y) L1(0,ε; 1) for any ε > 0. Certainly, if u(y) 1 ∞ H ∈ H ∈ W2 (0, ; ) then u(y) L2(0,ε; 1) and u(y) L1(0,ε; 1). By the deﬁnition of a generalized∞ H solution, the∈ equationH ∈ H

Fu′′(y)= iFu′(y)+ Su(y) − − o is satisﬁed in the sense of W 1(ε, ; ). The right hand side belongs to − ∞ H L2(ε, ; 1), hence, so does the left hand side. Suppose∞ H− that u(y) 0 as y 0. For λ + we have k k→ → ∈C ∞ d iλy 0= Tω idy u(y)e dy = ε (6) R iλε = e (Fu′(ε) i(λF G)u(ε))+ T (λ)ˆu (λ)=0, − − ω ε where ∞ iλy uˆε(λ)= u(y)e dy. Zε

We consider (6) as an equality in 1. Since u(y) is locally integrable at zero as a function with values in we canH take− the limit as ε 0 and obtain H1 → Tω(λ)ˆu0(λ)= Fu′(0) = g 1. − ∈H− 1 Therefore, uˆ0(λ)= Tω− (λ)g. Let us consider the function

1 1 F (λ)= Tω− (λ)g,g . λ + r0 It follows from (10) that F (λ) is bounded in + and has ﬁnitely many poles on R C with positive residues provided r0 is suﬃciently large. Repeating the arguments of Theorem 4.9 we obtain F (λ) 0. Hence, uˆ(λ) 0 and u(y) 0. ≡ ≡ ≡ 12.9 Application to the Lame system of the elasticity theory Small oscillations of an elastic medium are described by the system of equations (see the books of Landau and Lifshitz [LL] or Kupradze et.al. [Ku]) ∂2w ρ + Lw =0, ∂t2 156

where w = w(t, x)=(w1,w2,w3) is the displacement vector, ρ = ρ(x) is the density of the medium, L is the operator matrix with the entries

ˆ 2 2 2 ∂ Lkj(D)=(λ +ˆµ)DkDj + δkjµˆ(D1 + D2 + D3), Dk = i , ∂xk and λ,ˆ µˆ are the Lame constants. We suppose that the space variable x = (x1,x2,x3) belongs to the wave-guide domain Q = [0, ) Ω where Ω is ∞ × iωt a bounded domain in the plane (x2,x3). Separating the time variable w = ue we obtain the stationary equation with given frequency ω

(L ω2ρ)u =0. (1) − We have to impose with this equation boundary and initial conditions. We pose on the lateral surface of the half-cylinder Q homogeneous conditions, since Q is a wave-guide domain. For simplicity let us consider the Dirichlet boundary conditions

u(x1,x2,x3) (x ,x ) ∂Ω =0 x1 0. (2) | 2 3 ∈ ∀ ≥ At the base of Q we assume that

u(0,x2,x3)= ϕ(x), (3) where ϕ(x) is a given function. We rewrite equation (1) in the form

d d2u du T i u = F + iG +(H ω2R)u =0, (4) ω dy − dy2 dy − where y = x1,

λˆ +2ˆµ 0 0 0 D2 D3 F = 0µ ˆ 0 , G = i(λˆ +ˆµ) mathcalD 0 0 ,    2  0 0ˆµ mathcalD3 0 0     µ elta 0 0 D ˆ 2 ˆ H = 0µ ˆ elta +(λ +ˆµ)D2 (λ +ˆµ)D2D3 ,  D ˆ ˆ 2  0 (λ +ˆµ)D2D3 µˆ elta +(λ +ˆµ)D3  D  R = ρ(x)I, elta = (D2 + D2), D − 2 3 157 and I is the identity matrix. We suppose that the operators F ,G,H act in the 3 Hilbert space =[L2(Ω)] . We have toH specify a domain of the main operator H. Taking into account boundary conditions (2) we deﬁne

3 (H)= v v W 2(Ω) ,v =0 , D | ∈ 2 |∂Ω n o o 3 3 (G)= v v W 1(Ω) ,v =0 =: W1(Ω) , D | ∈ 2 |∂Ω 2 n o k 3 R2 where W2 (Ω) are the Sobolev spaces of vector functions on Ω . We notice that the operator H is positive, since ⊂ 3 (Hv,v)= Hv(x)v(x)dx =µ ˆ D v 2 + D v 2 + k 2 jk k 3 jk j=1 ! ZΩ X +(λˆ +ˆµ) D v + D v 2, k 2 2 3 3k where f 2 = f 2dx. Taking into account boundary condition (2) and the k k Ω | | Friedrichs inequalityR we obtain (Hv,v) ε v 2 with some ε> 0. The operator G is symmetric, as ≥ k k

(Gv,v)=2(λˆ +ˆµ)Re(iD2v2 + iD3v3,v1). The operator F , obviously, is uniformly positive and bounded provided ρ(x) ε> 0 is a measurable bounded function on Ω. ≥ 3 It is well-known (see [Ag] or [Tr], Ch. 5) that H + cI is invertible in [L2(Ω)] provided c 0 and Ω is a smooth domain. Therefore, H is a self-adjoint operator if Ω is smooth.≥ This is not always true, if Ω, for instance, has corner points (see examples in the paper of Kondratiev and Shkalikov [KoS]). In this case let us consider the Friedrichs extention HF of the operator H. It is known (see [RN], Ch. 8) that it is the only extension which possesses the property (H1/2)= (G). D F D Denoting HF = H we remark that all the assumptions on the operator coeﬃ- cients claimed at the beginning of Section 1 are fulﬁlled. We note that in the case of a non-smooth domain Ω there is no precise information on (H ), however, we do know that D F o 3 (H1/2)= (G)= W1(Ω) . D F D 3 158

Actually, the domain of HF (= H) is not involved in our considerations, the know- 1/2 ledge of (H )= 1 is the only important information which we need. NowD let us proveH that the pencil corresponding to equation (4) is regular elliptic in the case of a smooth domain and strongly elliptic otherwise.

Proposition 12.6. The pencil Tω(λ) generated by the Lame system and the Dirich- let boundary conditions is strongly elliptic.

Proof. We have

(T (λ)v,v)= λ2 (λˆ +2ˆµ) v 2 +ˆµ( v 2 + v 2) + ω k 1k k 2k k 3k h i +2λ(λˆ +ˆµ)Re [(iD v + iD v ,v )]+(λˆ +ˆµ) D v + D v 2+ 2 2 3 3 1 k 2 2 3 3k 3 2 2 2 1/2 2 +ˆµ D2vj + D3vj ω ρ v (5) j=1 k k k k − k k ≥ P 3 2 2 2 2 2 1/2 2 µλˆ vj + D2vj + D3vj ω ρ v ≥ j=1 k k k k k k − k k ≥ P ε[λ2(v,v)+(Hv,v)] ω2(ρv,v), λ R, v 2 = v 2 + v 2. ≥ − ∈ k k k 1k k 2k o 1 3 3 It is known (see [Tr], Ch 4.10) that the embedding I :[W2(Ω)] [L2(Ω)] is compact for any bounded domain Ω (we pay attention that if we→ consider, say, Neuman boundary conditions, then we have to assume in addition that Ω is a Lipshitzian domain). By virtue of Proposition 1.4 we obtain that Tω(λ) is strongly elliptic. We remark that Proposition 6.1 can be also proved in the case of an unbounded domain Ω if we assume ρ(x) 0 as x . → | |→∞ Proposition 12.7. The pencil Tω(λ) is regular elliptic if a domain Ω is smooth. Moreover, estimate (3) holds asymptotically outside any double sector containing the imaginary axis and the Keldysh-Agmon condition holds.

Proof. (Cf.[KO2]). Denoting iDk = ξk, let us calculate the principal characteristic symbol of the Lame system (the− principal symbol does not depend on ω and we can 159 assume ω =0). We have

λˆ +2ˆµ 0 0 0 ξ2 ξ3 det T (λ)= det λ2 0µ ˆ 0 + λ(λˆ +ˆµ) ξ 0 0 + 0     2  0 0ˆµ ξ3 0 0      µˆ ξ 2 0 0 | | + 0µ ˆ ξ 2 +(λˆ +ˆµ)ξ2 (λˆ +ˆµ)ξ ξ =µ ˆ2(λˆ +2ˆµ)(λ2 + ξ 2)3,  | | 2 2 3  | | 0 (λˆ +ˆµ)ξ ξ µˆ ξ 2 +(λˆ +ˆµ)ξ2 2 3 | | 3  2 2 2  where ξ = ξ2 + ξ3. Hence,| | the ellipticity condition in the sense of [AN] and [AV] holds for all λ not belonging to the imaginary axis. It is well known (see, e.g., [LM]) that the Dirichlet boundary condition satisﬁes the Lopatinskii condition for all elliptic systems. Hence, the problem (1), (2) is regular elliptic and according to the results of [AN] and [AV] estimate (3) holds outside arbitrary small sector containing the imaginary axis. Since Tω(λ) is a seladjoint pencil, estimate (3) implies

1 1 HT − (λ) + T − (λ)H const, || ω || || ω || ≤ and, by virtue of the interpolation theorem, we have

1/2 1 1/2 H T − (λ)H const, λ >r , (6) || ω || ≤ | | 0 at any ray in with exception of the imaginary exis. According to the Weyl asymptotic formulaC for eigenvalues of the elliptic operators, we have the estimate (2) with p =1. Hence, if Ω is a smooth domain then the Keldysh-Agmon condition for the Lame system is valid. In the case of a non-smooth domain we are able to prove the validity of the Keldysh-Agmon condition only under additional constraints on the Lame constants.

Proposition 12.8. Let Ω be a bounded domain in R2. If µ>ˆ √2λˆ then estimate(9) is satisﬁed in a double sector Λϕ with some ϕ > π/4 and the Keldysh-Agmon condition holds.

iπ/4 Proof. Let us estimate the quadratic form (Tω(λ)v,v) at the ray λ = e ζ, ζ> 0. Suppose ω =0. Bearing in mind (5) we obtain

iπ/4 iπ/4 Re e− T (e ζ)v,v 0 ≥ 160

3 √2 µˆ ζ2 v 2 + D v 2 + D v 2 + 2 k jk k 2 jk k 3 jk j=1 ! X √2 + (λˆ +ˆµ) ζ2 v 2 2√2 ζ D v + D v v + D v + D v 2 . 2 k 1k − | | k 2 2 3 3k k 1k k 2 2 3 3k Taking into account the inequality

2(ac + bc) √2(a2 + b2 + c2), a,b,c> 0 ≤ we can estimate the second summand as follows √2 (λˆ +ˆµ)(2 √2) ζ2 v 2 + D v 2 + D v 2 . ≥− − 2 k 1k k 2 2k k 3 3k Therefore,

Re eiπ/4 T (eiπ/4ζ)v,v ε v + ζ2 v ω2(ρv,v). (7) ω ≥ k k1 k k − with some ε> 0 provided µ>ˆ √2λˆ. Obviously, a similar estimate (if π/4 is replaced iθ by θ) holds at any ray λ = re in a double sector Λϕ provided 0 <ϕ π/4 is small enough. − According to Theorem 1.7 estimate (7) gives the estimate of the resolvent (6). Since (2) holds in our case with p =1, we see that the Keldysh –Agmon condition is satisﬁed. For simplicity we formulate the main result of this section not in the whole generality.

Theorem 12.1. Let Ω be a bounded domain in R2 and assume that either Ω is smooth or Lame constants satisfy the condition µˆ > √2λˆ. Then for any function o 3 ϕ(x) W1(Ω) there is a unique classical solution u(y) in the half-cylinder Q = ∈ 2 R+ Ω of the stationary Lame system (4), (2) with given non-resonant frequency ω, × such that this solution satisﬁes the Mandelstam radiation principle as x1 = y and the initial condition is understood in the following sense →∞

lim u(y,x2,x3) ϕ(x2,x3) 1 =0. y 0 → k − k Proof. It follows from results of Section 5. 161

Our conjecture (which we can not prove at the moment) is that the condition µˆ > 2λˆ in Theorem 6.4 is superﬂuous. This condition is used only in the proof of the existence. Apparently it is essential for the validity of the Keldysh–Agmon condition and, hence, for the half range completeness. However, it has not to be essential for the existence of a solution. The reason is that for suﬃciently small frequencies ω the pencil Tω(λ) is positive and Theorem 3.4 can be applied to prove the existence.

Bibliography for Section 12

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[Ag] S. Agmon, Lectures on elliptic boundary value problems, New York, 1965.

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[AN] S. Agmon and L. Nirenberg, Properties of solutions of ordinary diﬀer- ential equations in Banach space. Comm. Pure Appl. Math. Vol 16 (1963), 121-239.

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[ZK] A. S. Zilbergleit and Yu. I. Kopilevich, Spectral theory of regular wave-guides. Ioffe Inst. of Physics, Leningrad, 1983. 165 13 Scattering of waves by periodic gratings and factorization problems 13.1 Introduction In this paper we consider two scattering problems for the Helmholtz equation. The study of the first one was originated by Lord Rayleigh [Rl, R2]. The problem is to give an analysis of the scattering of a monochromatic plane wave incident on a grating with a periodic curve in R2 (Rayleigh considered a sinusoidal grating profile). The second problem is a three-dimensional analogue of the first one: to give an analysis of the scattering of a space wave by a periodic surface in R3. The scattering of acoustic and electromagnetic waves by periodic gratings plays a significant role in physics and engineering, which caused a vast literature devoted to these problems. The works are mostly connected with the scattering in R2, because even in this case the problem is quite non-trivial and involves an intricate and lengthy analysis. A mathematical core of the problem is to prove existence and uniqueness of the solution which describes the scattering. In the plane case and in the case of non- resonant frequencies the proof was given by Badyukov [Bl, B2], who reduced the problem to an integral equation of the Fredholm type using the Hankel function expansion for the kernel of the Helmholtz equation. Wilcox and Guilliot [WG] independently obtained similar results using Rayleigh-Bloch wave expansions (which, essentially, coincide with those in [Bl, B2]). Alber [1] and Wilcox [?] developed an alternative method for solving the scattering problem based on analytic continuations. Further developments, numerical studies, historical remarks and references can be found in the book edited by Petit [25], in the monographs of Gunter [6], Wilcox [32], [33], Galashnikova and Il’inskii [7], Nazarov and Plamenevskii [24], in the papers of Babich [2], Il’inskii and Mikheev [14], Beljaev, Mikheev and Shamaev [5]. For the case of resonant frequencies, we have not found in the literature rigorous results on the solvability of the scattering problems. In this case an additional problem arises: how to select outgoing waves and to pose the radiation conditions? It turns out that the formulation of the radiation condition for the plane scattering problem in the case of resonant frequencies remains the same as in the non-resonant case. However, this circumstance is rather incidental from the mathematical point of view. It is explained by the fact that the Jordan chains of the spectral problem corresponding to the Helmholtz equation have the simplest structure: their lengths 166 equal 2. This is not true if the scattering problem is considered for the system of elasticity (see [18]). The aim of this paper is to propose a new approach to treat scattering problems. This approach is quite general and allows to consider scattering problems which were not treated before. It is based on the possibility to reformulate these problems in terms of abstract ordinary differential equations with operator coefficients on a Hilbert space. It turns out that the solvability of a scattering problem is equivalent to the solvability of the appropriate differential equation on the semiaxis with the radiation conditions at + . To solve the last problem we apply the factorization theorems for the operator∞ symbol of the corresponding equation. This paper can be considered as a continuation of the paper [Sh2], but it can be read independently. An outstanding role in development of the factorization theory is played by the works of I. Gohberg and his co-authors. A particular mention deserves his pioneering work with M. Krein [11]. Further developments and references can be found in the monograph of Gohberg and Feldman [8], Gohberg and Krupnik [10], Gohberg, Lancaster and Rodman [13], Markus [?], Gohberg, Goldberg and Kaashoek [9]. In our case the factorization problem has to be solved for selfadjoint operator pencils with unbounded operator coefficients. This leads to new difficulties. In particular, even if a linear right divisor λ Z of a pencil is found, one has to investigate − the properties of Z: does this operator generate a holomorphic C0-semigroup? The study of factorization of operator pencils with unbounded coefficients was originated in the authors’ paper [19]. Here we give a short review of results on the factorization of elliptic pencils which are essentially used in the sequel. In contrast with the previous works, an abstract approach of this paper does not use specific properties of the Helmholtz equation and can be applied to scattering problems in electrodynamics described by the Maxwell equation (see [7]) or the system of elasticity. The Helmholtz equation itself can be modified; the frequency k2 can be replaced by a periodic function k2c2(x) that corresponds to the scattering in a non-homogeneous medium. All our arguments remain valid; the only change is to replace the exponents by the eigenfunctions of the Sturm-Liouville operator with potential k2c2(x) and with quasi-periodic boundary conditions. We remark also that the resonant case is not an obstacle for the method. The plan of the paper is the following. In Section 1 we pose the scattering problem for the plane Helmholtz equation and formulate the radiation condition. In Section 2 we give a review of results on factorization of elliptic pencils, properties of divisors and solvability of the corresponding operator equations on the semiaxis with the radiation conditions at + . In the subsequent sections we give the detailed ∞ 167 analysis of the scattering problems for the two- and three-dimensional cases. To our best knowledge, the space problem has not been considered in the literature before.

13.2 Scattered Waves and the Radiation Condition for the Two-dimensional Helmholtz Equation Let (x, y) be the coordinates in R2 and Γ be a 2π-periodic curve given by a smooth function y = a(x). Let ik(x sin ϕ+y cos ϕ) vϕ(x, y)= e− (1) be a monochromatic wave incident on the grating Γ. The reﬂection of this wave generates the scattered waves which are to be found. The number ϕ coincides with the angle between the axis Oy and the direction of the wave (see Figure 1). The wave vϕ(x, y) satisﬁes the Helmholtz equation

∆u + k2u =0, u = u(x, y) (2) and the quasi-periodic boundary conditions

it iν u(0, y)= e − u(2π, y) iν (3) ux′ (0, y)= e− ux′ (2π, y) where ν =2πk sin ϕ. Figure 1 Naturally, the scattered waves also satisfy equation (1.2) and boundary conditions (1.3). We have to declare the law of reﬂection. Assume that the wave vϕ reaches all points of the grating. This means that cot ϕ > max a′(x). The full reﬂection means u(x, y) = v (x, y) (4) |Γ ϕ |Γ or u(x, a(x)) = vϕ(x, a(x)), (5) where u(x, y) is a solution of the scattering problem in the domain

Ω= x, y x R,y>a(x) , { | ∈ } i.e., a solution of equation (2) subject to boundary conditions (3). The problem given in the unbounded domain Ω by equation (2) and initial condition (4) is not well-posed, since the frequency k2 > 0 belongs to the continuous 168 spectrum of the Laplace operator in Ω , with the Dirichlet boundary condition on ∂Ω. To extract physically reasonable solutions in such cases one claims additional conditions. It is well known that for the Helmholtz equation on the exterior of a bounded domain the condition ∂u iku = o(r) as r = x2 + y2 ∂r − →∞ guarantees existence and uniqueness of a solution.p This is the so-called Sommerfeld radiation condition. For unbounded domains with periodic boundaries the radiation conditions have a more intricate form. To formulate them, we remark that the elementary quasi- periodic solutions of equation (2) have the representation v iλny iµnx 2 2 u±(x, y)= e± e , µ = + n, λ = k µ . (6) n n 2π n − n Here n Z, and the main branch of the square rootp function is chosen, i.e., ∈ λ > 0 for k > µ and Im λ > 0 for µ > k. The solutions u−(x, y) corre- n | n| n | n| n sponding to the non-real values λn grow exponentially in Ω as y and have no physical sense in scattering problems. The solutions corresponding→ ∞ to the real numbers λn are called propagating waves and play the most important role. The + solutions un (x, y)(un−(x, y)) which correspond to λn > 0(λn < 0) are called outgoing (incoming) waves. The physical sense prompt us that the scattered waves must include only outgoing waves and exponentially decaying waves corresponding + to solutions un (x, y), Im λn > 0. Actually, it was Rayleigh [[26], [27] who assumed that scattered waves consist only of outgoing and decaying waves. The problem of a choice of physically reasonable propagating waves has been widely discussed in the literature since 30’s. In particu lar, Mandelstam noticed that the Rayleigh hypothesis does not work for some equations of electrodynamics and proposed to choose the waves with positive group velocity (see details in [30]). For the Helmholtz equation the Rayleigh and the Mandelstam hypotheses coincide. Now we can formulate the scattering problem as follows: to ﬁnd a solution of equation (2) subject to quasi-periodic conditions (3), initial condition (4) and the radiation condition

iµnx iλny u(x, y)= cne e + o(1) (7)

k6µn6k − X where the sum contains only outgoing waves corresponding to λn > 0 and o(1) is a decaying function as y . Here c are the amplitudes of the outgoing waves. →∞ n 169

They must be determined by initial condition (4). Further we will see that o(1) in (7) is represented as a convergent series of the exponentially decaying waves (although this is not necessarily true in the three-dimensional problem). 2 2 A frequency k is called resonant (in some books it is called cut-oﬀ) if λn = 2 2 Z Z k µn =0 for some n . If ν/π / then the equality λn =0 may hold for the only− value n Z In this∈ case the pair∈ of elementary solutions 0 ∈ ikx ikx e , ye , µ0 = v/2π + n0 = k > 0, corresponds to the wave number λ = 0. If ν/π Z and the equality k2 µ2 = n ∈ − n 0holds for some n Z, then it holds for two values n0,n1 Z. In this case the resonant elementary∈ solutions have the form ∈

ikx ikx ikx ikx e , ye ; e− , ye− .

The ﬁrst functions of these pairs of solutions are degenerated waves (independent of y); the second ones are associated solutions. There is one-to-one correspondence between elementary solutions of problem (2), (3) and Jordan chains of the pencil (operator symbol) corresponding to this problem. It follows from the subsequent general results that in the presence of a Jordan chain of even length only the ﬁrst half of the functions of this chain has to be taken into consideration. In our case the lengths of Jordan chains equal 2. Therefore, in this case only the eigenfunctions must be involved in the group of the scattered waves participating in the radiation condition. Hence the radiation condition in the resonant case is given as before by formula (7).

13.3 Factorization of elliptic operator pencils and solvability of the corresponding equations on the semi-axis In this section we deal with an operator pencil

T (λ)= λ2F + λG + H V − on a Hilbert space H. It is always assumed that the "main" operator H is self-adjoint and uniformly positive while the other ones are symmetric. Deﬁnition 2.1 T (λ) is called elliptic if the following conditions are fulﬁlled: i) F is bounded and uniformly positive (0 F ); ii) H = H∗ 0 and V is ≪ 1 ≪ ∞ ≫ a symmetric H-compact operator, i.e., VH− is compact in H; iii) G is symmetric 170

1/2 and (G) H ; iv) T (λ) > 0 for all λ R with λ >r0 provided r0 is large enough.D ⊃ D ∈ | | Deﬁnition 2.2 An elliptic pencil T (λ) is called strongly elliptic if there exist a number ε> 0 and a symmetric H-compact operator V ′ such that

2 T (λ) > ε λ I + H V ′ λ R. (8) − ∀ ∈ Deﬁnition 2.3 An elliptic pencil L(λ)is called regular elliptic if

1 2 1 HT − (λ) + λ T − (λ) 6 const λ R, λ >r . (9) | | ∀ ∈ | | 0 1/2 Let Hθ be the scale of the Hilbert spaces generated by the operator H , i.e., θ/2 θ/2 Hθ = H and x θ = H x . Recall that the abstract Sobolev space W s(a, bD; H) consists ofk functionsk f(t) deﬁned on (a, b) R, taking values in H, ⊂ and having a ﬁnite norm

b 2 2 2 (s) s/2 f s = f (t) + H f(t) dt. k k a Z

(see details in [21, Ch. 1]). Deﬁnition 2.4 An elliptic pencil T (λ) is called strongly regular if for all functions v(t) W 2[0, ; H] subject to the condition v(0) = 0, the following estimate holds ∈ ∞ d T i v(t) > ε v , ε> 0 (10) χ − dt k k2 L2

2 where Tκ(λ)= λ F + κ λG + H and ε does not depend on v and κ [0, 1]. It is known [29, A §3] that estimate (10) implies (9), i.e., a strongly∈ regular elliptic pencil is regular elliptic but not vice versa. We remark that for usual elliptic operators estimate (8) is equivalent to the Garding inequality (see [30]), estimate (9) is known as the Agmon-Nirenberg or the Agranovich–Vishik estimate for regular elliptic problems with parameter on smooth bounded domains (in this context H = n L2(Ω), where Ω is a smooth bounded domain in R ), and estimate (10)is known as the Bermstein-Ladyzhenskaya inequality (see [29, |§3, 6]). More details on the motivation of the above deﬁnitions can be found in the papers [29], [30]. The veriﬁcation of estimate (10) is not trivial even for concrete pencils. We shall use the following result. Proposition 2.5 Let T (λ) be elliptic. Suppose that F = F0 + F1, G = G0 + G1, H = H0 + H1, where F1, G1,H1 are symmetric operators such that 171

1/2 1 F , G H− ,H H− are compact, and H 0. If the estimate 1 1 1 0 ≫ 1 1/2 1/2 F − 2 Gy 6 (2 ε) H y , y H (11) 0 − 0 ∈ D 0 holds with some ε> 0, then T (λ) is strongly regular. Proof: It can be found in [29, §9]. Let the operator H have discrete spectrum. It follows from the theorem on holomorphic operator functions (see [12, Ch. 1]and [29, §1.4] for the version of this theorem for pencils with unbounded coeﬃcients) that the spectrum of an elliptic pencil T (λ) in this case is also discrete. It is known [16] that the principal part of the Laurent expansion of the resolvent 1 T − (λ) admits a representation

N pk pk s s , z − x · k k (λ µ)pk+1 s k=1 s=0 − X X − where 0 pk xk,...,xk , k =1,...,N (12) is a canonical system of eigen and associated vectors of the pencil T (λ), and

0 pk zk,...,zk , k =1,...,N is the adjoint canonical system of the eigen and associated vectors of T (λ) corresponding to the eigenvalue µ¯. The following result is essential in the sequel. Proposition 2.6 Canonical system (12) corresponding to a real eigenvalue µ, can be chosen so that

xs = ε zs, k =1,...,N, s =0,...,p , k k k ∀ k 0 pk where εk = 1 and εk = sign L′(µ)xk,xk . Proof: See± [19, Lemma 1.2]. Other definitions of the sign characteristics are given in [13]. Actually, the sign characteristics are important only for Jordan chains of odd length. Further, we assume for convenience that the sign characteristics of Jordan chains of even length equal zero. Canonical systems that possess the properties formulated in Proposition 2.6 are called normal. Normal canonical systems and sign characteristics are important in the analysis of the factorization problem. We shall give here short historical 172 comments concerning the problem of the factorization of operator polynomials with respect to the real axis. Krein and Langer [17] studied pencils of the form L(λ)= I + λB + λ2C, where B and C are bounded self-adjoint operators and C > 0. They proved that L(λ) possesses a right divisor of the form (λZ I), whose spectrum is located in the closed upper half-plane. The real spectrum− of this divisor was investigated by Kostyuchenko and Orazov [18]. The factorization of higher order operator polynomials was out carried by Langer [20]; the detailed analysis of the real spectrum of divisors of polynomials with Hermitian matrix coefficients was done by Gohberg, Lancaster and Rodman [13]. We should mention that the problem of factorization of non-negative operator pencils (and operator functions) on the real axis has its own history (see the book of Rosenblum and Rovnjak [28]). In the paper [19] the authors proposed a new analytic approach to the factorization of quadratic pencils, investigated the properties of a linear operator Z participating in the factorization and proved the first factorizatio n theorem for pencils with unbounded coefficients. Further developments of the theory was carried out by Shkalikov [29]-[31]. Now, let us define the half of the eigen and associated vectors of an operator pencil T (λ). Let canonical system (12) be normal. Its half consists of the vectors

0 1 lk xk,xk,...,xk , k =1,...,N (13) where l =(p 1+ ε ) /2 and ε are the sign characteristics (we assumes ε =0 for k k − k k k Jordan chains of even length). We imply that in the case lk = 1 the corresponding set in (13) is empty. The set of all canonical systems of T (λ)−corresponding to the eigenvalues from the open upper half-plane and of the halves of canonical systems corresponding to the real eigenvalues, is called thehalf of eigen and associated vectors of T (λ). We point out a particular important case (connected with the scattering problem): if the lengths of Jordan chains corresponding to the real eigenvalues do not exceed 2, then the half contains the canonical systems corresponding to the eigenvalues from the upper half-plane and only the eigenvectors xk corresponding to λk R subject to the condition (T ′ (λk) xk,xk) > 0. To∈ formulate the basic results we shall introduce the class of operator pencils whose resolvents are meromorphic functions of ﬁnite order having polynomial growth on some rays in the complex plane. Deﬁnition 2.7 We say T (λ) belongs to the class K if i) the eigenvalues of the operator H satisfy the estimate p λj(H) > cj 173 with some constants c and p. ii) either p > 2 or p < 2 and there exist rays zγ = λ arg λ = θ , j = j { | j} 1,...,N,θj+1 >θj, in the upper half-plane such that

max(θ ,θ θ , π θ ) < 2π/p 1 j+1 − j − N and 1/2 1 1/2 m H T − (λ)H 6 c λ for λ γ (14) | | ∈ j with some constants c, m, provided λ is large enough. It is proved in [30] that the inequality| |

(T (λ)x,x) > ε r2(x,x)+(Hx,x) , λ γ, λ = r>r | | ∀ ∈ | | 0 is sufficient for the validity of estimate (14) with m = 0 on the ray γ. The last inequality is easier to verify for concrete problems. In particular, (14) holds on the p real line for strongly elliptic pencils. If T (λ) is elliptic and the condition λj(H) > cj 1 holds, then the resolvent T − (λ) is a meromorphic function of order 6 p/2 (see [29, §2]). Applying the Phragmen–Lindelof theorem we find: if an elliptic pencil T (λ) 1 is strongly regular, belongs to the class K and F (λ) = T − (λ)f,f is an entire function for some f H, then F (λ) 0. Hence conditions i) and ii) in Definition 2.7 are needed to prove∈ the completeness≡ theorems for eigenv ectors (see details in [30]). Now let us specify the understanding of solutions of the equation

d T i u(t)= Fu′′(t)+ iGu′(t)+(H + V )u(t)=0. (15) dt − A function u(t) W 1(a, b; H) is said to be a generalized solution of equation (15) if for all functions∈ v(t) W 1(a, b; H) subject to the conditions v(a)= v(b)=0, the equality ∈

1/2 1/2 (Fu′,v′)+ i (u′, Gv)+ (I + V ′) H u,H v =0 1/2 1/2 holds, where V ′ = H− VH− and the scalar product is taken in L2(a, b; H). Details clarifying this deﬁnition see in [30]. A function u(t) W 2(a, b; H) is called a classical solution of equation (15) on ∈ (a, b) if (15) holds as equality of functions in L2(a, b; H). 174

We say that a classical (generalized) solution of equation (15) satisfies the radiation condition at + if ∞ iλkt u(t)= cke xk + u0(t) (16) ε >0 Xk + where u0(t) is a classical (generalized) solution on R satisfying the condition u0(t) 1 0 as t . Here the first term in (16) is a finite sum of elemen- taryk solutionsk → corresponding→ ∞ to the real eigenvalues of nonnegative type, and for simplicity we have assumed that the lengths of Jordan chains corresponding to the real eigenvalues do not exceed 2. In the general case the sum has to contain all elementary solutions corresponding to the halves of Jordan chains (13). Let us formulate the basic results on elliptic pencils. Theorem 2.8 The half of eigen and associated vectors of a self-adjoint elliptic pencil T (λ) is minimal (i.e., there exists a biorthogonal system) in the spaces Hθ, 0 6 θ 6 1. It is complete in the same spaces if T (λ) is either strongly or regular elliptic and belongs to the class K. If T (λ) is strongly regular and belongs to the class K, then the half is a complete system in for 0 <6 θ 6 3/2. Theorem 2.9 Let T (λ) be either strongly elliptic or regular elliptic pencil and belong to the class K. Then T (λ)=(λ Z ) F (λ Z), (17) − 1 − where the operator Z possesses the properties: 1/2 1. (Z)= H1 and Z λ = KH , where K is bounded and boundedly invertible inD H, provided λ /−σ(Z); ∈ 2. the spectrum of Z lies in the closed upper half-plane and the system of its eigen and associated vectors coincides with the half of those of T (λ) ;

3. iZ generates a holomorphic semigroup in the spaces Hθ, 0 6 θ 6 1. If in addition T (λ) is strongly regular, then property 3) remains valid in the spaces Hθ for 0 6 θ 6 3/2. Theorem 2.10 Let T (λ) be either strongly or regular elliptic self-adjoint pencil belonging to the class K. Then for any vector f Hθ, 0 6 θ 6 1, there exists a unique function u(t) which is a generalized solution∈ of equation (15) on (ε, ) for any ε> 0 and satisﬁes the radiation condition (16) and the initial condition∞

lim u(t) f θ =0. (18) t 0+ → k − k 175

If T (λ)is strongly regular, then the same is true for 0 6 θ 6 3/2. Moreover, u(t) is a classical solution for t> 0 and is represented by the formula

1 itλ 1 u(t)= e (Z λ)− dλ, (19) 2πi − Zγ where a contour γ contains the spectrum of the operator Z and lies asymptotically in the upper half-plane. Theorem 2.11 Let T (λ) be a strongly regular self-adjoint pencil. Then there exists a unique classical solution of equation (15) on the semi-axis satisfying radiation condition (16) and initial condition (18) for θ =3/2. The most important fact of the last theorem is that existence and uniqueness of solutions of the half-range Cauchy problem on the semiaxis R+ is true for operator pencils not necessarily belonging to the class K. The proofs of Theorems 2.9-2.11 can be found in [29], [30] (see also [19], where the ﬁrst results of this kind were obtained).

13.4 Existence and Uniqueness of the Solution of the Plane Scattering Problem It suﬃces to deﬁne a solution of the scattering problem in the semi-strip

Ω = x, y 0 6 x 6 2π, a(x) 6 y < 0 { | ∞} The corresponding solution in the whole half-plane y > a(x) is restored by quasi- periodic conditions (3). The substitution of ξ = x, η = y a(x) (20) − maps the semi-strip Ω0 onto the standard semi-strip Ω0′ (see Figure 2). Taking into account that ξx =1, ξy =0, ξxx =0, ξyy = 0;

η = a′(x), η =1, η = a′′(x), η =0, x − y xx − yy and 2 2 uxx = uξξξx +2uξηξxηx + uηηηx + uξξxx + uηηxx, 2 2 uyy = uξξξy +2uξηξyηy + uηηηy + uξξyy + uηηyy, 176 we ﬁnd that the Helmholtz equation is transformed into the following one

2 2 u 2u a′(ξ)+ u a′(ξ) +1 u a′′(ξ)+ k u =0 (21) ξξ − ξη ηη − η The form of this equation is more intricate but its coeﬃcients do not depend on η and the advantage is that ξ and η belong to the domain Ω0′ , where the separation of variables can be realized. Looking for solutions of the form u(ξ,η)= eiληf(ξ) and taking into account quasi-periodic conditions (3) we come to the spectral problem

2 2 2 f ′′ k f + λi (2a′(ξ)f ′ + a′′(ξ)f)+ λ a′(ξ) +1 f =0 (22) − − with the boundary conditions

iv iv f(0) = e− f(2π), f ′(0) = e− f ′(2π). (23) Since functions (6) represent a complete set of elementary quasi-periodic solutions of the Helmholtz equation, the functions

iλna(x) iµnx fn±(x)= un±(x, a(x)) = e± e v (24) µ = + n, λ = k2 µ2 n 2π n − n form a complete set of all eigenfunctions of the spectralp problem (22) and (23), which correspond to the eigenvalues λn. The set of eigenfunctions of the problem (22), (23) can also be found by a straightforward calculation. Substituting f(ξ)= z(ξ)eiλa(ξ) in (22) and (23) we ﬁnd 2 2 z′′ + λ k z =0, − − iv iv z(0) = e− z(2π), z′(0) = e− z′(2π). Since eiµnx form a complete set of eigenfunctions of this problem, we obtain that set (24) possesses the same property with respect to problem (22), (23). 2 2 It is easily seen that in the non-resonant case all eigenvalues λn = k µn are simple provided ν/π / Z. The location of λ for values ν close to 0 is shown− in n p Figure 3. ∈ If ν/π Z but ν/2π / Z, then two eigenfunctions ∈ ∈ ia(x)√k2 (n+1/2)2 ia(x)√k2 (n+1/2)2 e − sin(n +1/2)x, e − cos(n +1/2)x

2 2 correspond to the eigenvalues λn = k (n +1/2) . Finally, if ν/2π Z then the pair of eigenfunctions − ∈ p ia(x)√k2 n2 ia(x)√k2 n2 e − cos nx, e − sin nx, 177

2 2 correspond to all eigenvalues λn = √k n = k. The extremal real eigenvalues − 6 ± ika(x) λ0 = k are simple; the corresponding eigenfunctions are e± . ± In the± resonant case zero is the eigenvalue of pencil (22), (23) of algebraic multiplicity 2 or 4. If ν/π / Z, then the only eigenfunction f (x) = ei(n0+ν/2π)x ∈ n0 corresponds to this eigenvalue (here no is defined by the equality k = n0 + ν/2π), and there is an associated function that coincides with fn0(x) (we omit here elemen- i(n +ν/2π)x tary calculations). If ν/π Z, then two eigenfunctions e± 0 correspond to 0, and there are associate functions∈ coinciding with the previous ones. + According to the general definition (see Section 2) the functions fn (x) n∞= defined in (24) form the half of the root functions of pencil (22), (23).{ The} same−∞ is true in the resonant case, since the lengths of Jordan chains do not exceed 2. It is worth mentioning that Rayleigh [26] calculated these waves in the case of a vertical incident wave (ν =0) assuming that a(x) is an even function with respect to x = 0 and x = π. In this case a solution u(x, y) of the scattering problem the same property u′(0, y)= u′(π, y)=0. If conditions (23) are replaced by f ′(0) = f ′(π)=0, then the functions

ia(x)√k2 n2 fn(x)= e − cos nx, n =0, 1,... (25) form the half of the root functions of the corresponding pencil. System (25) is called the Rayleigh system. Suppose that the Rayleigh system is minimal, say in the space L2(0, π). In this case a solution of the scattering problem can be represented by a formal series ∞ i√k2 n2 u(x, y)= (v ,f ∗) f (x)e − (y a(x)) ϕ n n − n=0 X ika(x) + where vϕ = e and fn∗ is a biorthogonal system with respect to fn . However, we are not aware of papers{ } where the minimality or the completeness{ of the} Rayleigh system (or the generalized Rayleigh system defined by (24)) is proved. Moreover, the minimality and the completeness do not guarantee the convergence of the series to the solution u(x, y). Hence a rigorous justification of the Fourier method for the Rayleigh problem seems to be a hard task (see Theorem 3.4 below). We intend to apply the results of Section 2 to solve the scattering problem. Let us represent pencil (22), (23) in the abstract form. For s = 1 (s = 2) denote S s by WU [0, 2π] the subspace of the Sobolev space W2 [0, 2π] consisting of functions satisfying the first boundary condition (23) (both conditions (23)). The intermediate spaces θ 2 6 6 WU [0, 2π]= WU [0, 2π],L2[0, 2π] θ , 0 θ 2 178 are defined by interpolation (see [21, Ch.1]). In the space L2[0, 2π], let us define the operators

2 Hf = f ′′ + f, (H)= WU [0, 2π] − D 1 Gf = i (2a′(x)f ′ + a′′(x)f) , (G)= WU [0, 2π] 2 D Ff = a′(x) +1 f, (F )= L [0, 2π]. D 2 2 Further it is assumed that a(x) W2 [0, 2π]. Now, problem (22), (23) is represented in the form ∈

T (λ)f =0,T (λ)= λ2F + λG + H V, V = k2 +1 I − Proposition 3.1 The pencil T (λ) is self-adjoint and strongly elliptic. Proof: It is obvious that H = H∗ 0. Integrating by parts we ﬁnd ≫

i (2a′(x)f ′ + a′′(x)f,f)= i (a′(x)f ′,f) i (f, a′(x)f ′) . (26) − Hence the quadratic form (Gf,f) is real and G is symmetric. Since

2 1/2 (Hf,f)=(f ′,f ′)+(f,f)= H f

1/2 1 we have H = WU [0, 2π] (G). Finally, let us prove estimate (8). We shall use the inequalityD ⊂ D 2 (f,f) > M − (a′f, a′f) , where M = max a′(x) 06x62π | | Bearing in mind (26), we obtain for λ R ∈

2 (T (λ)f,f) > (f ′,f ′) k (f,f) 2 λ Im (f ′, a′f) − − | | 2 2 2 + λ (a′f, a′f)+ λ (f,f) > f ′ 2 λ f ′ a′f k k − | |k kk k 1 2 2 2 1 2 2 2 2 2 + 1+ M − a′f λ + M − λ f k f 2 k k 2 k k − k k 1 1 2 − 2 1 2 2 2 2 2 > 1 1+ M − f ′ + M − λ f k f − 2 k k 2 k k − k k " # 1 > 2M 2 +1 − (Hf,f)+ λ2(f,f) k2 +1 (f,f). (27) − 179

This proves the proposition. Proposition 3.2The pencil T (λ) is regular elliptic and, moreover, strongly regular elliptic. Proof: Estimate (9) can be obtained by a straightforward calculation of the 1 resolvent kernel of the integral operator T − (λ). First, one has to prove by standard means that there is a pair of solutions of equation (22) having the asymptotics

(a(x) ix)λ 1 f ±(λ, x)= e ± 1+ O λ− , λ →∞ if λ, is located in one of the quadrants that are formed by the real and the imaginary axes. Then these solutions have to be substituted in the well-known formulas for the Green function (see [23, Ch.1]). A detailed proof of estimates of the type (9) for ordinary diﬀerential pencils of arbitrary order can be found in the work of Pliev [25]. To prove the strong regularity we recall Proposition 2.5. Since the operator 1/2 2 a′′(x)H− is compact in the space H = L2 (provided a W2 ), it suﬃces to obtain the estimate ∈ 2 1/2 2 F − a′(x)y′ 6 (1 ε) (Hy,y), ε> 0 (28) − for functions y ( H). We have ∈ D 1/2 2 1/2 F − a′(x) = 1+ a′(x) − a′(x) < 1 ε − for some ε> 0. Therefore,

2 1/2 2 2 2 F − a′(x)y′ 6 (1 ε) y′ = (1 ε) (Hy,y). − k k −

The proposition is proved.

Proposition 3.3 The sign characteristics of the eigenfunctions of T (λ) are deﬁned by the relations

ε± = sign T ′ ( λ ) f ±,f ± = sign( λ ) , λ R n ± n n n ± n n ∈ + Hence the functions fn (x) n∞= defined in (24) form the half of the eigen and associated functions{ of T (λ)}. −∞ Proof: For λn > 0 we find + + + + + + (T ′ (λn) fn ,fn )=(Gfn ,fn )+2λn (Ffn ,fn ) + + + + + + = µ (a′f ,f )+ i (a′′f ,f )+2λ (f ,f ) − n n n n n n n n = µn (a′, 1) + i (a′′, 1)+4πλn =4πλn 180 as the functions a and a′ are periodic. + Theorem 3.4 The generalized Rayleigh system fn (x) n∞= defined in (24) is θ { } −∞ minimal and complete in the spaces WU [0, 2π] if 0 6 θ 6 3/2. For any function g W θ the Fourier series ∈ U ∞ + iλnη u(x,η)= (g,fn∗) fn (x)e , (29) n= X−∞ θ converges for η>η in the norm of W provided η is large enough (here f ∗ the 0 U 0 { n} biorthogonal system in L2 with respect to fn ). Moreover, the function u(x,η)) admits a holomorphic continuation in a sector{ } arg η < ε for sufficiently small ε, and there exists | | s lim u(x,η)= g η +0 − → θ (the limits is understood in the norm of W2 ) Proof: The completeness and the minimality is the consequence of Theorem 2.8. The convergence of series (29) follows from representation (19) if there is a sequence of semicircles λ = r in the upper half-plane such that | | k →∞ 1 η Im λ η sin ϕ κ (Z λ)− 6 e 0| | = e 0| | k , λ = r (30) − | | k

Let us prove (30). Without loss of generality suppose that Z is invertible (equivalently, T (0) is invertible). It follows from Theorem 2.9 that

1/2 1/2 2 1/2 1 1 Z = KH , Z = H k +1 H− K− F − , 1 − 1 where K and K− are bounded. Hence

1 1 (Z λ)− = T − (λ)(Z λ) F. (31) − 1 − 1 Recall that the Green function of the integral operator T − (λ) (see Proposition 3.2) is a meromorphic function of order 1 and of ﬁnite type. By virtue of the Titchmarsh theorem for any C exceeding the type there is a sequence rk such 1 →∞ that T − (λ) 6 exp(C λ ) for λ = rk. Hence estimate (30) outside a double sector Λ containing the| real| axis| follows| from (31) . Since T (λ) is strongly elliptic, ϕ0 estimate (30) inside a small double sector follows from [30, Theorem 1.7]. Hence, (30) is proved and series (29) converges for η>C. The last assertion of the theorem follows from the fact that the operator Z is a generator of a holomorphic semigroup θ in the spaces WU . 181

We remark that series (29) does not converge for all η > 0 and arbitrary functions g, i.e., the system in question does not form a basis for the Abel summability method of order 1. Let us clarify our claim for Rayleigh system (25) assuming in addition that a(x) is holomorphic. 2 2 Proposition 3.5Let λn = √k n , fn(x) be deﬁned by (25) and a(x) be holomorphic on R. If the series −

∞ iλnη u(x,η)= (g,fn∗) fn(x)e (32) n=0 X converges in L2 for all η > 0, then g(x) is holomorphic at all points x (0, π) except the points where a(x) attains the maximum. ∈ (M ε)n Proof: Let M = max a(x). For any ε > 0 we have f > e − for all k nk sufficiently large n. Assuming that series (32) converges in L2 for η = ε we get (2ε M)n the estimate |( (g,fn∗) < e − (under our assumption the norms of functions in series (32) tend| to 0 as| n ). If a(ξ) < M and ε is small enough, then there is a neighborhood U of ξ such→∞ that series (32) converges uniformly for all x U and 0 6 η 6 ε. Since the terms of the series are holomorphic functions of x,∈ the sum is also holomorphic at ξ. According to Theorem 3.4 this sum coincides with g(x). Thus g(x) is holomorphic at ξ. Remark 3.6 It follows from the proof of Proposition 3.5 that g(x) admits a holomorphic continuation in the domain Λ= z Re a(z) M < 0 { | − } if the series (32) is summable by the Abel method of order 1 (i.e., converges for all η > 0). We do not know if the converse assertion is also true. Let us formulate the basic result of this section. Theorem 3.7 There is the only solution u(x, y) of scattering problem (2)- (4), 2 2 2 2 (7). If CR = x, y x + y < R then u(x, y) W2,loc (Ω CR) for any R > 0. For large y the solution| u(x, y) is represented by∈ the Fourier∩ series with respect to + the generalized Rayleigh system fn (x) . Proof: It suffices to put η ={y a(}x) and recall Theorems 2.11 and 3.4. − 13.5 Scattering by Two-periodic Surfaces in M3 Let a smooth function z = a(x, y) be 2π-periodic with respect to both variables x, y. This function defines the surface (the grating) S in R3. Let ik(x cos α+y cos β+z cos γ) v(x,y,z)= e− 182 be the wave incident onto this surface with directing vector ϕ¯ = (cos α, cos β, cos γ), ϕ¯ =1. | | The wave v satisfies the Helmholtz equation

∆u + k2u =0, u = u(x,y,z) (33) and quasi-periodic conditions

iν iν u(0,y,z)= e− u(2π,y,z), ux′ (0,y,z)= e− ux′ (2π,y,z), iδ iδ , (34) u(x, 0, z)= e− u(x, 2π, z), uy′ (x, 0, z)= e− uy′ (x, 2π, z) where ν =2πk cos α,δ =2πk cos β. Scattered waves do the same. It is assumed that the wave v reaches all the points of the surface (there are no shadows). The problem is to find a quasi-periodic solution of (33) that is represented (in some sense) as a superposition of scattered waves and satisfies the "full reflection" condition

u(x,y,a(x, y)) = v(x,y,a(x, y)). (35)

To define the scattered waves let us find all quasi-periodic solutions of equation (33) in the cylinder with the base K2π = [0, 2π] [0, 2π]. Separating the variables x and y, we find that the elementary solutions have× the representation

iµnx iρj y iλnj z unj± (x,y,z)= e e e± , (36) where ν δ µn = + n, ρj = + j 2π 2π . (37) λ = k2 µ2 ρ2, n,j Z nj − n − j ∈ q The branch of the square root is chosen so that either λnj > 0 or Im λnj > 0. The set of scattered waves consists of outgoing propagating waves and decaying waves. + It coincides with the system unj . We can not guarantee that the reflected solution is a finite or infinite superposition of the scattered waves. Therefore, we are looking for solutions satisfying the radiation condition at z , namely →∞

i(µnx+ρj y+λnj z) u(x,y,z)= cnje + o(1). (38) 2 26 2 µn+Xρj k Here o(1) 0 as z and c are unknown constants to be determined. → →∞ nj 183

The substitution of ξ = x, η = y, ζ = z a(x, y) − maps the half-cylinder Ω= x,y,z 0 6 x 6 2π, 0 6 y 6 2π, a(x, y) 6 z < { | ∞} onto a usual half-cylinder whose base is the square K2π. This substitution transforms equation (33) to the form

uξξ + uηη + uζζ 2uξaξ′ 2uηζaη′ 2 −2 − 2 +u a′ + a′ u a′′ u a′′ + k u =0 ζζ ξ η − ζ ξ − ζ η Separating the variable ζ by putting u = f(ξ,η)eiλζ and taking into account boundary conditions (34) , we ﬁnd T (λ)f = F λ2 + λG + H k2 +1 I f =0, (39) − where the operators H, G, F are deﬁned as follows

2 Hf = fξξ′′ fηη′′ + f, (H)= WU [K2π] − − D 1 Gf = i 2aξ′ fξ′ +2aη′ fη′ + aξξ′′ + aηη′′ f , (G)= WU [K2π] 2 2 D Ff = 1+ a′ + a′ f, (F )= L (K ) ξ η D 2 2π s s Here WU [K2π] is the subspace of the Sobolev space W2 [K2π] consisting of functions subject to the quasi-periodic boundary conditions. It is easily seen that the eigenfunctions of pencil (39) coincide with the traces of elementary solutions (36) on the surface z = a(x, y), i.e.,

iλnj a(x,y) iµnx iρj y fnj± (x, y)= e± e e . (40) This can be checked independently by a straightforward calculation if one puts in (39) f(ξ,η)= v(ξ,η)eiλa(ξ,η). Then the function v satisﬁes the equation ∆v + k2 λ2 v =0 − iµnx iρjy and quasi-periodic boundary conditions. This holds for functions vnj = e e , where µn and ρj are deﬁned by (37). We remark that the multiplicities of the 184 eigenvalues λ = k2 µ2 ρ2 may grow as k . For example, if ν = δ =0 nj − n − j →∞ 2 and k = 50, thenq zero is the eigenvalue of the geometric multiplicity 16 (λnj = 0 for n = 1, 4, 6, 7 and j = 7, 6, 4, 1, respectively) and of algebraic multiplicity± 32± (all± Jordan± chains have± ± length± 2± and associated functions coincide with eigenfunctions). Let us prove that T (λ) is an elliptic pencil. The properties

H = H∗ 0, 0 F , G G∗, ≫ ≪ ≪∞ ⊂ (G) W 1 (K )= H1/2 , D ⊂ U 2π D are trivial to check. Further,

2 2 (T (λ)f,f)= f ′ + f ′ 2 f ′, a′ f 2λ Im f ′ , a′ f ξ η − ξ ξ − η η +λ2 1+ a 2 + a 2 f,f k2(f,f) > 0, ξ′ η′ − 2 2 R 2 provided λ > k , λ . Since the embedding I : WU (K2π) L2 (K2π) is compact, the last estimate∈ implies that T (λ) is strongly elliptic. It is→ important in the sequel to ﬁnd explicitly a double angle where estimate(8) holds. Proposition 4.1 Let

M1 = max ax′ (x, y) , M2 = max ay′ (x, y) , x,y K2π | | x,y K2π ∈ ∈

1 ϕ = arctg . 2 2 1+ M1 + M2 If arg λ = θ<ϕ, then for λ = r>r the following estimate holds | ± | | | p 0 Re(T (λ)f,f) > ε r2(f,f)+(Hf,f) (41) Proof: Let λ = reiθ and let α, β be positive numbers such that α + β =1. We have

2 iθ 2 2iθ 2 2 2 Re f ′ 2re Im f ′, a′ f + r e α + a′ (f,f) > ε f ′ + r (f,f) ξ − ξ ξ ξ ξ (42) for some ε = ε(θ), provided α cos2 θ cos2θ 1+ < 0. (43) − M 2 1 185

Similarly,

2 iθ 2 2iθ 2 Re fη′ 2re Im fη′ , aη′ f + r e β + aξ′ (f,f) − (44) 2 2 >h ε f ′ + r (f,f ) i η provided β cos2 θ cos2θ 1+ < 0. (45) − M 2 2 The inequalities (43) and (45) are equivalent to the following ones α β 2 6 2 tg θ 2 , tg θ< 2 . M1 + α M2 + β

2 2 M1 M2 For α = 2 2 , β = 2 2 last inequalities are equivalent to the condition θ <ϕ M1 +M2 M1 +M2 | | . Summing inequalities(42) and (44) with the chosen α and β we obtain estimate (41). Analyzing the proof of Proposition 4.1 one can understand that the bound for ϕ is precise, i.e., estimate (41) does not hold generally inside the angle ϕ< arg λ < π ϕ. 1 − Seemingly, T − (λ) has an exponential growth inside this angle. The eigenvalue asymptotics of the Laplace operator on a bounded domain is known, hence, in our p problem we have λj(H) = cj with p = 1. Therefore, we can guarantee that the pencil T (λ) belong to the class K and we can claim (by virtue of Theorem 2.8) + the completeness of the traces fnj(x, y) of the scattered waves only in the case ϕ<π/4. The problem whether the system f + is complete in the case ϕ > π/2 nj is open. Nevertheless, the following basic result is true. Theorem 4.2 There exists the only solution u (x,y,z) of scattering problem (33) -(35), (38). Proof: Elliptic pencil (39) is strongly regular. One can prove this fact repeating the arguments of Proposition 2.5. Putting z = ζ + a(x, y) and recalling Theorem 2.11 we obtain the assertion of the theorem.

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[33] C.H. Wilcox, Scattering Theory for Diﬀraction Gratings, Springer Verlag, New York, 1984.

[34] C.H. Wilcox and J.C. Guilliot, Scattering theory for acoustic diffraction gratings preliminary report, Notices AMS 25, A356 (1978). 189 14 On the stability of a top with a cavity filled with a viscous fluid 14.1 Introduction We consider small oscillations of a rotating top with a cavity entirely filled with an incompressible viscous fluid. There are no restrictions on the distribution of mass in the body and on the form of the cavity. In the nonperturbed state, the velocity field of the fluid is that of a rigid body rotating together with the top about the vertical axis that coincides with one of the principal axes of inertia of the system. The shell has a fixed point. The system is moving in a gravitational field. The center of gravity lies on the principal axis of the nonperturbed rotation. The main goal of this work is to obtain a stability criterion for the system and to investigate spectral properties of the evolution operator corresponding to the linear equations. This problem has a long history. In the first place, the investigations of Sobolev [1], Rumyantsev [2], and Chernous’ko [3], including sufficient conditions for stability and instability of the top, must be noted (see also the monographs [4],[5]). Rumyantsev [2],[4] showed that the top is stable if both

n = a a k/ω2 and n = a a k/ω2 (1) 1 0 − 1 − 2 0 − 2 − are positive. Here ω is the angular velocity of the top. The other constants are defined in §1. In particular, if the top has a symmetry axis of order greater than one, then a1 = a2 and the positiveness of the number n1 = n2 is sufficient for the stability. In the absence of gravity, this sufficient condition means that the axis of the nonperturbed rotation is the axis of the greatest central moment of inertia. Rumyantsev obtained this condition with the help of the Lyapunov second method, following Sobolev, who obtained the same condition for the case of an ideal fluid (i.e., zero viscosity). He showed that the evolution operator corresponding to the linear problem is self-adjoint in a Hilbert space with indefinite metric. The metric is definite if n1 = n2 > 0. On the other hand, Chernous’ko [3] obtained conditions of instability in the problem. He found asymptotic expansions for solutions of the linear equations of motion of the top with respect to the powers of the Reynolds number and showed that the top is not stable if the sods of the nonperturbed rotation is the axis of the minimal or middle moment of inertia. Later, Smirnova [23] obtained this result without certain symmetry assumptions of Chernous’ko. She also showed that the rotation about the axis of greatest moment of inertia is the unique stable rotation if the viscosity of the fluid is sufficiently small and the cavity is toroidal 190

[6]. Chernous’ko [7] obtained this result for the top with a spherical cavity. In [4], Rumyantsev obtained some implicit conditions of instability. In this paper, we show that the Unear equation for the problem can be represented in the form with an evolution operator that is dissipative in a space with indefinite metric (a Pontryagin space). Hence, we can use nontrivial results of the theory of such operators and obtain necessary and sufficient conditions of stability. Moreover, we obtain the exact value of the instability index of the problem. The assumptions on the symmetry of the top, the absence of gravity, and on the large or small viscosity turn out to be superfluous. Furthermore, unlike the previous works, we.investigate the stability of the infinite-dimensional system as a whole but not its finite-dimensional “rigid” part. One of the main results of the paper is that the instability index of the problem does not depend on the nonzero viscosity and on the form of the cavity. It is equal to the number of negative eigenvalues of the matrix n 0 1 . This fails to be true if the viscosity is zero (see [1],[8]). 0 n2 The methodology of our approach is close to that of the work of Sobolev [1] However, in the derivation of the equations of motion of the system, Sobolev used the coordinate system rotating with respect to the inertial frame at the same angular velocity as the non-perturbed top. These equations turn out to be complicated for analysis. They admit a representation in an operator form only in the case of zero viscosity (see §5). Following [4], we write out the equations of motion of the top in a coordinate system attached to the rigid shell and obtain linear equations that admit a convenient operator form (cf. [8],[9]). The work is divided into five sections. In §1 we more exactly define the setting of the problem and write out the equations of motion of the system. In §2 the linear equations are represented in the form of an operator equation in a Hilbert space. We show that the evolution operator T of the system is a maximal dissipative operator in the Pontryagin space. In §3 we find the number of eigenvalues of the operator T in the open lower half-plane and prove that this number coincides with the instability index. In §4 the basis property of the eigenfunctions of the operator T is investigated and the description of the spectrum for large viscosity is given. In §5 we consider a symmetric top. The family of invariant subspaces of the evolution operator T is given. The subspaces that contain unstable motions of the top are singled out. The correspondence between the operator T and Sobolev’s operator is given. subsection*Notations. Spaces and operators of fluid mechanics. Suppose that 3 a rigid body occupies a bounded region Ω1 in R and contains a cavity Ω inside, that is, Ω Ω . The cavity is completely filled with fluid. Further, assume that Ω ⊂ 1 191

3 is a domain in R . The rigid body itself occupies the domain Ω1 Ω and is called the shell. Speaking of a body with frozen fluid, we imply the following:\ the region ft) as a whole is considered as a rigid body; the density of this body is equal to that of the original one in Ω Ω and is equal to the density of the fluid in Ω. 1 \ The body is rotating about a fixed origin of coordinates. Let eo, e1, e2 be the unit vectors of the principal axes of inertia of the body with frozen fluid. Consider the orthonormal frame Oe0e1e2 rigidly attached to the body and a fixed orthonormal frame Of0f1f2. In the non-perturbed state, e0 coincides with f0. The motion of the system “(body+fluid)’ ’ is described by three variables (z,w,v(x)), namely, the coordinates z of f0, the angular velocity w of the shell, and the velocity field v(x) of the fluid. Here x = (x ,x ,x ) Ω are the coordinates of a point. The four 0 1 2 ∈ variables z, w, v(x), and x axe vectors in the frame Oe0e1e2. The finiteness of kinetic energy implies that the values of the velocity field v 3 3 belong to the subspace J0(Ω) L2(Ω). The subspace J0(Ω) is the closure in L2 of the set of smooth solenoidal vectors⊂ compactly supported in Ω (i.e., w =0). We have the Weyl decomposition ÷

L3(Ω) = J (Ω) G(Ω) 2 0 ⊕ Here G(Ω) consists of the functions grad p(x), where p(x) is a scalar locally square integrable function whose ﬁrst generalized derivatives belong to L2(Ω). The following operators are well known in ﬂuid mechanics (see [10] and [5]). The vector

B∗v = [x,v(x)]dx ZΩ is called the gyrostatic moment of the fluid. Here [ , ] denotes the vector product in 3 3 · · R . Obviously, B∗ : J0(Ω) R is a bounded operator and the adjoint operator is → 3 defined by Bw = P0([w,x]), where P0 denotes the orthogonal projection of L2(Ω) onto J0(Ω). It is easily shown that rot Bw 2w. It is known [4] that if the cavity ≡3 is simply connected, then the operator B : R J0(Ω) is uniquely determined by this identity. → Let A be the Laplace operator acting in the subspace J0(Ω). The domain of this operator consists of smooth functions v J (Ω) compactly supported in Ω. Since ∈ 0 the operator R = P0∆ is nonnegative, we can define its Friedrichs extension. As before, denote it by R. This operator is called the Stokes operator. It is difficult to describe the domain of R if the boundary is not smooth, but it is easy to find the domain of its quadratic form (the domain of R1/2). Namely, it coincides with ◦ ◦ J0(Ω) H1 (Ω), where H1 (Ω) is the closure of the set of smooth functions compactly ∩ 192

supported in Ω with respect to the metric in the Sobolev space H1(Ω). It is known [10] that the Stokes operator is uniformly positive and has a discrete spectrum. In the sequel, we work with the complex space C3 instead of the real space R3. 3 Consider the operators A and A0 on C deﬁned by the following quadratic forms:

2 2 (Aw,w)= [w,x] dρ(x), (A0w,w)= [w,x] dρ(x), Ω k k Ω Ω k k Z 1 Z 1\ where ρ(x) is the distribution of mass in the shell Ω1 Ω and the distribution of mass of the fluid in Ω. The corresponding matrices are\ the inertia tensors of the body with frozen fluid and the shell, respectively. According to the definition of the frame Oe0e1e2, we have A = diag a0, a1, a2 , where the aj are the moments of inertia of the body with frozen fluid{ with respect} to the principal axes of inertia Oxj. Further, we assume that A0 > 0, that is, the shell is not weightless. Consider the operator H in C3 defined by 00 0 iHx =[x, e0] or H = 0 0 i (2)  0 i −0    The operator G = 2P H − 0 acting in the space J0(Ω) is called the gyroscopic operator. Obviously, G is a bounded self-adjoint operator. By I denote the identity operator in C3 and in J0(Ω). By v we denote the viscosity of the fluid. Throughout the following, unless otherwise specified, we assume that v =0. By k we denote the gravity force divided 6 by the cosine of the angle between f0 and e0. We have k = glm, where l is the distance from the fixed point to the center of gravity of the system, m is the mass of the system, and g is the gravitational acceleration. We can assume, without loss of generality, that the density of the fluid is equal to 1.

Equations of motion of the top with ﬂuid. The evolution of the system is described by the equations (e.g., see [4]) z˙ =[z,w]

Aw˙ + B∗v˙ =[Aw + B∗v,w]+ k [z, e0] (3) p + ν∆v =v ˙ +(v )v + 2[w,v]+[w, [w,x]]+[w,x ˙ ], div v =0 ∇ ·∇ 193

The ﬁrst equation is the kinematic relation, and z is called the Poisson vector. The second equation describes the evolution of the kinetic moment of the system with respect to the point O. The third equation is the Navier–Stokes equation of motion of the ﬂuid in the coordinate system attached to the rigid body. Applying the projector P0 to this equation and using the relation P p =0, P [w, [w,x]]=0, P (u )u = P ([v, rot v]), 0∇ 0 0 ·∇ 0 we obtain the equation

Bw˙ +v ˙ = P0([v, (2w + rot v)]) + νP0∆v. Let us linearize the ﬁrst two equations of the motion of the system and the obtained third equation at the following stationary solution of system (3):

z0 = e0, w0 = ωe0, v0 =0, where ω is a constant (the angular velocity of the non-perturbed rotation). Let us substitute the shifts z0 + z, w0 + w, v0 + v for z, w, v into (1) and retain only the linear terms. Taking into account (2), we obtain the linearized equations of motion 1 z˙ = iω Hz ω− Hw −1 Aw˙ + B∗v˙ = iω kω− Hz H (a0I A) w + HB∗v . (4) − 1 − Bw˙ +v ˙ = iω 2P H + iνω− P ∆ v − 0 0 It is more convenient to investigate an operator form of these equations.

14.2 Analysis of the Operator Equation Corresponding to the Linear Equations of Evolution of the System The operator form of the linear equations. Consider the Hilbert space H = C3 C3 J (Ω). × × 0 The elements of this space are columns u = (z,w,v)t. Here t denotes the trans- position. The ﬁrst two entries of a column lie in C3 and the last one belongs to 0(Ω). To each operator in H, a 3 3 matrix is naturally assigned. The entries × 3 of this matrix are operators acting in the spaces C and J0(Ω) and between them. Obviously, system (4) admits the operator form Wu˙ = iωMu, u H, (5) ∈ 194 where W and M are the operator matrices

1 I 0 0 H ω− H 0 − w = 0 A B∗ , M = k0 1 H (A a0I) HB∗  0 BI   0− 0− D      Here the operator D acting in J0(Ω) is deﬁned by ν D = G + R, ω where G is the gyroscopic operator and R is the Stokes operator. All the operators occurring in the preceding are bounded with the exception of the Stokes operator R. Let (M)= C3 C3 (R) D × × D be the domain of M. Since the operator M does not possess any special properties, it is diﬃcult to investigate Eq. (5). This operator is neither symmetric nor dissipative. Our idea is to “guess” an operator S such that Eq.(5) becomes more symmetric after applying the operator S to it. Namely, SM becomes dissipative and SW remains self-adjoint. Let

ω2(A + N) ωI 0 10 0 − S = ωA I 0 , where N = 0 n1 0  −ωB 0 I   0 0 n  − 2     with n1 and n2 deﬁned in (1). Proposition 14.1.

2 ω (A + N) ωA ωB∗ − − J := SW = ωA A B∗,  −ωB B I  2 − 1 ω a0 2k H ωHA + kω− H  ωHB∗ − 1 − − L := SM = ωAH + kω− H AH + HA a0H HB∗  − ωBH BH− D  −   Proof. The proof is by straightforward calculation using the formula NH = 2 a I A kω− I H. 0 − − Proposition 14.2. If N is nonsingular, then the operator S acting in H is a bounded, boundedly invertible operator. 195

Proof. The proposition follows from the relation

ω2(A + N) ωI I ωI ω2N 0 I 0 = . ωA− I 0 −I 0 I ωA I − −

Proposition 14.3. The operator W acting in H is uniformly positive.

Proof. Since W is a finite-dimensional perturbation of the identity operator in H, by virtue of the Fredholm alternative it suffices to show that W > 0. It follows from the definitions of A, A0, and B that

(Wu, u)=(z, z)+(A0w,w)+ ([w,x]+ v(x), [w,x]+ v(x))dx. ZΩ Hence W > C. Suppose that Wu = 0; then z = 0 and w = 0 (we recall that A > 0). Thus v(x) 0. 0 ≡ Proposition 14.4. Let N be nonsingular. Then J is a bounded invertible self- adjoint operator acting in H. Its negative spectrum consists of ﬁnitely many eigenvalues. The number of negative eigenvalues of the operator J coincides with that of the matrix N.

Proof. Consider the following operators in H:

ω2N 0 0 I 0 0 J0 = 0 A B∗B 0 , S0 = ωI I 0  0− 0 I   −ωB B I  −     By virtue of Proposition 3, W is a uniformly positive operator. Hence, so is the operator A B I 0 A B B 0 I B ∗ = ∗ ∗ BI BI −0 I 0 I Therefore A B∗B >> 0 and π (J0) = π (N), where π denotes the number of negative eigenvalues− of the operators.− Further,− it can be− easily checked that J = S0∗J0S0. Hence, J is invertible. Moreover, π (J0) = π (J) = π (N). This concludes the proof. − − − 196

We recall that an operator C is called dissipative if Im(Cx,x) > 0 for all x ∈ (C). If the open lower half-plane C− (or at least one point µ C−) belongs to theD resolvent set of the operator C, then C is called maximal dissipative.∈ Further, we use the notion of a dissipative operator in a Pontryagin space. Suppose that J is a bounded invertible operator and π = κ < . Then the Hilbert space H with the indeﬁnite scalar product (Jx,x) is− called a∞ Pontryagin space and is denoted by Πκ = J, H . An operator C is called dissipative in if Im(JCx,x) > 0 for all { } x (C). If there is a point µ C− (but not obligatorily the whole half-plane) ∈ D ∈ that belongs to the resolvent set of C, then C is called maximal dissipative in Πκ. It is known (see [11, Chap. 2, Theorem 2.10]) that C is maximal dissipative in Πκ if and only if JC is maximal dissipative in H. The main goal of the previous transformations is to prepare the proof of the following result.

Theorem 14.1. Suppose that N is nonsingular; then Eq.(5) of the evolution of the system is equivalent to the equation

u˙ = iωTu (6)

1 3 3 where the operator T = J − L with the domain C C (R) is maximal dissipative × ×D in the Pontryagin space Πκ = J, H , where κ = π (N). { } − Proof. It follows from the representation

1 L = V + D, D = diag I,I,D , D = G + iνω− R { } where V is a self-adjoint, ﬁnite-dimensional operator, G = G is the gyroscopic operator, and R = R >> 0 is the Stokes operator, that Im(JTx,x∗ ) = Im(Lx,x) > 0 for all x (T ). ∗ Obviously,∈ DD is maximal dissipative in H. Since V is bounded and self-adjoint, L is also maximal dissipative. Hence, since T has no nontrivial J-dissipative exten- sions, T is maximal J-dissipative in Πκ. The result also follows from the discreteness of the spectrum of T , which is proved inthe sequel. Remark. Observe the following useful fact. If

2 1 1 1 ω N − A(ωN)− (ωN)− B∗ 1 1 1 S1 = (ωN) (A + N)N N B  − − − ∗  0 0 I   197 then 2 1 1 1 ω N − A(ωN)− (ωN)− B∗ 1 1 1 J1 := WS1 = A(ωN)− A(A + N)N − (A + N)N − B∗  1 1 1  B(ωN)− B(A + N)N − BN − B∗ + I  1  0 ω− H 0 1 − L1 := MS1 = ω− H a0H 0 .  − 0− 0 D    2 1 ω− Aω− Moreover, it follows from the non-singularity of the matrix 1 that ω− A + N S1 is invertible. Hence, after the substitution u = S1f, Eq.(5) becomes ˙ 1 J1f = iωL1f, or r˙ = iωT1f, T1 = J1− L1 where J1 is self-adjoint and L1 is maximal dissipative in H. In addition, J0, J, and J1 are congruent. Namely,

J1 = S1∗JS1, where J = S0∗J0S0. (7)

Thus, J1 generates a Pontryagin metric and π (J1) = π (J) = π (J0) = π (N). The substitution makes the operator M more convenient− for− the investigation,− − since 3 3 L1 is the direct sum of two operators acting in C C and J0(Ω), respectively. This fact is used in the sequel. Also, note that the operators× L and Li are congruent: L1 = S1∗LS1.

14.3 A Stability Criterion and the Instability Index The absence of eigenvalues of the operator T in the open lower half-plane is a necessary condition for the stability of Eq. (6). The following theorem describes the spectrum of T in the closed lower half-plane.

Theorem 14.2. The spectrum of T is discrete;λ = 0 is the only real point of the spectrum. The subspace Ker T is J-positive and coincides with the linear span of t t x0 = (e0, 0, 0) and x1 = (0, e0, 0) There are no associated vectors corresponding to the zero eigenvalue. There are exactly κ = π (N) eigenvalues of T in the open − lower half-plane. In particular, if κ =0, then C− is a subset of the resolvent set of T. 198

1 Proof. Let D = diag(I,I,D). Obviously, it follows from the compactness of − in 1 J0(Ω) that D− is a compact operator in H. We have

1 1 T λI = J− I +(V λJ)D− D, V = L D, − − − where V is ﬁnite-dimensional. Since T is maximal dissipative, its resolvent set is not empty. Hence, it follows from the representation written out above and the theorem on a holomorphic operator function (see [12, Chap. 1]) that the spectrum of T is discrete. Let us show that there are no nonzero real eigenvalues of T. Since the spectrum of T is discrete, it suﬃces to show that Ker(µJ L)=0 whenever 0 = µ R. Suppose that − 6 ∈

(µJ L)u =0, u =(z,w,v)t f, 0 = µ R, − ∈ 6 ∈ Then 0 = Im((µJ L)u,u)=(Rv,v); hence v =0. Using the non-singularity of S and the equality −S(µW M)= µJ L, we get (µWM) z,w, 0 . From the deﬁnitions of W and M, we successively− obtain− { }

Bw =0, w =0, Hz =0, and (µI H)z =0. − Therefore, u = 0. Hence T has no real nonzero eigenvalues. Let us show that zero is an eigenvalue and calculate the corresponding eigenvector. Suppose that 1 Tu = 0. Using T = W− M, we get Mu = 0. Using the matrix representation t of M, we prove that the null space of M is the linear span of x0 = (e0, 0, 0) and x = (0, e , 0)t. Let x = αx + βx . Hence, if x =0, then 1 0 0 1 6 2 2 2 1 (Jx,x)= ω− α + a (a + 1) β 2+2a ω− Re(αβ) > 0, | | 0 0 | | 0 i. e., the null space of T is J-positive. Thus, there are no associated vectors corresponding to x Ker T (see [11, Chap. 2]). Let us use the fundamental∈ theorem on the existence of invariant subspaces of maximal dissipative operators in Pontryagin spaces to prove a statement on the number of eigenvalues of T in the lower halfplane. Essentially, this theorem is due to Pontryagin [13] (although it is stated there only for self-adjoint operators). In the following form (and even more generally) it was obtained by Krein and Langer [14] and Azizov [15]. Theorem on invariant subspaces. Suppose that T is a maximal dissipative operator in a Pontryagin space Πκ = J, H . Then there exists a T-invariant J- { } non-positive κ-dimensional subspace H− such that the spectrum of the restriction 199

C T H− lies in the closed lower half-plane and in the open half-plane − it coincides with| the spectrum of T. We proved that the whole real line except zero belongs to the resolvent set of T. Zero is an eigenvalue of this operator, but the corresponding subspace is J- positive. Hence, the intersection of this subspace with the J-non-positive invariant subspaceH− is zero. Therefore, the spectrum of the restriction T − lies in the open |H lower half-plane C−. But according to the previous theorem, in C− the spectrum of the restriction coincides with the spectrum of T. Thus, the number of eigenvalues of T in C− is equal to the number κ = dim H− = π (N). − The dimension of the quotient space of the solution space for Eq. (6) by the linear space of bounded solutions is called the instability index (i.e., the number of linearly independent unbounded solutions modulo bounded ones). Let ν(T) be the instability index of Eq. (6). Obviously,ν(T) is greater than or equal to the number of eigenvalues of T in C−. Hence, using Theorem 2, we have ν(T) π (N). In the general situation, the instability index can be greater than the number≥ of− eigenvalues of the operator in the half-plane C− (e.g., see [16]). However, in our case equality takes place.

Theorem 14.3. Suppose that n1n2 > 0; then the instability index ν(T) of Eq. (6) is equal to π (N). In particular, the problem is stable if and only if the numbers n1 − and n2 are positive. Proof. If N > 0, then J is uniformly positive. Hence, the metric (Jx,x) in H is equivalent to the original one. The operator T is maximal dissipative with respect to the metric (Jx,x). Hence, it generates a contraction semigroup (see [17, Chap. 9]). Therefore, every solution of Eq. (6) (understood as an equality in H) satisﬁes the condition J1/2u(t) J1/2u(0) whenever t> 0. Hence, the problem is stable. Supposek that π k≤k(N) > 0. Byk virtue of the inequality ν(T) π (N), the problem is unstable. Let− us calculate the index. Let H be the one-≥ or two-dimensional− − linear span of the eigenvectors corresponding to the eigenvalues from C−. Let us show that H is a J-negative subspace. Suppose− that x H and u(t) is an elementary solution of Eq. (6) such that ∈ − iωλ0t u(0) = x (if π (n)=1, then u(t)= e x, where λ0 C−). Using (6), we get − ∈ d (Ju, u)=(Ju˙ , u)+(Ju, u˙ )=(iωJTu, u)+(u,iωJTu)= dt = 2ω Im(JTu, u)= 2ν(Rv,v) − − 200

It is clear that u(t) 0 as t . Integrating the last equality, we get → → −∞ 0 (Jx, x)= 2ν (Rv(t),v(t)) dt. − Z−∞ Since R> 0, we have (Jx,x) 0. Suppose that (Jx,x)=0; then v(t) 0. Using the matrix representation of Eq.≤ (5), we easily get w˙ (t) 0 and z˙(t) ≡0. Hence, the vector x =(z,w, 0)t belongs to the null space of T, i.e.,≡ corresponds≡ to the zero eigenvalue. This implies x =0. 1 1 Let us consider the operator J− T∗J = J− L∗. It follows from the matrix 1 1 representation of L that (J− T∗J) = (T). Obviously, the operator J− T∗J D D 1 is maximal J-dissipative. By H∗ we denote the invariant subspace of J− T∗J, that is, the linear span of the eigenvectors− corresponding to the eigenvalues from the upper half-plane. Arguing as above, we see that H∗ is J-negative. Since H∗ is finite-dimensional, it is uniformly J negative. By− virtue of the Pontryagin− − theorem ([13, Theorem 1]), the J-orthogonal complement H+ of H∗ is uniformly J- − positive. Obviously, the space H+ is T-invariant and the spectrum of the restriction T+ = T lies in the closed upper half-plane. Hence, H is the direct sum of two |H+ T-invariant uniformly J-definite subspaces H and H+. Hence, it is clear that Eq. − (6) in H+ is stable. Thus, the instability index ν(T) is equal to dim H = π(N). − In addition to Theorem 3, note that the instability of Eq. (6) implies the instability of the top, i.e., the rigid part of the system. Namely, the following result is valid. Theorem 14.4. Suppose that u(t)=(z(t),w(t),v(t))t is an unbounded solution of Eq. (6); then z(t) is also unbounded. Proof. It follows from the proof of Theorem 3 that H is the direct sum of two T-invariant uniformly J-definite subspaces: H = H++˙ H+. In particular, any unbounded solution u(t) of Eq. (6) has the form

u(t)= u+(t)+ u (t), where (Ju (t), u (t)) as t + . − − − → −∞ → ∞ Let t 0 t u (t)=(z(t),w(t),v(t) , u (t) := S0u (t)=(z(t),w0(t),v0(t)) . − − − Hence, using (7), we get

0 0 (Ju(t), u(t))=(J0u (t), u (t)) = − 2 − = ω (Nz(t), z(t))+((A B∗B)w (t),w (t))+(v (t),v (t)). − 0 0 0 0 201

Since the last two summands on the right side are positive and the sum tends to as t + , we see that z(t) is an unbounded function. −∞ → ∞ 1 Remark. The upper bound for the number of eigenvalues of T1 = J1− L1 with negative imaginary part was obtained in [9]. It can be shown that the statements of Theorems 2 and 3 concerning the number of eigenvalues in C− and the instability index are consequences of the results of [19], [20]. But we prefer to give an independent proof here.

14.4 Further Properties of the Evolution Operator T of the System. The Spectrum at Large Viscosity The numerical range of T and properties of its eigenfunctions. The range of the quadratic form (Fx,x) for x (F ) and x = 1 is called the numerical range of an operator F . It is clear that∈ D the spectrumk k of an operator is a subset of its numerical range. Let us recall the concept of a basis for the Abel summability method of order a, which is due to Lidskii [20]. Suppose that F has a discrete spectrum and all of its eigenvalues except for ﬁnitely many lie in the sector arg λ π/2 <θ<π. To be concise, suppose that the eigenvalues are simple. We say| that− the| system y of eigenfunctions of F corresponding to the eigenvalues { k} λk is a basis for the Abel summability method of order α, α < 2θ/π, if for any {f }H the series α ∈ ( iλk) t f(t)= e − (f, zk)yk strongly converges in H for all t >X0 (it is permitted to put groups of terms in brackets independently of f and t) and f(t) strongly tends to f as t 0. Here z → { k} is the set of eigenfunctions of the adjoint operator. It is biorthogonal to yk , and α α { } ( iλk) is the main branch of λ . Obviously, every basis for the Abel summability method− is a complete system in H. As before, by T we denote the restriction of the generator T of Eq. (6) to (see Theorem 2). ±

Theorem 14.5. The numerical range of T lies in the half-strip Re λ < c , + | | 0 Im λ > c1 for some constants c0, c1> 0. The eigenvalues λk = λk(T) satisfy the condition− 3π2k 2/3 ν λ = i (1 + o(1)), k =1, 2, 3,... (8) k mes(Ω) ω 202 and lie in a half-strip of the same form. The spectrum is symmetric about the imaginary cuds. The set of all eigenvectors of T is a basis for the Abel summability method of order a whenever α > 1/2. In particular, if α = 1 and t > 0, then any solution of Eq. (6) is represented by a convergent series in the eigenfunctions of T. The operator iT generates a holomorphic semigroup in H.

Proof. Suppose that Q+ is the orthogonal projection onto H+ and J+ = Q+JQ+. Since H+ is uniformly J-positive, it follows that J+ is uniformly positive in H+. If x H , then ∈ + 1 1 1 1 (Tx, x)=(J+Tx, J+− x)=(JTx, J+− x)=(Lx, J+− x)=(Q+LQ+x, J+− x). (9) Note that if C is uniformly positive, then the real form (Fx,x) is bounded (semibounded) if and only if so is the form (Fx,Cx). It follows from the matrix representation of L that the real part of (Lx, x) is bounded and the imaginary part is semibounded. Hence, so is form (9). Therefore, the numerical range of T+ lies in a half-strip. The following estimate for the resolvent holds outside the numerical range W(T+): 1 (λ T )− l/d(λ, W (T )), (10) k − + k≤ + where d is the distance from λ to W (T+. In particular, if λ asymptotically lies outside a small sector containing the imaginary axis, then the resolvent is of maximal 1 decay ( c λ − ). It follows from the well- known results of semigroup theory (see ≤ | | [17, Chap. 9]) that iT+ is a generator of a holomorphic semigroup in H+. Since T = T+ + T , so is iT. Here T is ﬁnite dimensional. Let R = diag(− I,I,R), where−R is the Stokes operator. We have

1 1 1 J = I + K, L =(iνω− I + V)R, V =(L iνω− R)R− , − 1 where K is finite-dimensional and V is compact. Hence, T = iνω− (I + V1)R, where V1 is a compact operator. The eigenvalues of R and R coincide, except for one, λ = 1. The asymptotics for the eigenvalues of R is known [5]. It is given in (8) up to the coefficient ν/ω. The Keldysh–Gokhberg–Krein theorem (see [12, Chap. 5]) implies a similar asymptotics for the compact perturbation T of R. Since the matrices A, N, iH are real, it follows that the spectrum of the pencil λJ L or of T is symmetric about the imaginary axis. − Now it suffices to prove that the set of eigenvectors of the operator T+ acting in H+ is a basis for the Abel summability method. We proved that the estimate 203

(10) for the resolvent holds outside a half-strip. Obviously, the s-numbers of the 1 1 p operators T− and R− satisfy the condition sk(T) k , p = 2/3 (see [?, Chap. 2]). Thus, it readily follows from the result of [23] that∼ the system is a basis for the 1 Abel summability method of order a>p− =1/2. The spectrum of T for large viscosity. It is natural that the spectrum of the evolution operator approaches the spectrum of the small oscillations of the body with frozen ﬂuid as ν . The equation of motion→∞ for the small oscillations of the body with frozen ﬂuid is the projection of Eq. (5) onto the subspace C3 C3 0 of H. Hence, the spectrum of the corresponding problem coincides with that× of× the matrix pencil

1 I 0 H ω− H (s) (s) V (λ)= λ 1 =: λW M 0 A − kω− H H (A a0I) − − where W (s) and M (s) are the projections of W and M onto the “rigid” part C3 C3 of H. Obviously, zero is a double eigenvalue of the pencil V (λ). Consider× the operators S1, J1, L1, deﬁned in Remark 1. Note that the eigenvalues of the pencils V(λ) = λW M and V1(λ) = λJ1 L1 coincide. The eigenvalues of their projections V(−λ) and −

(s) V1(λ)= V (λ)S1 = 2 1 1 1 ω N − A(ωN)− 0 ω− H (s) (s) = λ 1 1 1 − =: λJ1 L1 A(ωN)− A(A + N)N − − ω− H a0H − − − onto C3 C3 also coincide. × Theorem 14.6. Suppose that n1n2 =0 and ν . Then zero is a double eigenvalue of T; four eigenvalues approach6 the nonzero→∞ eigenvalues of the pencil V (λ), which corresponds to the rigid part of the system. The degree of approximation is 1/2 1/4 estimated by O(ν− ) if the four eigenvalues are simple and by O(ν− ) if they are double. The spectrum of V (λ) and V1(λ) is symmetric about the real and imaginary axes. All other eigenvalues approach inﬁnity and satisfy

1 (f) λ = νω− λ (l + o(l)) asν , (11) k k →∞ A B where the λ(f) are the nonzero eigenvalues of the pencil λ ∗ k BI − 0 0 i in C3 J (Ω). 0 R × 0 204

Proof. Since V1(λ) is self-adjoint, it follows that the spectrum of V1(λ) and V (λ) is symmetric about the real axis. Since the matrices A, N, iH are real, it follows that the spectrum of V1(λ) is symmetric about the imaginary axis. 1/2 Let µ = ω/ν and ρ = µ . Multiplying the pencil λJ1 L1 by the operator 1/2 1/2 − diag(I,I,µ R− ) on the left and on the right, we see that the eigenvalues of T coincide with those of the pencil

(s) (s) S (λ)= λ J1 ρC1∗ L1 0 (12) ρ ρC ρ2C − 0 ρ2G + iI 1 where C and C1 are some unbounded operators. Using the Rellich–Kato theorem (see [17, Chap. 7]), we see that if ρ is small (ν is large), then the eigenvalues of the pencil Sρ(λ) lie in neighborhoods of the eigenvalues of S0(λ), i.e., in neighborhoods of zero, inﬁnity, and four nonzero eigenvalues of V1(A). In the general case, the nonzero eigenvalues are simple and their perturbations depend analytically on ρ. Hence, the degree of approximation is proportional to ρ = µ1/2. If the non-perturbed eigenvalues are double, then they are represented by the Puiseux series in powers of ρ1/2. Therefore, the degree of approximation is estimated by O(µ1/4). It follows from the symmetry of the spectrum that there are no triple or quadruple eigenvalues. Note that ρ(λW M)= ρλW i diag0, 0, R + K(ρ), − − where K(ρ)= O(ρ). Using the Rellich-Kato theorem again, we get (11). The real eigenvalues of V(λ) are of particular interest. How do the eigenvalues of T approach them as ν ? The following theorem gives the answer. ∞ Theorem 14.7. Suppose that λ0 is a simple real eigenvalue of the pencil V(λ) and t h0 = (z0,w0) is the corresponding eigenvector. Then the eigenvalue λ of T close to λ0 satisﬁes

2 1 ωλ B∗R− Bw0,w0 λ = λ + ciµ + O µ3/2 , c = 0 (13) 0 (s) J h0, h0 Here J (s) is the restriction of J to C3 C3. Hence, λ is in the upper or lower (s) × half-plane when (J h0, h0) is positive or negative, respectively.

Proof. It follows from the representation (12) for Sρ(λ) and the Rellich-Kato theorem that the eigenvalue λ and the corresponding eigenvector fdepend analytically 205 on ρ. Namely,

2 2 λ = λ0 + c1ρ + c2ρ + ..., f = f0 + ρf1 + ρ f2 + ... Let f = (g ,r )t, j = 1, 2, 3, where the g are the projections of f , onto C3 C3. j j j j j × Let us write out lower-order terms in the equation Sρ(λ)f = 0 with respect to the powers of ρ. Writing out the ﬁrst two terms, we get

λ J (s)g L(s)g =0, r =0 0 1 0 − 1 0 0 λ C g ir =0, c J (s)g + λ J (s)g L(s)g C .r =0 0 1 0 − 1 1 1 0 0 1 1 − 1 1 − 1 0 (s) Hence c1(J1 g0,g0)=0. It is known that any eigenvector of the pencil V1(λ) (s)1 corresponding to a simple real eigenvalue is J -deﬁnite. Hence c1 = 0. Further, the coeﬃcient of ρ2 is equal to

(s) (s) (s) c J g + λ C∗r + λ J g L g =0. 2 1 0 0 1 1 0 1 2 − 1 2 Hence, (s) 2 c (J g ,g ) iλ (C∗C g ,g )=0. (14) 2 1 0 0 − 0 1 1 0 0 t (s) Obviously, the eigenvector h0 = (z0,w0) of V (λ) is equal to S1 g0. From the (s) (s) (s) representation of C1 in (12), using the equalities WS1 = J1 and W S1 = J1 , we get 1/2 (s) 1/2 C1g0 = R− W h0 = R− Bw0. (s) (s) (s) (s) Now it follows from the formula J1 = (S1 )∗J S1 (cf. (7)) and Eq. (14) that c2 = ic in (13).

(s) h Remark. The sign of J1 g0,g0 = (J0 , h0) determines the type of eigenvalues λ0 of the self- adjoint pencil V1(λ). It follows from the symmetry that the eigenvalues λ0 and λ0 are of the same type. If there are exactly two real eigenvalues, then they are− of positive type, because otherwise there would be three eigenvalues of T in the lower half-plane for large ν, which is impossible. Suppose that there are four real eigenvalues, i.e., n1n2 > 0; then these eigenvalues are of positive type if n1 and n2 are positive. Otherwise, one symmetric pair has the negative type. In particular, the conclusion of Theorem 2 concerning the number of eigenvalues of T in the lower half-plane readily follows from Theorem 7 and from the fact that there are no nonzero real eigenvalues. 1 Remark. The operator B∗R− B is evaluated in [3] for some forms of the cavities. Formula (13) is obtained in [3] and [23] for k =0, i.e., when there is no gravity. 206

We have supposed that n1n2 = 0. However, it is possible to investigate the behavior of the eigenvalues of T without6 this assumption. Also, it is interesting to observe the transition of the eigenvalues from the upper half-plane to the lower half-plane as ω passes through a critical value.

Proposition 14.5. Take ω such that n1 = 0 and n2 = 0. Then zero is a triple semisimple eigenvalue of the operator T = T(w). If e6 is suﬃciently small, then T(ω + ε) has a simple eigenvalue λ(ε) satisfying

2kε 2 λ(ε)= 2 1 + O ε as ε 0. ω (D− Be1, Be1) →± 1 Here e1 is the unit vector deﬁned in §1. The number (D− Be1, Be1) is pure imaginary.

Proof. Arguing as in the proof of Theorem 2, we see that zero is a triple semi-simple eigenvalue. It follows from (7) that the eigenvalues of T and those of the pencil 1 1 λJ L coincide. Here L = (S∗)− LS− . It is easy to write out the expansions 0 − 0 0 0 0 of J0 and L0 in powers of ε. Using the ﬁrst two terms of the expansions, it is easy to show that (GB1, Be1)=0. Thus, λ(ε) passes through zero along the imaginary axis (here we use the symmetry). Arguing as in Theorem 7, we can conclude the proof. Therefore, we omit the details.

14.5 The Symmetric Top Invariant subspaces. If symmetry occurs, then it is possible to single out a family of T-invariant subspaces and to test the stability in them. Let Ox0 be the axis of a pth-order symmetry. In other words, the top is invariant with respect to the rotation by the angle of 2π/p about the axis Ox0. Further, assume that p> 2. If the top is a solid of revolution, then we set p = . Let us introduce the operators ∞

10 0 Us 0 0 Us = 0 cos2πs/p sin2πs/p , Us = 0 Us 0  0 sin2πs/p cos2πs/p   0 0 V  − s     where Vs, is deﬁned by Vsv = Us(v(U sx)). If p = , then we substitute 2π for p in these equalities. − ∞ 207

Proposition 14.6. If Ox0 is the axis of a pth-order symmetry for the top, then

AUs = UsA, HUs = UsH, B∗Vs = VsB∗, VsD = DVs (15) for all s Z (if p = , then s R). The operator T commutes with Us . The operators∈ ∞ ∈

p 1 − 2π 1 2πilj/p 1 ijx Qj = e Ul Qj = e Uxdx if p = p 2π 0 ∞ Xl=0 Z p 1 are orthoprojections in H, and j=0− Qj = I (if p = , then j∞= Qj = I). The ∞ −∞ subspaces Hj = Qj(H) are T-invariant. P P Proof. Equations (15) follow from the deﬁnition (see [8, Lemma 1]). The other conclusions are direct consequences of these equations. Note that if p = , then ∞ we use the notion of the Pettis integral in the deﬁnition of Qj.

It is readily seen that Qj = Qj p. In the sequel, it is convenient to consider − Qp 1 instead of Q1. − Theorem 14.8. If there is a pth-order symmetry (2

S (H ) Ce Ce J (Ω) =: H′ 0 0 ⊂ 0 × 0 × 0 0 208

1 and J0 is positive in H0′ . Hence, J is positive in H0. Therefore, T = J− L is maximal dissipative with respect to the inner product (Jx, x) in H0, which is equivalent to the original one. Thus, Eq. (6) is stable in H0. Since Tu = Tu (the bar denotes complex conjugation) and H 1 is the complex − − conjugate of H 1, it follows that the instability index of (6) in H1 is equal to that − in H 1. If n1 < 0, then the instability index in H1 is neither zero nor greater than one (otherwise,− we would have a contradiction with Theorem 3). Hence, it is equal to 1.

Connection with Sobolev’s equation. In [1], Sobolev investigated the case of a symmetric top and ideal ﬂuid (ν =0). He introduced the operator F in C C J (Ω) deﬁned by × × 0 F(z,w,v)= z1,w1,v1 .

1 1 1 1 1 1 Here the numbers z , w and the function v = (v0,v1,v2) J0(Ω) satisfy the equations ∈

1 ∂p 1 ∂p ∂χ¯ v0 = i ,v2 = 2ωiv1 + i +2ωw ∂x0 − ∂x2 ∂x1 1 ∂p ∂χ¯ 1 1 v1 =2ωiv2 + i 2ωw ,z = ωw, div v = div v =0 ∂x1 − ∂x2 ∂χ ∂χ v1 ∂Ω=0,A w1 = A w + A z + v v dΩ n| 1 2 3 2 ∂x − 1 ∂x ZΩ 1 2 The function χ is deﬁned by

∂χ ∆χ =0, = x (cos nx + i cos nx ) (x + ix ) cos nx . ∂n 0 2 1 − 2 1 0 ∂Ω

It can be readily shown that χ satisfies the equality ∂χ ∂χ t 1 P , , 0 =(BH GB)e, e = (0, 1,i)t (16) o ∂x −∂x − √ 2 1 2 1 and F is the restriction of T0 = S0TS0− to H (see [8, 9]). Hence, if ν =0, then we can consider T0 instead of Sobolev’s operator F. We do not know the definition of Sobolev’s operator if the fluid is viscous. The obstacle is that we cannot substitute 209

D for G in (16) because the vector Be does not belong to the domain of the Stokes operator R. Indeed, suppose that Be (R); then Be =[e, x]+ p. Hence, ⊂ D ∇ RBe = ∆([e, x])+ ∆p + q = (∆p + q). ∇ ∇ ∇

Here p and q are smooth functions. Thus, P0RBe = RBe = 0. This contradicts the condition >> 0. The content of this section is base on the paper by A. G. Kostyuchenko, A. A. Shkalikov and M. Yu. Yurkin,

Bibliography for Section 14

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