inverse sturm-liouville problems and their applications

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INVERSE STURM-LIOUVILLE PROBLEMS AND THEIRAPPLICATIONS

G. Freiling and V. Yurko

PREFACE

This book presents the main results and methods on inverse spectral problems for Sturm-Liouville differential operators and their applications. Inverse problems of spectral analysisconsist in recovering operators from their spectral characteristics. Such problems often ap-pear in mathematics, mechanics, physics, electronics, geophysics, meteorology and otherbranches of natural sciences. Inverse problems also play an important role in solving nonlin-ear evolution equations in mathematical physics. Interest in this subject has been increasing

permanently because of the appearance of new important applications, and nowadays theinverse problem theory develops intensively all over the world.

The greatest success in spectral theory in general, and in particular in inverse spectralproblems has been achieved for the Sturm-Liouville operator

y := −y + q (x)y, (1)

which also is called the one-dimensional Schrodinger operator. The first studies on thespectral theory of such operators were performed by D. Bernoulli, J. d‘Alembert, L. Euler,J. Liouville and Ch. Sturm in connection with the solution of the equation describing thevibration of a string. An intensive development of the spectral theory for various classesof differential and integral operators and for operators in abstract spaces took place in theXX-th century. Deep ideas here are due to G. Birkhoff, D. Hilbert, J. von Neumann, V.Steklov, M. Stone, H. Weyl and many other mathematicians. The main results on inversespectral problems appear in the second half of the XX-th century. We mention here theworks by R. Beals, G. Borg, L.D. Faddeev, M.G. Gasymov, I.M. Gelfand, B.M. Levitan, I.G.Khachatryan, M.G. Krein, N. Levinson, Z.L. Leibenson, V.A. Marchenko, L.A. Sakhnovich,

E. Trubowitz, V.A. Yurko and others (see Section 1.9 for details). An important role in theinverse spectral theory for the Sturm-Liouville operator was played by the transformationoperator method (see [mar1], [lev2] and the references therein). But this method turned outto be unsuitable for many important classes of inverse problems being more complicated thanthe Sturm-Liouville operator. At present time other effective methods for solving inversespectral problems have been created. Among them we point out the method of spectralmappings connected with ideas of the contour integral method. This method seems to beperspective for inverse spectral problems. The created methods allowed to solve a number

of important problems in various branches of natural sciences.



2

In recent years there appeared new areas for applications of inverse spectral problems. Wemention a remarkable method for solving some nonlinear evolution equations of mathemat-ical physics connected with the use of inverse spectral problems (see [abl1], [abl3], [zak1]and Sections 4.1-4.2 for details). Another important class of inverse problems, which of-ten appear in applications, is the inverse problem of recovering differential equations from

incomplete spectral information when only a part of the spectral information is availablefor measurement. Many applications are connected with inverse problems for differentialequations having singularities and turning points, for higher-order differential operators, fordifferential operators with delay or other types of ”aftereffect”.

The main goals of this book are as follows:

• To present a fairly elementary and complete introduction to the inverse problem theoryfor ordinary differential equations which is suitable for the ”first reading” and accessible

not only to mathematicians but also to physicists, engineers and students. Note thatthe book requires knowledge only of classical analysis and the theory of ordinary lineardifferential equations.

• To describe the main ideas and methods in the inverse problem theory using the Sturm-Liouville operator as a model. Up to now many of these ideas, in particular those whichappeared in recent years, are presented in journals only. It is very important that themethods, provided in this book, can be used (and have been already used) not only for

Sturm-Liouville operators, but also for solving inverse problems for other more compli-cated classes of operators such as differential operators of arbitrary orders, differentialoperators with singularities and turning points, pencils of operators, integro-differentialand integral operators and others.

• To reflect various applications of the inverse spectral problems in natural sciences andengineering.

The book is organized as follows. In Chapter 1, Sturm-Liouville operators (1) on a finite

interval are considered. In Sections 1.1-1.3 we study properties of spectral characteristics andeigenfunctions, and prove a completeness theorem and an expansion theorem. Sections 1.4-1.8 are devoted to the inverse problem theory. We prove uniqueness theorems, give algorithmsfor the solution of the inverse problems considered, provide necessary and sufficient conditionsfor their solvability, and study stability of the solutions. We present several methods forsolving inverse problems. The transformation operator method, in which the inverse problemis reduced to the solution of a linear integral equation, is described in Section 1.5. In Section1.6 we present the method of spectral mappings, in which ideas of the contour integral methodare used. The central role there is played by the so-called main equation of the inverseproblem which is a linear equation in a corresponding Banach space. We give a derivation of the main equation, prove its unique solvability and provide explicit formulae for the solutionof the inverse problem. At present time the contour integral method seems to be the mostuniversal tool in the inverse problem theory for ordinary differential operators. It has awide area for applications in various classes of inverse problems (see, for example, [bea1],[yur1] and the references therein). In the method of standard models, which is described inSection 1.7, a sequence of model differential operators approximating the unknown operatorare constructed. In Section 1.8 we provide the method for the local solution of the inverse



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problem from two spectra which is due to G. Borg [Bor1]. In this method, the inverseproblem is reduced to a special nonlinear integral equation, which can be solved locally.

Chapter 2 is devoted to Sturm-Liouville operators on the half-line. First we consider non-selfadjoint operators with integrable potentials. Using the contour integral method we study

the inverse problem of recovering the Sturm-Liouville operator from its Weyl function. Thenlocally integrable complex-valued potentials are studied, and the inverse problem is solvedby specifying the generalized Weyl function. In Chapter 3 Sturm-Liouville operators on theline are considered, and the inverse scattering problem is studied. In Chapter 4 we providea number of applications of the inverse problem theory in natural sciences and engineering:we consider the solution of the Korteweg-de Vries equation on the line and on the half-line,solve the inverse problem of constructing parameters of a medium from incomplete spec-tral information, and study boundary value problems with aftereffect, inverse problems inelasticity theory and others.

There exists an extensive literature devoted to inverse spectral problems. Instead of tryingto give a complete list of all relevant contributions, we mention only monographs, surveyarticles and the most important papers, and refer to the references therein. In Section 1.9we give a short review on literature on the inverse problem theory.



4

CONTENTS

1. Sturm-Liouville operators on a finite interval 5

1.1. Behavior of the spectrum 5

1.2. Properties of eigenfunctions 141.3. Transformation operators 191.4. Uniqueness theorems 231.5. Gelfand-Levitan method 241.6. The method of spectral mappings 491.7. The method of standard models 741.8. Local solution of the inverse problem 771.9. Review of the inverse problem theory 97

2. Sturm-Liouville operators on the half-line 102

2.1. Properties of the spectrum. The Weyl function 1022.2. Recovery of the differential equation from the Weyl function 1202.3. Recovery of the differential equation from the spectral data 1312.4. An inverse problem for a wave equation 1482.5. The generalized Weyl function 1592.6. The Weyl sequence 163

3. Inverse scattering on the line 168

3.1. Scattering data 1683.2. The main equation 1833.3. The inverse scattering problem 1883.4. Reflectionless potentials. Modification of the discrete spectrum 201

4. Applications of the inverse problem theory 204

4.1. Solution of the Korteweg-de Vries equation on the line 2044.2. Korteweg-de Vries equation on the half-line. Nonlinear reflection 2084.3. Constructing parameters of a medium from incomplete spectral information 2164.4. Discontinuous inverse problems 2364.5. Inverse problems in elasticity theory 2524.6. Boundary value problems with aftereffect 257

4.7. Differential operators of the Orr-Sommerfeld type 2684.8. Differential equations with turning points 273

References 288



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I. STURM-LIOUVILLE OPERATORS ON A FINITEINTERVAL

In this chapter we present an introduction to the spectral theory for Sturm-Liouville

operators on a finite interval. Sections 1.1-1.2 are devoted to the so-called direct problemsof spectral analysis. In Section 1.1 we introduce and study the main spectral characteristicsfor Sturm-Liouville boundary value problems. In particular, the theorem on the existenceand the asymptotic behavior of the eigenvalues and eigenfunctions is proved. In Section 1.2properties of the eigenfunctions are investigated. It is proved that the system of the eigen-functions is complete and forms an orthogonal basis in L2(0, π). An expansion theorem inthe uniform norm is provided. We also investigate oscillation properties of the eigenfunc-tions, and prove that the n -th eigenfunction has exactly n zeros inside the interval (0, π) .

In Section 1.3 we study the so-called transformation operators which give us an effective toolin the Sturm-Liouville spectral theory.

Sections 1.4-1.8 are devoted to constructing the inverse problem theory for Sturm-Liouvilleoperators on a finite interval. In Section 1.4 we give various formulations of the inverseproblems and prove the corresponding uniqueness theorems. In Sections 1.5-1.8 we presentseveral methods for constructing the solution of the inverse problems considered. The vari-ety of ideas in these methods allows one to use them for many classes of inverse problemsand their applications. Section 1.5 deals with the Gelfand-Levitan method [gel1], in whichthe transformation operators are used. The method of spectral mappings, considered inSection 1.6, leans on ideas of the contour integral method. By these two methods we obtainalgorithms for the solution of the inverse problems, and provide necessary and sufficient con-ditions for their solvability. In Section 1.7 we describe the method of standard models whichallows one to obtain effective algorithms for the solution of a wide class of inverse problems.The method for the local solution of the inverse problem, which is due to Borg [Bor1], isprovided in Section 1.8. A short historical review on the inverse problems of spectral analysisfor ordinary differential operators is given in Section 1.9.

1.1. BEHAVIOR OF THE SPECTRUM

Let us consider the boundary value problem L = L(q (x), h , H ) :

y := −y + q (x)y = λy, 0 < x < π, (1.1.1)

U (y) := y(0)−

hy(0) = 0, V (y) := y(π) + Hy(π) = 0. (1.1.2)

Here λ is the spectral parameter; q (x), h and H are real; q (x) ∈ L2(0, π). We shallsubsequently refer to q as the potential. The operator is called the Sturm-Liouville operator.

We are interested in non-trivial solutions of the boundary value problem (1.1.1)-(1.1.2).

Definition 1.1.1. The values of the parameter λ for which L has nonzero solutionsare called eigenvalues, and the corresponding nontrivial solutions are called eigenfunctions.The set of eigenvalues is called the spectrum of L.



6

In this section we obtain the simplest spectral properties of L, and study the asymptoticbehavior of eigenvalues and eigenfunctions. We note that similarly one can also study othertypes of separated boundary conditions, namely:

(i) U (y) = 0, y(π) = 0;

(ii) y(0) = 0, V (y) = 0;

(iii) y(0) = y(π) = 0.

Let C (x, λ), S (x, λ), ϕ(x, λ), ψ(x, λ) be solutions of (1.1.1) under the initial conditions

C (0, λ) = 1, C (0, λ) = 0, S (0, λ) = 0, S (0, λ) = 1,

ϕ(0, λ) = 1, ϕ(0, λ) = h, ψ(π, λ) = 1, ψ(π, λ) =

−H.

For each fixed x, the functions ϕ(x, λ), ψ(x, λ), C (x, λ), S (x, λ) are entire in λ. Clearly,

U (ϕ) := ϕ(0, λ) − hϕ(0, λ) = 0, V (ψ) := ψ(π, λ) + Hψ(π, λ) = 0. (1.1.3)

Denote∆(λ) = ψ(x, λ), ϕ(x, λ), (1.1.4)

where

y(x), z (x)

:= y(x)z (x)

−y(x)z (x)

is the Wronskian of y and z. By virtue of Liouville‘s formula for the Wronskian [cod1, p.83],ψ(x, λ), ϕ(x, λ) does not depend on x. The function ∆(λ) is called the characteristic

function of L. Substituting x = 0 and x = π into (1.1.4), we get

∆(λ) = V (ϕ) = −U (ψ). (1.1.5)

The function ∆(λ) is entire in λ, and it has an at most countable set of zeros λn.

Theorem 1.1.1. The zeros

λn

of the characteristic function coincide with the eigen-

values of the boundary value problem L. The functions ϕ(x, λn) and ψ(x, λn) are eigenfunctions, and there exists a sequence β n such that

ψ(x, λn) = β nϕ(x, λn), β n = 0. (1.1.6)

Proof. 1) Let λ0 be a zero of ∆(λ). Then, by virtue of (1.1.3)-(1.1.5), ψ(x, λ0) =β 0ϕ(x, λ0), and the functions ψ(x, λ0), ϕ(x, λ0) satisfy the boundary conditions (1.1.2).Hence, λ0 is an eigenvalue, and ψ(x, λ0), ϕ(x, λ0) are eigenfunctions related to λ0.

2) Let λ0 be an eigenvalue of L, and let y0 be a corresponding eigenfunction. ThenU (y0) = V (y0) = 0. Clearly y0(0) = 0 (if y0(0) = 0 then y0(0) = 0 and, by virtue of the uniqueness theorem for the equation (1.1.1) (see [cod1, Ch.1]), y0(x) ≡ 0 ). Withoutloss of generality we put y0(0) = 1. Then y0(0) = h, and consequently y0(x) ≡ ϕ(x, λ0).Therefore, (1.1.5) yields ∆(λ0) = V (ϕ(x, λ0)) = V (y0(x)) = 0. We have also proved thatfor each eigenvalue there exists only one (up to a multiplicative constant) eigenfunction. 2

Throughout the whole chapter we use the notation

αn := π

0 ϕ

2

(x, λn) dx. (1.1.7)



7

The numbers αn are called the weight numbers, and the numbers λn, αn are called the spectral data of L.

Lemma 1.1.1. The following relation holds

β nαn =−

∆(λn), (1.1.8)

where the numbers β n are defined by (1.1.6), and ∆(λ) =d

dλ∆(λ).

Proof. Since

−ψ(x, λ) + q (x)ψ(x, λ) = λψ(x, λ), −ϕ(x, λn) + q (x)ϕ(x, λn) = λnϕ(x, λn),

we get

ddx

ψ(x, λ), ϕ(x, λn) = ψ(x, λ)ϕ(x, λn) − ψ(x, λ)ϕ(x, λn) = (λ − λn)ψ(x, λ)ϕ(x, λn),

and hence with the help of (1.1.5),

(λ − λn) π

0ψ(x, λ)ϕ(x, λn) dx = ψ(x, λ), ϕ(x, λn)

π0

= ϕ(π, λn) + Hϕ(π, λn) + ψ(0, λ) − hψ(0, λ) = −∆(λ).

For λ → λn, this yields π

0ψ(x, λn)ϕ(x, λn) dx = −∆(λn).

Using (1.1.6) and (1.1.7) we arrive at (1.1.8). 2

Theorem 1.1.2. The eigenvalues λn and the eigenfunctions ϕ(x, λn), ψ(x, λn) are real. All zeros of ∆(λ) are simple, i.e. ∆(λn) = 0. Eigenfunctions related to different

eigenvalues are orthogonal in L2(0, π).Proof. Let λn and λk ( λn = λk ) be eigenvalues with eigenfunctions yn(x) and yk(x)

respectively. Then integration by parts yields π

0yn(x) yk(x) dx =

π

0yn(x)yk(x) dx,

and henceλn

π

0yn(x)yk(x) dx = λk

π

0yn(x)yk(x) dx,

or π

0yn(x)yk(x) dx = 0.

Further, let λ0 = u+iv,v = 0 be a non-real eigenvalue with an eigenfunction y0(x) = 0.Since q (x), h and H are real, we get that λ0 = u − iv is also the eigenvalue with theeigenfunction y0(x). Since λ0 = λ0, we derive as before

y02L2

=

π

0y0(x)y0(x) dx = 0,



8

which is impossible. Thus, all eigenvalues λn of L are real, and consequently the eigen-functions ϕ(x, λn) and ψ(x, λn) are real too. Since αn = 0, β n = 0, we get by virtue of (1.1.8) that ∆(λn) = 0. 2

Example 1.1.1. Let q (x) = 0, h = 0 and H = 0, and let λ = ρ2. Then one can

check easily that

C (x, λ) = ϕ(x, λ) = cos ρx, S (x, λ) =sin ρx

ρ, ψ(x, λ) = cos ρ(π − x),

∆(λ) = −ρ sin ρπ, λn = n2 (n ≥ 0), ϕ(x, λn) = cos nx, β n = (−1)n.

Lemma 1.1.2. For |ρ| → ∞, the following asymptotic formulae hold

ϕ(x, λ) = cos ρx + O 1

|ρ| exp(|τ |x) = O(exp(|τ |x)),

ϕ(x, λ) = −ρ sin ρx + O(exp(|τ |x)) = O(|ρ| exp(|τ |x)),

(1.1.9)

ψ(x, λ) = cos ρ(π − x) + O 1

|ρ| exp(|τ |(π − x))

= O(exp(|τ |(π − x))),

ψ(x, λ) = ρ sin ρ(π − x) + O(exp(|τ |(π − x))) = O(|ρ| exp(|τ |(π − x))),

(1.1.10)

uniformly with respect to x ∈ [0, π]. Here and in the sequel, λ = ρ2, τ = Imρ, and o

and O denote the Landau symbols.Proof. Let us show that

ϕ(x, λ) = cos ρx + hsin ρx

ρ+ x

0

sin ρ(x − t)

ρq (t)ϕ(t, λ) dt. (1.1.11)

Indeed, the Volterra integral equation

y(x, λ) = cos ρx + h

sin ρx

ρ + x

0

sin ρ(x

−t)

ρ q (t)y(t, λ) dt

has a unique solution (for the theory of Volterra integral equations see [Kre1]). On the otherhand, if a certain function y(x, λ) satisfies this equation, then we get by differentiation

y(x, λ) + ρ2y(x, λ) = q (x)y(x, λ), y(0, λ) = 1, y(0, λ) = h,

i.e. y(x, λ) ≡ ϕ(x, λ), and (1.1.11) is valid.Differentiating (1.1.11) we calculate

ϕ(x, λ) = −ρ sin ρx + h cos ρx + x

0cos ρ(x − t)q (t)ϕ(t, λ) dt. (1.1.12)

Denote µ(λ) = max0≤x≤π

(|ϕ(x, λ)| exp(−|τ |x)). Since | sin ρx| ≤ exp(|τ |x) and | cos ρx| ≤exp(|τ |x), we have from (1.1.11) that for |ρ| ≥ 1, x ∈ [0, π],

|ϕ(x, λ)| exp(−|τ |x) ≤ 1 +1

|ρ

|h + µ(λ)

x

0|q (t)| dt

≤ C 1 +

C 2

|ρ

|µ(λ),



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and consequently

µ(λ) ≤ C 1 +C 2|ρ|µ(λ).

For sufficiently large |ρ|, this yields µ(λ) = O(1), i.e. ϕ(x, λ) = O(exp(|τ |x)). Substi-tuting this estimate into the right-hand sides of (1.1.11) and (1.1.12), we arrive at (1.1.9).Similarly one can derive (1.1.10).

We note that (1.1.10) can be also obtained directly from (1.1.9). Indeed, since

−ψ(x, λ) + q (x)ψ(x, λ) = λψ(x, λ), ψ(π, λ) = 1, ψ(π, λ) = −H,

the function ϕ(x, λ) := ψ(π−x, λ) satisfies the following differential equation and the initialconditions

−ϕ(x, λ) + q (π − x)ϕ(x, λ) = λϕ(x, λ), ϕ(0, λ) = 1, ϕ(0, λ) = H.

Therefore, the asymptotic formulae (1.1.9) are also valid for the function ϕ(x, λ). From thiswe arrive at (1.1.10). 2

The main result of this section is the following theorem on the existence and the asymp-totic behavior of the eigenvalues and the eigenfunctions of L.

Theorem 1.1.3. The boundary value problem L has a countable set of eigenvalues

λn

n≥0. For n

≥0,

ρn = λn = n + ωπn + κnn , κn ∈ l2, (1.1.13)

ϕ(x, λn) = cos nx +ξ n(x)

n, |ξ n(x)| ≤ C, (1.1.14)

where

ω = h + H +1

2

π

0q (t) dt.

Here and everywhere below one and the same symbol

κn

denotes various sequences

from l2, and the symbol C denotes various positive constants which do not depend on x, λand n.

Proof. 1) Substituting the asymptotics for ϕ(x, λ) from (1.1.9) into the right-handsides of (1.1.11) and (1.1.12), we calculate

ϕ(x, λ) = cos ρx + q 1(x)sin ρx

ρ+

1

2ρ

x

0q (t)sin ρ(x − 2t) dt + O

exp(|τ |x)

ρ2

,

ϕ(x, λ) = −ρ sin ρx + q 1(x)cos ρx +

1

2 x

0 q (t)cos ρ(x − 2t) dt + Oexp(

|τ

|x)

ρ ,

(1.1.15)where

q 1(x) = h +1

2

x

0q (t) dt.

According to (1.1.5), ∆(λ) = ϕ(π, λ) + Hϕ(π, λ). Hence by virtue of (1.1.15),

∆(λ) = −ρ sin ρπ + ω cos ρπ + κ(ρ), (1.1.16)



10

where

κ(ρ) =1

2

π

0q (t)cos ρ(π − 2t) dt + O

1

ρexp(|τ |π)

.

2) Denote Gδ = ρ : |ρ − k| ≥ δ, k = 0, ±1, ±2, . . ., δ > 0. Let us show that

| sin ρπ| ≥ C δ exp(|τ |π), ρ ∈ Gδ, (1.1.17)

|∆(λ)| ≥ C δ|ρ| exp(|τ |π), ρ ∈ Gδ, |ρ| ≥ ρ∗, (1.1.18)

for sufficiently large ρ∗ = ρ∗(δ ).

Let ρ = σ + iτ. It is sufficient to prove (1.1.17) for the domain

Dδ = ρ : σ ∈ [−1

2,

1

2], τ ≥ 0, |ρ| ≥ δ .

Denote θ(ρ) = | sin ρπ| exp(−|τ |π). Let ρ ∈ Dδ. For τ ≤ 1, θ(ρ) ≥ C δ. Since sin ρπ =(exp(iρπ) − exp(−iρπ))/(2i), we have for τ ≥ 1, θ(ρ) = |1 − exp(2iσπ)exp(−2τ π)|/2 ≥1/4. Thus, (1.1.17) is proved. Further, using (1.1.16) we get for ρ ∈ Gδ,

∆(λ) = −ρ sin ρπ

1 + O1

ρ

,

and consequently (1.1.18) is valid.

3) Denote Γn = λ : |λ| = (n + 1/2)2.

By virtue of (1.1.16),

∆(λ) = f (λ) + g(λ), f (λ) = −ρ sin ρπ, |g(λ)| ≤ C exp(|τ |π).

According to (1.1.17), |f (λ)| > |g(λ)|, λ ∈ Γn, for sufficiently large n (n ≥ n∗). Then byRouche‘s theorem [con1, p.125], the number of zeros of ∆(λ) inside Γn coincides with thenumber of zeros of f (λ) =

−ρ sin ρπ, i.e. it equals n+1. Thus, in the circle

|λ

|< (n+1/2)2

there exist exactly n + 1 eigenvalues of L : λ0, . . . , λn. Applying now Rouche‘s theoremto the circle γ n(δ ) = ρ : |ρ − n| ≤ δ , we conclude that for sufficiently large n, in γ n(δ )there is exactly one zero of ∆(ρ2), namely ρn =

√ λn. Since δ > 0 is arbitrary, it follows

thatρn = n + εn, εn = o(1), n → ∞. (1.1.19)

Substituting (1.1.19) into (1.1.16) we get

0 = ∆(ρ2n) = −(n + εn)sin(n + εn)π + ω cos(n + εn)π + κn,

and consequently−n sin εnπ + ω cos εnπ + κn = 0. (1.1.20)

Then sin εnπ = O 1

n

, i.e. εn = O

1

n

. Using (1.1.20) once more we obtain more precisely

εn =ω

πn+

κn

n, i.e. (1.1.13) is valid. Substituting (1.1.13) into (1.1.15) we arrive at (1.1.14),

where

ξ n(x) = h +

1

2 x

0 q (t) dt − x

ω

π − xκn sin nx +

1

2 x

0 q (t)sin n(x − 2t) dt + O1

n. (1.1.21)



11

Consequently |ξ n(x)| ≤ C, and Theorem 1.1.3 is proved. 2

By virtue of (1.1.6) with x = π,

β n = (ϕ(π, λn))−1.

Then, using (1.1.7), (1.1.8), (1.1.14) and (1.1.21) one can calculate

αn =π

2+

κn

n, β n = (−1)n +

κn

n, ∆(λn) = (−1)n+1 π

2+

κn

n. (1.1.22)

Since ∆(λ) has only simple zeros, we have sign ∆(λn) = (−1)n+1 for n ≥ 0.

Denote by W N 2 the Sobolev space of functions f (x), x ∈ [0, π], such that f ( j)(x), j =

0, N − 1 are absolutely continuous, and f (N )(x) ∈ L2(0, π).

Remark 1.1.1. If q (x) ∈ W N 2 , N ≥ 1, then one can obtain more precise asymptoticformulae as before. In particular,

ρn = n +N +1 j=1

ω j

n j+

κn

nN +1, ω2 p = 0, p ≥ 0, ω1 =

ω

π,

αn =π

2+

N +1 j=1

ω+ j

n j+

κn

nN +1, ω+

2 p+1 = 0, p ≥ 0, αn > 0.

(1.1.23)

Indeed, let q (x) ∈ W 12 . Then integration by parts yields

1

2

x

0q (t)cos ρ(x − 2t) dt =

sin ρx

4ρ(q (x) + q (0)) +

1

4ρ

x

0q (t)sin ρ(x − 2t) dt,

1

2ρ

x

0q (t)sin ρ(x − 2t) dt =

cos ρx

4ρ2(q (x) − q (0)) − 1

4ρ2

x

0q (t)cos ρ(x − 2t)dt.

(1.1.24)It follows from (1.1.15) and (1.1.24) that

ϕ(x, λ) = cos ρx +

h +1

2

x

0q (t)dt

sin ρx

ρ+ O

exp(|τ |x)

ρ2

.

Substituting this asymptotics into the right-hand sides of (1.1.11)-(1.1.12) and using (1.1.24)and (1.1.5), one can obtain more precise asymptotics for ϕ(ν ) (x, λ) and ∆(λ) than (1.1.15)-(1.1.16):

ϕ(x, λ) = cos ρx + q 1(x)

sin ρx

ρ + q 20(x)

cos ρx

ρ2 −1

4ρ2 x

0 q (t)cos ρ(x − 2t) dt + Oexp(

|τ

|x)

ρ3 ,

ϕ(x, λ) = −ρ sin ρx+q 1(x)cos ρx+q 21(x)sin ρx

ρ+

1

4ρ

x

0q (t)sin ρ(x−2t) dt+O

exp(|τ |x)

ρ2

,

∆(λ) = −ρ sin ρπ + ω cos ρπ + ω0sin ρπ

ρ+

κ0(ρ)

ρ, (1.1.25)

where

q 1(x) = h +

1

2 x

0 q (t) dt, ω0 = q 21(π) + Hq 1(π),



12

q 2 j (x) =1

4

q (x) + (−1) j+1q (0)

+

(−1) j+1

2

x

0q (t)q 1(t) dt, j = 0, 1,

κ0 (ρ) =1

4

π

0q (t)sin ρ(π − 2t) dt + O

exp(|τ |π)

ρ

.

From (1.1.25), by the same arguments as above, we deduce

ρn = n + εn, −n sin εnπ + ω cos εnπ +κn

n= 0.

Henceρn = n +

ω

πn+

κn

n2, κn ∈ l2.

Other formulas in (1.1.23) can be derived similarly.

Theorem 1.1.4. The specification of the spectrum λnn≥0 uniquely determines the characteristic function ∆(λ) by the formula

∆(λ) = π(λ0 − λ)∞

n=1

λn − λ

n2. (1.1.26)

Proof. It follows from (1.1.16) that ∆(λ) is entire in λ of order 1/2, and consequentlyby Hadamard‘s factorization theorem [con1, p.289], ∆(λ) is uniquely determined up to a

multiplicative constant by its zeros:

∆(λ) = C ∞

n=0

1 − λ

λn

(1.1.27)

(the case when ∆(0) = 0 requires minor modifications). Consider the function

∆(λ) := −ρ sin ρπ = −λπ∞

n=1

1 − λ

n2

.

Then∆(λ)

∆(λ)= C

λ − λ0

λ0πλ

∞n=1

n2

λn

∞n=1

1 +

λn − n2

n2 − λ

.

Taking (1.1.13) and (1.1.16) into account we calculate

limλ→−∞

∆(λ)

∆(λ)= 1, lim

λ→−∞

∞n=1

1 +

λn − n2

n2 − λ

= 1,

and hence

C = πλ0

∞n=1

λn

n2.

Substituting this into (1.1.27) we arrive at (1.1.26). 2

Remark 1.1.2. Analogous results are valid for Sturm-Liouville operators with othertypes of separated boundary conditions. Let us briefly formulate these results which will beused below. The proof is recommended as an exercise.



13

(i) Consider the boundary value problem L1 = L1(q (x), h) for equation (1.1.1) with theboundary conditions U (y) = 0, y(π) = 0. The eigenvalues µnn≥0 of L1 are simple andcoincide with zeros of the characteristic function d(λ) := ϕ(π, λ), and

d(λ) =∞

n=0

µn − λ

(n + 1/2)2

. (1.1.28)

For the spectral data µn, αn1n≥0, αn1 := π

0ϕ2(x, µn) dx of L1 we have the asymptotic

formulae: √ µn = n +

1

2+

ω1

πn+

κn

n, κn ∈ l2, (1.1.29)

αn1 =π

2+

κn1

n, κn1 ∈ l2, (1.1.30)

where ω1 = h + 12 π

0q (t) dt.

(ii) Consider the boundary value problem L0 = L0(q (x), H ) for equation (1.1.1) withthe boundary conditions y(0) = V (y) = 0. The eigenvalues λ0

nn≥0 of L0 are simple andcoincide with zeros of the characteristic function ∆0(λ) := ψ(0, λ) = S (π, λ) + HS (π, λ).The function S (x, λ) satisfies the Volterra integral equation

S (x, λ) =sin ρx

ρ+

x

0

sin ρ(x − t)

ρq (t)S (t, λ) dt, (1.1.31)

and for |ρ| → ∞,

S (x, λ) =sin ρx

ρ+ O

1

|ρ|2exp(|τ |x)

= O

1

|ρ| exp(|τ |x)

,

S (x, λ) = cos ρx + 1

|ρ| exp(|τ |x)

= O(exp(|τ |x)),

∆0(λ) = cos ρπ + 1

|ρ|exp(

|τ

|π), τ = Imρ.

(1.1.32)

Moreover,

∆0(λ) =∞

n=0

λ0n − λ

(n + 1/2)2.

λ0

n = n +1

2+

ω0

πn+

κn

n, κn ∈ l2,

where ω0 = H +1

2 π

0

q (t) dt.

(iii) Consider the boundary value problem L01 = L0

1(q (x)) for equation (1.1.1) with theboundary conditions y(0) = y(π) = 0. The eigenvalues µ0

nn≥1 of L01 are simple and

coincide with zeros of the characteristic function d0(λ) := S (π, λ), and

d0(λ) = π∞

n=1

µ0n − λ

n2.

µ0

n = n +

ω01

πn +

κn

n , κn ∈ l2,



14

where ω01 =

1

2

π

0q (t) dt.


λn < µn < λn+1, n

≥0. (1.1.33)

i.e. the eigenvalues of two boundary value problems L and L1 are alternating.

Proof. As in the proof of Lemma 1.1.1 we get

d

dxϕ(x, λ), ϕ(x, µ) = (λ − µ)ϕ(x, λ)ϕ(x, µ), (1.1.34)

and consequently,

(λ − µ) π

0ϕ(x, λ)ϕ(x, µ) dx = ϕ(x, λ), ϕ(x, µ)π

0

= ϕ(π, λ)ϕ(π, µ) − ϕ(π, λ)ϕ(π, µ) = d(λ)∆(µ) − d(µ)∆(λ).

For µ → λ we obtain

π

0ϕ2(x, λ) dx = d(λ)∆(λ) − d(λ)∆(λ) with ∆(λ) =

d

dλ∆(λ), d(λ) =

d

dλd(λ).

In particular, this yields αn = −∆(λn)d(λn), (1.1.35)

1

d2(λ)

π

0ϕ2(x, λ) dx = − d

dλ

∆(λ)

d(λ)

, −∞ < λ < ∞, d(λ) = 0.

Thus, the function∆(λ)

d(λ)is monotonically decreasing on R \ µn| n ≥ 0 with

limλ→µn±0

∆(λ)

d(λ)=

±∞.

Consequently in view of (1.1.13) and (1.1.29), we arrive at (1.1.33). 2

1.2. PROPERTIES OF EIGENFUNCTIONS

1.2.1. Completeness and expansion theorems. In this subsection we prove

that the system of the eigenfunctions of the Sturm-Liouville boundary value problem Lis complete and forms an orthogonal basis in L2(0, π). This theorem was first proved bySteklov at the end of XIX-th century. We also provide sufficient conditions under which theFourier series for the eigenfunctions converges uniformly on [0, π]. The completeness andexpansion theorems are important for solving various problems in mathematical physics bythe Fourier method, and also for the spectral theory itself. In order to prove these theoremswe apply the contour integral method of integrating the resolvent along expanding contoursin the complex plane of the spectral parameter (since this method is based on Cauchy‘s

theorem, it sometimes called Cauchy‘s method).



15

Theorem 1.2.1. (i) The system of eigenfunctions ϕ(x, λn)n≥0 of the boundary value problem L is complete in L2(0, π) .

(ii) Let f (x), x ∈ [0, π] be an absolutely continuous function. Then

f (x) =∞

n=0

anϕ(x, λn), an =1

αn π

0f (t)ϕ(t, λn) dt, (1.2.1)

and the series converges uniformly on [0, π].(iii) For f (x) ∈ L2(0, π), the series (1.2.1) converges in L2(0, π), and

π

0|f (x)|2 dx =

∞n=0

αn|an|2 (Parseval‘s equality) . (1.2.2)

There are several methods to prove Theorem 1.2.1. We shall use the contour integralmethod which plays an important role in studying direct and inverse problems for variousdifferential, integro-differential and integral operators.

Proof. 1) Denote

G(x,t,λ) = − 1

∆(λ)

ϕ(x, λ)ψ(t, λ), x ≤ t,

ϕ(t, λ)ψ(x, λ), x ≥ t,

and consider the functionY (x, λ) =

π

0G(x,t,λ)f (t) dt

= − 1

∆(λ)

ψ(x, λ)

x

0ϕ(t, λ)f (t) dt + ϕ(x, λ)

π

xψ(t, λ)f (t) dt

.

The function G(x,t,λ) is called Green‘s function for L. G(x,t,λ) is the kernel of the inverseoperator for the Sturm-Liouville operator, i.e. Y (x, λ) is the solution of the boundary valueproblem

Y − λY + f (x) = 0, U (Y ) = V (Y ) = 0; (1.2.3)

this is easily verified by differentiation. Taking (1.1.6) into account and using Theorem 1.1.2we calculate

Resλ=λn

Y (x, λ) = − 1

∆(λn)

ψ(x, λn)

x

0ϕ(t, λn)f (t) dt + ϕ(x, λn)

π

xψ(t, λn)f (t) dt

=

−

β n˙

∆(λn)

ϕ(x, λn) π

0f (t)ϕ(t, λn) dt,

and by virtue of (1.1.8),

Resλ=λn

Y (x, λ) =1

αnϕ(x, λn)

π

0f (t)ϕ(t, λn) dt. (1.2.4)

2) Let f (x) ∈ L2(0, π) be such that

π

0 f (t)ϕ(t, λn) dt = 0, n ≥ 0.



16

Then in view of (1.2.4), Resλ=λn

Y (x, λ) = 0, and consequently (after extending Y (x, λ) con-

tinuously to the whole λ - plane) for each fixed x ∈ [0, π], the function Y (x, λ) is entirein λ. Furthermore, it follows from (1.1.9), (1.1.10) and (1.1.18) that for a fixed δ > 0 andsufficiently large ρ∗ > 0 :

|Y (x, λ)| ≤ C δ|ρ| , ρ ∈ Gδ, |ρ| ≥ ρ∗.

Using the maximum principle [con1, p.128] and Liouville‘s theorem [con1, p.77] we concludethat Y (x, λ) ≡ 0. From this and (1.2.3) it follows that f (x) = 0 a.e. on (0, π). Thus (i)is proved.

3) Let now f ∈ AC [0, π] be an arbitrary absolutely continuous function. Since ϕ(x, λ)and ψ(x, λ) are solutions of (1.1.1), we transform Y (x, λ) as follows

Y (x, λ) = − 1

λ∆(λ)

ψ(x, λ)

x

0(−ϕ(t, λ) + q (t)ϕ(t, λ))f (t) dt

+ϕ(x, λ) π

x(−ψ(t, λ) + q (t)ψ(t, λ))f (t) dt

.

Integration of the terms containing second derivatives by parts yields in view of (1.1.4),

Y (x, λ) =

f (x)

λ −1

λZ 1(x, λ) + Z 2(x, λ), (1.2.5)

where

Z 1(x, λ) =1

∆(λ)

ψ(x, λ)

x

0g(t)ϕ(t, λ) dt + ϕ(x, λ)

π

xg(t)ψ(t, λ) dt

, g(t) := f (t),

Z 2(x, λ) =1

∆(λ)

hf (0)ψ(x, λ) + Hf (π)ϕ(x, λ)

+ψ(x, λ) x

0q (t)ϕ(t, λ)f (t)) dt + ϕ(x, λ)

π

xq (t)ψ(t, λ)f (t)) dt

.

Using (1.1.9), (1.1.10) and (1.1.18), we get for a fixed δ > 0, and sufficiently large ρ∗ > 0 :

max0≤x≤π

|Z 2(x, λ)| ≤ C

|ρ| , ρ ∈ Gδ, |ρ| ≥ ρ∗. (1.2.6)

Let us show thatlim|ρ|→∞ρ∈Gδ

max0≤x≤π |

Z 1(x, λ)

|= 0. (1.2.7)

First we assume that g(x) is absolutely continuous on [0, π]. In this case another integrationby parts yields

Z 1(x, λ) =1

∆(λ)

ψ(x, λ)g(t)ϕ(t, λ)

x0

+ ϕ(x, λ)g(t)ψ(t, λ)π

x

−ψ(x, λ)

x

0

g(t)ϕ(t, λ) dt

−ϕ(x, λ)

π

x

g(t)ψ(t, λ) dt.



17

By virtue of (1.1.9), (1.1.10) and (1.1.18), we infer

max0≤x≤π

|Z 1(x, λ)| ≤ C

|ρ| , ρ ∈ Gδ, |ρ| ≥ ρ∗.

Let now g(t) ∈ L(0, π). Fix ε > 0 and choose an absolutely continuous function gε(t)such that π

0|g(t) − gε(t)| dt <

ε

2C +,

where

C + = max0≤x≤π

supρ∈Gδ

1

|∆(λ)||ψ(x, λ)|

x

0|ϕ(t, λ)| dt + |ϕ(x, λ)|

π

x|ψ(t, λ)| dt

.

Then, for ρ

∈Gδ,

|ρ

| ≥ρ∗, we have

max0≤x≤π

|Z 1(x, λ)| ≤ max0≤x≤π

|Z 1(x, λ; gε)| + max0≤x≤π

|Z 1(x, λ; g − gε)| ≤ ε

2+

C (ε)

|ρ| .

Hence, there exists ρ0 > 0 such that max0≤x≤π

|Z 1(x, λ)| ≤ ε for |ρ| > ρ0. By virtue of the

arbitrariness of ε > 0, we arrive at (1.2.7).

Consider the contour integral

I N (x) = 12πi

ΓN

Y (x, λ) dλ,

where ΓN = λ : |λ| = (N + 1/2)2 (with counterclockwise circuit). It follows from(1.2.5)-(1.2.7) that

I N (x) = f (x) + εN (x), limN →∞

max0≤x≤π

|εN (x)| = 0. (1.2.8)

On the other hand, we can calculate I N (x) with the help of the residue theorem [con1,p.112]. By virtue of (1.2.4),

I N (x) =N

n=0

anϕ(x, λn), an =1

αn

π

0f (t)ϕ(t, λn) dt.

Comparing this with (1.2.8) we arrive at (1.2.1), where the series converges uniformly on[0, π], i.e. (ii) is proved.

4) Since the eigenfunctions

ϕ(x, λn)

n

≥0 are complete and orthogonal in L2(0, π), they

form an orthogonal basis in L2(0, π), and Parseval‘s equality (1.2.2) is valid. 2

1.2.2. Oscillations of the eigenfunctions. Consider the boundary value problem(1.1.1)-(1.1.2) with q (x) ≡ 0, h = H = 0. In this case, according to Example 1.1.1, theeigenfunctions are cos nx, and we have that the n -th eigenfunction has exactly n zerosinside the interval (0, π). It is turns out that this property of the eigenfunctions holds alsoin the general case. In other words, the following oscillation theorem , which is due to Sturm,

is valid.



18

Theorem 1.2.2. The eigenfunction ϕ(x, λn) of the boundary value problem L has exactly n zeros in the interval 0 < x < π.

First we prove several auxiliary assertions.

Lemma 1.2.1. Let u j(x), j = 1, 2, x

∈[a, b] be solutions of the equations

u j + g j(x)u j = 0, g1(x) < g2(x), j = 1, 2, x ∈ [a, b]. (1.2.9)

Suppose that for certain x1, x2 ∈ [a, b], u1(x1) = u1(x2) = 0, and u1(x) = 0, x ∈ (x1, x2).Then there exists x∗ ∈ (x1, x2) such that u2(x∗) = 0. In other words, the function u2(x)has at least one zero lying between any two zeros of u1(x).

Proof. Suppose, on the contrary, that u2(x) = 0 for x ∈ (x1, x2). Without loss of generality we assume that u j(x) > 0 for x ∈ (x1, x2), j = 1, 2. By virtue of (1.2.9),

d

dx(u1u2 − u2u1) = (g2 − g1)u1u2,

and consequently x2

x1(g2 − g1)u1u2 dx = (u1u2 − u2u1)

x2x1

= u1(x2)u2(x2) − u1(x1)u2(x1). (1.2.10)

The integral in (1.2.10) is strictly positive. On the other hand, since u1(x1) > 0, u1(x2) < 0,

and u2(x) ≥ 0 for x ∈ [x1, x2], the right-hand side of (1.2.10) is non-positive. Thiscontradiction proves the lemma. 2

Corollary 1.2.1. Let g1(x) < −γ 2 < 0. Then each non-trivial solution of the equation u1 + g1(x)u1 = 0 cannot have more than one zero.

Indeed, this follows from Lemma 1.2.1 with g2(x) = −γ 2 since the equation u2 −γ 2u2 =0 has the solution u2(x) = exp(γx), which has no zeros.

We consider the function ϕ(x, λ) for real λ. Zeros of ϕ(x, λ) with respect to x are

functions of λ. We show that these zeros are continuous functions of λ.Lemma 1.2.2. Let ϕ(x0, λ0) = 0. For each ε > 0 there exists δ > 0 such that if

|λ − λ0| < δ, then the function ϕ(x, λ) has exactly one zero in the interval |x − x0| < ε.

Proof. A zero x0 of a solution ϕ(x, λ0) of differential equation (1.1.1) is a simple zero,since if we had ϕ(x0, λ0) = 0, it would follow from the uniqueness theorem for equation(1.1.1) (see [cod1, Ch.1]) that ϕ(x, λ0) ≡ 0. Hence, ϕ(x0, λ0) = 0, and for definiteness let usassume that ϕ(x0, λ0) > 0. Choose ε0 > 0 such that ϕ(x, λ0) > 0 for |x−x0| ≤ ε0. Thenϕ(x, λ0) < 0 for x

∈[x0

−ε0, x0), and ϕ(x, λ0) > 0 for x

∈(x0, x0 + ε0]. Take ε

≤ε0. By

continuity of ϕ(x, λ) and ϕ(x, λ), there exists δ > 0 such that for |λ−λ0| < δ, |x−x0| < εwe have ϕ(x, λ) > 0, ϕ(x0 − ε0, λ) < 0, ϕ(x0 + ε0, λ) > 0. Consequently, the functionϕ(x, λ) has exactly one zero in the interval |x − x0| < ε. 2

Lemma 1.2.3. Suppose that for a fixed real µ, the function ϕ(x, µ) has m zeros in the interval 0 < x ≤ a. Let λ > µ. Then the function ϕ(x, λ) has not less than m zeros in the same interval, and the k-th zero of ϕ(x, λ) is less than the k-th zero of ϕ(x, µ).

Proof. Let x1 > 0 be the smallest positive zero of ϕ(x, µ). By virtue of Lemma 1.2.1,

it is sufficient to prove that ϕ(x, λ) has at least one zero in the interval 0 < x < x1.



19

Suppose, on the contrary, that ϕ(x, λ) = 0, x ∈ [0, x1). Since ϕ(0, λ) = 1, we haveϕ(x, λ) > 0, ϕ(x, µ) > 0, x ∈ [0, x1); ϕ(x1, µ) = 0, ϕ(x1, µ) < 0. It follows from (1.1.34)that

(λ − µ)

x1

0ϕ(x, λ)ϕ(x, µ) dx = ϕ(x, λ), ϕ(x, µ)

x1

0= ϕ(x1, λ)ϕ(x1, µ) ≤ 0.

But the integral in the left-hand side is strictly positive. This contradiction proves thelemma. 2

Proof of Theorem 1.2.2. Let us consider the function ϕ(x, λ) for real λ. By virtue of (1.1.9), the function ϕ(x, λ) has no zeros for sufficiently large negative λ : ϕ(x, λ) > 0, λ ≤−λ∗ < 0, x ∈ [0, π]. On the other hand, ϕ(π, µn) = 0, where µnn≥0 are eigenvalues of the boundary value problem L1.

Using Lemmas 1.2.2-1.2.3 we get that if λ moves from −∞ to ∞, then the zeros of

ϕ(x, λ) on the interval [0, π] continuously move to the left. New zeros can appear onlythrough the point x = π. This yields:(i) The functions ϕ(x, µn) has exactly n zeros in the interval x ∈ [0, π).(ii) If λ ∈ (µn−1, µn), n ≥ 1, µ1 := −∞, then the function ϕ(x, λ) has exactly n zeros

in the interval x ∈ [0, π].According to (1.1.33),

λ0 < µ0 < λ1 < µ1 < λ2 < µ2 < . . .

Consequently, the function ϕ(x, λn) has exactly n zeros in the interval [0, π]. 2

1.3. TRANSFORMATION OPERATORS

An important role in the inverse problem theory for Sturm-Liouville operators is played bythe so-called transformation operator. It connects solutions of two different Sturm-Liouvilleequations for all λ. In this section we construct the transformation operator and study its

properties. The transformation operator first appeared in the theory of generalized transla-tion operators of Delsarte and Levitan (see [lev1]). Transformation operators for arbitrarySturm-Liouville equations were constructed by Povzner [pov1]. In the inverse problem the-ory the transformation operators were used by Gelfand, Levitan and Marchenko (see Section1.5 and the monographs [mar1], [lev2]).

Theorem 1.3.1. For the function C (x, λ) the following representation holds

C (x, λ) = cos ρx +

x

0K (x, t)cos ρt dt, λ = ρ2, (1.3.1)

where K (x, t) is a real continuous function, and

K (x, x) =1

2

x

0q (t) dt. (1.3.2)

Proof. It follows from (1.1.11) for h = 0, that the function C (x, λ) is the solution of the following integral equation

C (x, λ) = cos ρx + x

0

sin ρ(x − t)

ρq (t)C (t, λ) dt. (1.3.3)



20

Sincesin ρ(x − t)

ρ= x

tcos ρ(s − t) ds,

(1.3.3) becomes


0q (t)C (t, λ) x

tcos ρ(s − t) ds dt,

and hence


0

t

0q (τ )C (τ, λ)cos ρ(t − τ ) dτ

dt.

The method of successive approximations gives

C (x, λ) =∞

n=0

C n(x, λ), (1.3.4)

C 0(x, λ) = cos ρx, C n+1(x, λ) = x

0

t

0q (τ )C n(τ, λ)cos ρ(t − τ ) dτ

dt. (1.3.5)

Let us show by induction that the following representation holds

C n(x, λ) = x

0K n(x, t)cos ρtdt, (1.3.6)

where K n(x, t) does not depend on λ.

First we calculate C 1(x, λ) , using cos α cos β = 12 cos(α + β ) + cos(α − β ) :

C 1(x, λ) = x

0

t

0q (τ )cos ρτ cos ρ(t − τ ) dτ

dt

=1

2

x

0cos ρt

t

0q (τ ) dτ

dt +

1

2

x

0

t

0q (τ )cos ρ(t − 2τ ) dτ

dt.

The change of variables t − 2τ = s in the second integral gives

C 1(x, λ) = 12

x

0cos ρt

t

0q (τ ) dτ

dt + 1

4

x

0

t

−tq t − s

2

cos ρsds

dt.

Interchanging the order of integration in the second integral we obtain

C 1(x, λ) =1

2

x

0cos ρt

t

0q (τ ) dτ

dt +

1

4

x

0cos ρs

x

sq t − s

2

dt

ds

+1

4 0

−x

cos ρs x

−s

q t − s

2 dt ds =1

2 x

0

cos ρt t

0

q (τ ) dτ dt

+1

4

x

0cos ρs

x

s

q t − s

2

+ q

t + s

2

dt

ds.

Thus, (1.3.6) is valid for n = 1, where

K 1(x, t) =1

2

t

0q (τ ) dτ +

1

4

x

t

q s − t

2

+ q

s + t

2

ds

=

1

2 x+t2

0 q (ξ ) dξ +

1

2 x−t2

0 q (ξ ) dξ, t ≤ x. (1.3.7)



21

Suppose now that (1.3.6) is valid for a certain n ≥ 1. Then substituting (1.3.6) into(1.3.5) we have

C n+1(x, λ) = x

0

t

0q (τ )cos ρ(t − τ )

τ

0K n(τ, s)cos ρs ds dτ dt

= 12

x

0

t

0q (τ )

τ

0K n(τ, s)(cos ρ(s + t − τ ) + cos ρ(s − t + τ )) ds dτ dt.

The changes of variables s + t − τ = ξ and s − t + τ = ξ, respectively, lead to

C n+1(x, λ) =1

2

x

0

t

0q (τ )

t

t−τ K n(τ, ξ + τ − t)cos ρξ dξ dτ dt

+1

2

x

0

t

0q (τ )

2τ −t

τ −tK n(τ, ξ + t − τ )cos ρξ dξ dτ dt.

6

-

-t

t

ξ = t − τ

ξ

τ

6

τ t

−t

t

ξ

ξ = 2τ − t

ξ = τ − t

fig. 1.3.1

Interchanging the order of integration (see fig. 1.3.1) we obtain

C n+1(x, λ) = x

0K n+1(x, t)cos ρtdt,

where

K n+1(x, t) =1

2

x

t

ξ

ξ−tq (τ )K n(τ, t + τ − ξ ) dτ

+ ξ

ξ+t2

q (τ )K n(τ, t

−τ + ξ ) dτ +

ξ−t

ξ−t2

q (τ )K n(τ,

−t

−τ + ξ ) dτ dξ. (1.3.8)

Substituting (1.3.6) into (1.3.4) we arrive at (1.3.1), where

K (x, t) =∞

n=1

K n(x, t). (1.3.9)

It follows from (1.3.7) and (1.3.8) that

|K n(x, t)

| ≤(Q(x))n xn−1

(n − 1)!, Q(x) :=

x

0 |q (ξ )

|dξ.



22

Indeed, (1.3.7) yields for t ≤ x,

|K 1(x, t)| ≤ 1

2

x+t2

0|q (ξ )| dξ +

1

2

x−t2

0|q (ξ )| dξ ≤

x

0|q (ξ )| dξ = Q(x).

Furthermore, if for a certain n ≥ 1, the estimate for |K n(x, t)| is valid, then by virtue of (1.3.8),

|K n+1(x, t)| ≤ 1

2

x

t

ξ

ξ+t2

|q (τ )|(Q(τ ))n τ n−1

(n − 1)!dτ +

ξ

ξ−t2

|q (τ )|(Q(τ ))n τ n−1

(n − 1)!dτ

dξ

≤ x

0

ξ

0|q (τ )|(Q(τ ))n τ n−1

(n − 1)!dτ dξ ≤

x

0(Q(ξ ))n+1 ξ n−1

(n − 1)!dξ ≤ (Q(x))n+1 xn

n!.

Thus, the series (1.3.9) converges absolutely and uniformly for 0≤

t≤

x≤

π, and thefunction K (x, t) is continuous. Moreover, it follows from (1.3.7)-(1.3.9) that the smoothness

of K (x, t) is the same as the smoothness of the function x

0q (t) dt. Since according to

(1.3.7) and (1.3.8)

K 1(x, x) =1

2

x

0q (t) dt, K n+1(x, x) = 0, n ≥ 1,

we arrive at (1.3.2). 2

The operator T, defined by

T f (x) = f (x) + x

0K (x, t)f (t) dt,

transforms cos ρx, which is the solution of the equation −y = λy with zero potential, tothe function C (x, λ), which is the solution of equation (1.1.1) satisfying the same initialcondition (i.e. C (x, λ) = T (cos ρx) ). The operator T is called the transformation operator for C (x, λ). It is important that the kernel K (x, t) does not depend on λ.

Analogously one can obtain the transformation operators for the functions S (x, λ) andϕ(x, λ) :

Theorem 1.3.2. For the functions S (x, λ) and ϕ(x, λ) the following representations hold

S (x, λ) =sin ρx

ρ+ x

0P (x, t)

sin ρt

ρdt, (1.3.10)

ϕ(x, λ) = cos ρx + x

0

G(x, t)cos ρtdt, (1.3.11)

where P (x, t) and G(x, t) are real continuous functions with the same smoothness as x

0q (t) dt, and

G(x, x) = h +1

2

x

0q (t) dt. (1.3.12)

P (x, x) =1

2

x

0q (t) dt. (1.3.13)



23

Proof. The function S (x, λ) satisfies (1.1.31), and hence

S (x, λ) =sin ρx

ρ+ x

0

t

0q (τ )S (τ, λ)cos ρ(t − τ ) dτ dt.


S (x, λ) =∞

n=0

S n(x, λ), (1.3.14)

S 0(x, λ) =sin ρx

ρ, S n+1(x, λ) =

x

0

t

0q (τ )S n(τ, λ)cos ρ(t − τ ) dτ dt.

Acting in the same way as in the proof of Theorem 1.3.1, we obtain the following represen-tation

S n(x, λ) = x

0P n(x, t)sin ρt

ρdt,

where

P 1(x, t) =1

2

x+t2

0q (ξ ) dξ − 1

2

x−t2

0q (ξ ) dξ,

P n+1(x, t) =1

2

x

t

ξ

ξ−tq (τ )P n(τ, t + τ − ξ ) dτ

+ ξ

ξ+t2

q (τ )P n(τ, t − τ + ξ ) dτ − ξ−t

ξ−t2

q (τ )P n(τ, −t − τ + ξ ) dτ dξ,

and

|P n(x, t)| ≤ (Q(x))n xn−1

(n − 1)!, Q(x) :=

x

0|q (ξ )| dξ,

Hence, the series (1.3.14) converges absolutely and uniformly for 0 ≤ t ≤ x ≤ π, and wearrive at (1.3.10) and (1.3.13).

The relation (1.3.11) can be obtained directly from (1.3.1) and (1.3.10):

ϕ(x, λ) = C (x, λ) + hS (x, λ) = cos ρx + x

0K (x, t)cos ρtdt + h

x

0cos ρtdt+

h x

0P (x, t)

t

0cos ρτ dτ

dt = cos ρx +

x

0G(x, t)cos ρt dt,

whereG(x, t) = K (x, t) + h + h

x

tP (x, τ ) dτ.

Taking here t = x we get (1.3.12). 2

1.4. UNIQUENESS THEOREMS

Let us go on to inverse problems of spectral analysis for Sturm-Liouville operators. Inthis section we give various formulations of these inverse problems and prove the correspond-ing uniqueness theorems. We present several methods for proving these theorems. These



24

methods have a wide area for applications and allow one to study various classes of inversespectral problems.

1.4.1. Ambarzumian‘s theorem. The first result in the inverse problem theory is dueto Ambarzumian [amb1]. Consider the boundary value problem L(q (x), 0, 0), i.e.

−y + q (x)y = λy, y(0) = y(π) = 0. (1.4.1)

Clearly, if q (x) = 0 a.e. on (0, π), then the eigenvalues of (1.4.1) have the form λn =n2, n ≥ 0. Ambarzumian proved the inverse assertion:

Theorem 1.4.1. If the eigenvalues of (1.4.1) are λn = n2, n ≥ 0, then q (x) = 0 a.e.on (0, π).

Proof. It follows from (1.1.13) that ω = 0, i.e. π

0

q (x) dx = 0. Let y0(x) be an

eigenfunction for the first eigenvalue λ0 = 0. Then

y0 (x) − q (x)y0(x) = 0, y0(0) = y0(π) = 0.

According to Theorem 1.2.2, the function y0(x) has no zeros in the interval x ∈ [0, π].Taking into account the relation

y0 (x)

y0(x)=

y0(x)

y0(x)2

+

y0(x)

y0(x),

we get

0 = π

0q (x) dx =

π

0

y0 (x)

y0(x)dx =

π

0

y0(x)

y0(x)

2dx.

Thus, y0(x) ≡ 0, i.e. y0(x) ≡ const, q (x) = 0 a.e. on (0, π). 2

Remark 1.4.1. Actually we have proved a more general assertion than Theorem 1.4.1,namely:

If λ0 =1

π π

0

q (x) dx, then q (x) = λ0 a.e. on (0, π).

1.4.2. Uniqueness of the recovery of differential equations from the spectraldata. Ambarzumian‘s result is an exception from the rule. In general, the specificationof the spectrum does not uniquely determine the operator. In Subsections 1.4.2-1.4.4 weprovide three uniqueness theorems where we point out spectral characteristics which uniquelydetermine the operator.

Consider the following inverse problem:

Inverse Problem 1.4.1. Given the spectral data λn, αnn≥0, construct the potentialq (x) and the coefficients h and H of the boundary conditions.

The goal of this subsection is to prove the uniqueness theorem for the solution of InverseProblem 1.4.1.

We agree that together with L we consider here and in the sequel a boundary valueproblem L = L(q (x), h, H ) of the same form but with different coefficients. If a certainsymbol γ denotes an object related to L, then the corresponding symbol γ with tilde

denotes the analogous object related to L, and γ := γ − γ.



25

Theorem 1.4.2. If λn = λn, αn = αn, n ≥ 0, then L = L, i.e. q (x) = q (x) a.e.on (0, π), h = h and H = H. Thus, the specification of the spectral data λn, αnn≥0

uniquely determines the potential and the coefficients of the boundary conditions.

We give here two proofs of Theorem 1.4.2. The first proof is due to Marchenko [mar4] and

uses the transformation operator and Parseval‘s equality (1.2.2). This method also worksfor Sturm-Liouville operators on the half-line, and gives us an opportunity to prove theuniqueness theorem of recovering an operator from its spectral function. The second proof is due to Levinson [Lev1] and uses the contour integral method. We note that Levinson firstapplied the ideas of the contour integral method to inverse spectral problems. In Section 1.6one can see a development of these ideas for constructing the solution of the inverse problem.

Marchenko‘s proof. According to (1.3.11),


0G(x, t)cos ρtdt, ϕ(x, λ) = cos ρx + x

0G(x, t)cos ρt dt.

In other words,

ϕ(x, λ) = (E + G)cos ρx, ϕ(x, λ) = (E + G)cos ρx,

where

(E + G)f (x) = f (x) + x

0

G(x, t)f (t) dt, (E + G)f (x) = f (x) + x

0

G(x, t)f (t) dt.

One can consider the relation ϕ(x, λ) = (E + G)cos ρx as a Volterra integral equation (see[jer1] for the theory of integral equations) with respect to cos ρx. Solving this equation weget

cos ρx = ϕ(x, λ) + x

0H (x, t)ϕ(t, λ) dt,

where H (x, t) is a continuous function which is the kernel of the inverse operator:

(E + H ) = (E + G)−1, Hf (x) = x

0H (x, t)f (t) dt.

Consequently

ϕ(x, λ) = ϕ(x, λ) + x

0Q(x, t)ϕ(t, λ) dt, (1.4.2)

where Q(x, t) is a real continuous function.Let f (x) ∈ L2(0, π). It follows from (1.4.2) that

π

0f (x)ϕ(x, λ) dx = π

0g(x)ϕ(x, λ) dx,

whereg(x) = f (x) +

π

xQ(t, x)f (t) dt.

Hence, for all n ≥ 0,an = bn,

an

:= π

0f (x)ϕ(x, λ

n) dx, b

n:=

π

0g(x)ϕ(x, λ

n) dx.



26

Using Parseval‘s equality (1.2.2), we calculate

π

0|f (x)|2 dx =

∞n=0

|an|2

αn=

∞n=0

|bn|2

αn=

∞n=0

|bn|2

αn= π

0|g(x)|2 dx,

i.e. f L2 = gL2 . (1.4.3)

Consider the operator

Af (x) = f (x) + π

xQ(t, x)f (t) dt.

Then Af = g. By virtue of (1.4.3), Af L2 = f L2 for any f (x) ∈ L2(0, π). Conse-quently, A∗ = A−1, but this is possible only if Q(x, t)

≡0. Thus, ϕ(x, λ)

≡ϕ(x, λ), i.e.

q (x) = q (x) a.e. on (0, π), h = h, H = H.

Levinson‘s proof. Let f (x), x ∈ [0, π] be an absolutely continuous function. Considerthe function

Y 0(x, λ) = − 1

∆(λ)

ψ(x, λ)

x

0f (t)ϕ(t, λ) dt + ϕ(x, λ)

π

xf (t)ψ(t, λ) dt

and the contour integral

I 0N (x) =

1

2πi

ΓN Y 0

(x, λ) dλ.

The idea used here comes from the proof of Theorem 1.2.1 but here the function Y 0(x, λ)is constructed from solutions of two boundary value problems.

Repeating the arguments of the proof of Theorem 1.2.1 we calculate

Y 0(x, λ) =f (x)

λ− Z 0(x, λ)

λ,

where

Z 0(x, λ) =1

∆(λ)

f (x)[ϕ(x, λ)(ψ(x, λ) − ψ(x, λ)) − ψ(x, λ)(ϕ(x, λ) − ϕ(x, λ))]

+hf (0)ψ(x, λ) + Hf (π)ϕ(x, λ) + ψ(x, λ) x

0(ϕ(t, λ)f (t) + q (t)ϕ(t, λ)f (t)) dt

+ϕ(x, λ) π

x(ψ(t, λ)f (t) + q (t)ψ(t, λ)f (t)) dt

.

The asymptotic properties for ϕ(x, λ) and ψ(x, λ) are the same as for ϕ(x, λ) and ψ(x, λ).Therefore, by similar arguments as in the proof of Theorem 1.2.1 one can obtain

I 0N (x) = f (x) + ε0N (x), lim

N →∞max

0≤x≤π|ε0

N (x)| = 0.

On the other hand, we can calculate I 0N (x) with the help of the residue theorem:

I 0N (x) =N

n=0 −1

∆(λn)ψ(x, λn) x

0

f (t)ϕ(t, λn) dt + ϕ(x, λn) π

x

f (t)ψ(t, λn) dt.



27

It follows from Lemma 1.1.1 and Theorem 1.1.4 that under the hypothesis of Theorem 1.4.2we have β n = β n. Consequently,

I 0N (x) =N

n=0

1

αnϕ(x, λn)

π


If N → ∞ we get

f (x) =∞

n=0

1

αnϕ(x, λn)

π


Together with (1.2.1) this gives π

0f (t)(ϕ(t, λn) − ϕ(t, λn)) dt = 0.

Since f (x) is an arbitrary absolutely continuous function we conclude that ϕ(x, λn) =ϕ(x, λn) for all n ≥ 0 and x ∈ [0, π]. Consequently, q (x) = q (x) a.e. on (0, π), h =h, H = H. 2

Symmetrical case. Let q (x) = q (π − x), H = h. In this case in order to constructthe potential q (x) and the coefficient h, it is sufficient to specify the spectrum λnn≥0

only.

Theorem 1.4.3. If q (x) = q (π − x), H = h, q (x) = q (π − x), H = h and λn =

λn, n ≥ 0, then q (x) = q (x) a.e. on (0, π) and h = h.Proof. If q (x) = q (π−x), H = h and y(x) is a solution of (1.1.1), then y1(x) := y(π−x)

also satisfies (1.1.1). In particular, we have ψ(x, λ) = ϕ(π−x, λ). Using (1.1.6) we calculate

ψ(x, λn) = β nϕ(x, λn) = β nψ(π − x, λn) = β 2nϕ(π − x, λn) = β 2nψ(x, λn).

Hence, β 2n = 1.On the other hand, it follows from (1.1.6) that β nϕ(π, λn) = 1. Applying Theorem 1.2.2

we conclude that β n = (−1)n.

Then (1.1.8) impliesαn = (−1)n+1∆(λn).

Under the assumptions of Theorem 1.4.3 we obtain αn = αn, n ≥ 0, and by Theorem 1.4.2,q (x) = q (x) a.e. on (0, π) and h = h. 2

1.4.3. Uniqueness of the recovery of differential equations from two spectra.

G. Borg [Bor1] suggested another formulation of the inverse problem: Reconstruct a dif-ferential operator from two spectra of boundary value problems with a common differentialequation and one common boundary condition. For definiteness, let a boundary conditionat x = 0 be common.

Let λnn≥0 and µnn≥0 be the eigenvalues of L and L1 ( L1 is defined in Remark1.1.2) respectively. Consider the following inverse problem

Inverse Problem 1.4.2. Given two spectra λnn≥0 and µnn≥0, construct thepotential q (x) and the coefficients h and H of the boundary conditions.



28

The goal of this subsection is to prove the uniqueness theorem for the solution of InverseProblem 1.4.2.

Theorem 1.4.4. If λn = λn, µn = µn, n ≥ 0, then q (x) = q (x) a.e. on (0, π), h = hand H = H. Thus, the specification of two spectra λn, µnn≥0 uniquely determines the

potential and the coefficients of the boundary conditions.Proof. By virtue of (1.1.26) and (1.1.28), the characteristic functions ∆(λ) and d(λ)

are uniquely determined by their zeros, i.e. under the hypothesis of Theorem 1.4.4 we have

∆(λ) ≡ ∆(λ) d(λ) ≡ d(λ).

Moreover, it follows from (1.1.13) and (1.1.29) that

H = H, h +1

2 π

0

q (x) dx = 0, (1.4.4)

where h = h − h, q (x) = q (x) − q (x). Hence

ϕ(π, λ) = ϕ(π, λ), ϕ(π, λ) = ϕ(π, λ). (1.4.5)

Since

−ϕ(x, λ) + q (x)ϕ(x, λ) = λϕ(x, λ), −ϕ(x, λ) + q (x)ϕ(x, λ) = λϕ(x, λ),

we get π

0q (x)ϕ(x, λ)ϕ(x, λ) dx =

π0

(ϕ(x, λ)ϕ(x, λ) − ϕ(x, λ)ϕ(x, λ)), q := q − q.

Taking (1.4.4) and (1.4.5) into account we calculate π

0q (x)

ϕ(x, λ)ϕ(x, λ) − 1

2

dx = 0. (1.4.6)

Let us show that (1.4.6) implies q = 0. For this we use the transformation operator (1.3.11).However, in order to prove this fact one can also apply the original Borg method which ispresented in Section 1.8.

Using the transformation operator (1.3.11) we rewrite the function ϕ(x, λ)ϕ(x, λ) − 1

2as follows:

ϕ(x, λ)ϕ(x, λ) − 1

2=

1

2cos2ρx +

x

0(G(x, t) + G(x, t)) cos ρx cos ρtdt

+ x

0

x

0G(x, t)G(x, s)cos ρt cos ρs dtds = 1

2cos2ρx

+1

2

x

−x(G(x, t) + G(x, t))cos ρ(x − t) dt +

1

4

x

−x

x

−xG(x, t)G(x, s)cos ρ(t − s) dtds

(here G(x, −t) = G(x, t), G(x, −t) = G(x, t) ). The changes of variables τ = (x − t)/2 andτ = (s − t)/2, respectively, yield

ϕ(x, λ)ϕ(x, λ) −1

2 =

1

2 cos2ρx + 2 x

0 (G(x, x − 2τ ) +˜

G(x, x − 2τ ))cos2ρτ dτ



29

+ x

−xG(x, s)

(s+x)/2

(s−x)/2G(x, s − 2τ )cos2ρτ dτ

ds

,

and hence

ϕ(x, λ)ϕ(x, λ) − 1

2=

1

2cos2ρx +

x

0V (x, τ )cos2ρτ dτ

, (1.4.7)

whereV (x, τ ) = 2(G(x, x − 2τ ) + G(x, x − 2τ ))

+ x

2τ −xG(x, s)G(x, s − 2τ ) ds +

x−2τ

−xG(x, s)G(x, s + 2τ ) ds. (1.4.8)

Substituting (1.4.7) into (1.4.6) and interchanging the order of integration we obtain π

0 q (τ ) +

π

τ V (τ, x)q (x) dx

cos2ρτ dτ ≡ 0.

Consequently,

q (τ ) + π

τ V (τ, x)q (x) dx = 0.

This homogeneous Volterra integral equation has only the trivial solution, i.e. q (x) = 0 a.e.on (0, π). From this and (1.4.4) it follows that h = h, H = H. 2

Remark 1.4.2. Clearly, Borg‘s result is also valid when instead of λn and µn, thespectra λn and λ0

n of L and L0 ( L0 is defined in Remark 1.1.2) are given, i.e. it is

valid for the following inverse problem.Inverse Problem 1.4.3. Given two spectra λnn≥0 and λ0

nn≥0, construct thepotential q (x) and the coefficients h and H of the boundary conditions.

The uniqueness theorem for Inverse Problem 1.4.3 has the form

Theorem 1.4.5. If λn = λn, λ0n = λ0

n, n ≥ 0, then q (x) = q (x) a.e. on (0, π), h = hand H = H. Thus, the specification of two spectra λn, λ0

nn≥0 uniquely determines the potential and the coefficients of the boundary conditions.

We note that Theorem 1.4.4 and Theorem 1.4.5 can be reduced each to other by thereplacement x → π − x.

1.4.4. The Weyl function. Let Φ(x, λ) be the solution of (1.1.1) under the condi-tions U (Φ) = 1, V (Φ) = 0. We set M (λ) := Φ(0, λ). The functions Φ(x, λ) and M (λ)are called the Weyl solution and the Weyl function for the boundary value problem L, re-spectively. The Weyl function was introduced first (for the case of the half-line) by H. Weyl.For further discussions on the Weyl function see, for example, [lev3]. Clearly,

Φ(x, λ) = −ψ(x, λ)

∆(λ)= S (x, λ) + M (λ)ϕ(x, λ), (1.4.9)

M (λ) = −∆0(λ)

∆(λ), (1.4.10)

ϕ(x, λ), Φ(x, λ) ≡ 1, (1.4.11)

where ∆0(λ) is defined in Remark 1.1.2. Thus, the Weyl function is meromorphic with

simple poles in the points λ = λn, n ≥ 0.



30

Theorem 1.4.6. The following representation holds

M (λ) =∞

n=0

1

αn(λ − λn). (1.4.12)

Proof. Since ∆0(λ) = ψ(0, λ), it follows from (1.1.10) that |∆0(λ)| ≤ C exp(|τ |π).Then, using (1.4.10) and (1.1.18), we get for sufficiently large ρ∗ > 0,

|M (λ)| ≤ C δ|ρ| , ρ ∈ Gδ, |ρ| ≥ ρ∗. (1.4.13)

Further, using (1.4.10) and Lemma 1.1.1, we calculate

Resλ=λn

M (λ) =

−

∆0(λn)˙

∆(λn)

=

−

β n˙

∆(λn)

=1

αn

. (1.4.14)

Consider the contour integral

J N (λ) =1

2πi

ΓN

M (µ)

λ − µdµ, λ ∈ int ΓN .

By virtue of (1.4.13), limN →∞

J N (λ) = 0. On the other hand, the residue theorem and (1.4.14)

yield

J N (λ) = −M (λ) +

N n=0

1

αn(λ − λn) ,

and Theorem 1.4.6 is proved. 2

In this subsection we consider the following inverse problem:

Inverse Problem 1.4.4. Given the Weyl function M (λ), construct L(q (x), h , H ).

Let us prove the uniqueness theorem for Inverse Problem 1.4.4.

Theorem 1.4.7. If M (λ) = M (λ), then L = L. Thus, the specification of the Weyl

function uniquely determines the operator.Proof. Let us define the matrix P (x, λ) = [P jk (x, λ)] j,k=1,2 by the formula

P (x, λ)

ϕ(x, λ) Φ(x, λ)

ϕ(x, λ) Φ(x, λ)

=

ϕ(x, λ) Φ(x, λ)ϕ(x, λ) Φ(x, λ)

. (1.4.15)

Using (1.4.11) and (1.4.15) we calculate

P j1(x, λ) = ϕ( j−1)(x, λ)Φ(x, λ)

−Φ( j−1)(x, λ)ϕ(x, λ),

P j2(x, λ) = Φ( j−1)(x, λ)ϕ(x, λ) − ϕ( j−1)(x, λ)Φ(x, λ), (1.4.16)

ϕ(x, λ) = P 11(x, λ)ϕ(x, λ) + P 12(x, λ)ϕ(x, λ),

Φ(x, λ) = P 11(x, λ)Φ(x, λ) + P 12(x, λ)Φ(x, λ).

(1.4.17)

It follows from (1.4.16), (1.4.9) and (1.4.11) that

P 11(x, λ) = 1 +1

∆(λ)ψ(x, λ)(ϕ(x, λ)−

ϕ(x, λ))−

ϕ(x, λ)(ψ(x, λ)−

ψ(x, λ)),



31

P 12(x, λ) =1

∆(λ)

ϕ(x, λ)ψ(x, λ) − ψ(x, λ)ϕ(x, λ)

.

By virtue of (1.1.9), (1.1.10) and (1.1.18), this yields

|P 11(x, λ) − 1| ≤C δ

|ρ| , |P 12(x, λ)| ≤C δ

|ρ| , ρ ∈ Gδ, |ρ| ≥ ρ∗, (1.4.18)

|P 22(x, λ) − 1| ≤ C δ|ρ| , |P 21(x, λ)| ≤ C δ, ρ ∈ Gδ, |ρ| ≥ ρ∗. (1.4.19)

According to (1.4.9) and (1.4.16),

P 11(x, λ) = ϕ(x, λ)S (x, λ) − S (x, λ)ϕ(x, λ) + (M (λ) − M (λ)ϕ(x, λ)ϕ(x, λ),

P 12(x, λ) = S (x, λ)ϕ(x, λ) − ϕ(x, λ)˜S (x, λ) + (M (λ) −

˜M (λ))ϕ(x, λ)ϕ(x, λ).

Thus, if M (λ) ≡ M (λ) , then for each fixed x, the functions P 11(x, λ) and P 12(x, λ) areentire in λ. Together with (1.4.18) this yields P 11(x, λ) ≡ 1, P 12(x, λ) ≡ 0. Substitutinginto (1.4.17) we get ϕ(x, λ) ≡ ϕ(x, λ), Φ(x, λ) ≡ Φ(x, λ) for all x and λ, and consequently,L = L. 2

Remark 1.4.3. According to (1.4.12), the specification of the Weyl function M (λ)is equivalent to the specification of the spectral data λn, αnn≥0. On the other hand, byvirtue of (1.4.10) zeros and poles of the Weyl function M (λ) coincide with the spectra of

the boundary value problems L and L0, respectively. Consequently, the specification of the Weyl function M (λ) is equivalent to the specification of two spectra λn and λ0

n, .Thus, the inverse problems of recovering the Sturm-Liouville equation from the spectral dataand from two spectra are particular cases of Inverse Problem 1.4.4 of recovering the Sturm-Liouville equation from the given Weyl function, and we have several independent methodsfor proving the uniqueness theorems. The Weyl function is a very natural and convenientspectral characteristic in the inverse problem theory. Using the concept of the Weyl functionand its generalizations we can formulate and study inverse problems for various classes of

operators. For example, the inverse problem of recovering higher-order differential operatorsfrom the Weyl functions has been studied in [yur1]. We will also use the Weyl function inChapter 2 for the investigation of the Sturm-Liouville operator on the half-line.

1.5. GELFAND-LEVITAN METHOD

In Sections 1.5-1.8 we present various methods for constructing the solution of inverseproblems. In this section we describe the Gelfand-Levitan method ([gel1], [lev2], [mar1]) inwhich the transformation operators constructed in Section 1.3 are used. By the Gelfand-Levitan method we obtain algorithms for the solution of inverse problems and provide neces-sary and sufficient conditions for their solvability. The central role in this method is playedby a linear integral equation with respect to the kernel of the transformation operator (seeTheorem 1.5.1). The main results of Section 1.5 are stated in Theorems 1.5.2 and 1.5.4.

1.5.1. Auxiliary propositions. In order to present the transformation operator

method we first prove several auxiliary assertions.



32

Lemma 1.5.1. In a Banach space B, consider the equations

(E + A0)y0 = f 0,

(E + A)y = f,

where A and A0 are linear bounded operators, acting from B to B, and E is the identity operator. Suppose that there exists the linear bounded operator R0 := (E +A0)−1, this yields in particular that the equation (E + A0)y0 = f 0 is uniquely solvable in B. If

A − A0 ≤ (2R0)−1,

then there exists the linear bounded operator R := (E + A)−1 with

R = R0E +∞

k=1

((A0

−A)R0)k,

and R − R0 ≤ 2R02 A − A0.

Moreover, y and y0 satisfy the estimate

y − y0 ≤ C 0(A − A0 + f − f 0),

where C 0 depends only on

R0

and

f 0

.

Proof. We have

E + A = (E + A0) + (A − A0) =

E + (A − A0)R0

(E + A0).

Under the assumptions of the lemma it follows that (A−A0)R0 ≤ 1/2, and consequentlythere exists the linear bounded operator

R := (E + A)−1 = R0E + (A

−A0)R0

−1= R0E +

∞

k=1

((A0

−A)R0)k

.

This yields in particular that R ≤ 2R0. Using again the assumption on A − A0 weinfer

R − R0 ≤ R0 (A − A0)R01 − (A − A0)R0 ≤ 2R02A − A0.

Furthermore,y − y0 = Rf − R0f 0 = (R − R0)f 0 + R(f − f 0).

Hence y − y0 ≤ 2R02f 0A − A0 + 2R0f − f 0.

2

The following lemma is an obvious corollary of Lemma 1.5.1 (applied in a correspondingfunction space).

Lemma 1.5.2. Consider the integral equation

y(t, α) + b

a A(t,s,α)y(s, α) ds = f (t, α), a ≤ t ≤ b, (1.5.1)



33

where A(t,s,α) and f (t, α) are continuous functions. Assume that for a fixed α = α0 the homogeneous equation

z (t) + b

aA0(t, s)z (s) ds = 0, A0(t, s) := A(t,s,α0)

has only the trivial solution. Then in a neighbourhood of the point α = α0, equation (1.5.1)has a unique solution y(t, α), which is continuous with respect to t and α. Moreover, the

function y(t, α) has the same smoothness as A(t,s,α) and f (t, α).

Lemma 1.5.3. Let ϕ j(x, λ), j ≥ 1, be the solution of the equation

−y + q j (x)y = λy, q j(x) ∈ L2(0, π)

under the conditions ϕ j(0, λ) = 1, ϕ j(0, λ) = h j, and let ϕ(x, λ) be the function defined in

Section 1.1. If lim j→∞ q j − q L2 = 0, lim

j→∞h j = h, then

lim j→∞ max

0≤x≤πmax|λ|≤r

|ϕ j(x, λ) − ϕ(x, λ)| = 0.

Proof. One can verify by differentiation that the function ϕ j (x, λ) satisfies the followingintegral equation

ϕ j(x, λ) = ϕ(x, λ) + (h j − h)S (x, λ) + x

0 g(x,t,λ)(q j(t) − q (t))ϕ j (t, λ) dt,

where g(x,t,λ) = C (t, λ)S (x, λ) − C (x, λ)S (t, λ) is the Green function for the Cauchyproblem:

−y + q (x)y − λy = f, y(0) = y(0) = 0.

Fix r > 0. Then for |λ| ≤ r and x ∈ [0, π],

|ϕ j(x, λ)

−ϕ(x, λ)

| ≤C |

h j

−h

|+

q j

−q

L2 max

0≤

x≤

πmax

|λ|≤

r

|ϕ j (x, λ)

|.

In particular, this yieldsmax


|ϕ j (x, λ)| ≤ C,

where C does not depend on j. Hence

max0≤x≤π

max|λ|≤r

|ϕ j(x, λ) − ϕ(x, λ)| ≤ C |h j − h| + q j − q L2

→ 0 for j → ∞.

2

Lemma 1.5.4. Let numbers ρn, αnn≥0 of the form

ρn = n +ω

πn+

κn

n, αn =

π

2+

κn1

n, κn, κn1 ∈ l2, αn = 0 (1.5.2)

be given. Denote

a(x) =∞

n=0

cos ρnx

αn

− cos nx

α0n

, (1.5.3)



34

where

α0n =

π

2, n > 0,

π, n = 0.

Then a(x) ∈ W 12 (0, 2π).

Proof. Denote δ n = ρn − n. Since

cos ρnx

αn

− cos nx

α0n

=1

α0n

cos ρnx − cos nx

+ 1

αn

− 1

α0n

cos ρnx,

cos ρnx − cos nx = cos(n + δ n)x − cos nx = − sin δ nx sin nx − 2sin2 δ nx

2cos nx

=

−δ nx sin nx

−(sin δ nx

−δ nx)sin nx

−2sin2 δ nx

2

cos nx,

we havea(x) = A1(x) + A2(x),

where

A1(x) = −ωx

π

∞n=1

sin nx

n= −ωx

π· π − x

2, 0 < x < 2π,

A2(x) =∞

n=0 1

αn

− 1

α0

ncos ρnx +

1

π cos ρ0x − 1− x∞

n=1

κnsin nx

n

−∞

n=1

(sin δ nx − δ nx)sin nx − 2∞

n=1

sin2 δ nx

2cos nx. (1.5.4)

Since

δ n = O1

n

,

1

αn

− 1

α0n

=γ nn

, γ n ∈ l2,

the series in (1.5.4) converge absolutely and uniformly on [0, 2π], and A2(x) ∈ W 12 (0, 2π).Consequently, a(x)

∈W 12 (0, 2π). 2

By similar arguments one can prove the following more general assertions.

Lemma 1.5.5. Let numbers ρn, αnn≥0 of the form (1.1.23) be given. Then a(x) ∈W N +1

2 (0, 2π).

Proof. Indeed, substituting (1.1.23) into (1.5.3) we obtain two groups of series. Theseries of the first group can be differentiated N + 1 times. The series of the second groupare ∞

n=1

sin nx

n2k+1

,∞

n=1

cos nx

n2k

.

Differentiating these series 2k and 2k − 1 times respectively, we obtain the series

∞n=1

sin nx

n=

π − x

2, 0 < x < 2π.

2



35

Lemma 1.5.6. Let numbers ρn, αnn≥0 of the form (1.5.2) be given. Fix C 0 > 0. If certain numbers ρn, αnn≥0, αn = 0 satisfy the condition

Ω := ∞

n=0

((n + 1)ξ n)21/2 ≤ C 0, ξ n := |ρn − ρn| + |αn − αn|,

then

a(x) :=∞

n=0

cos ρnx

αn

− cos ρnx

αn

∈ W 12 (0, 2π), (1.5.5)

and

max0≤x≤2π

|a(x)| ≤ C ∞

n=0

ξ n, a(x)W 12

≤ C Ω,

where C depends on

ρn, αn

n≥0 and C 0.

Proof. Obviously,∞

n=0

ξ n ≤ C Ω.

We rewrite (1.5.5) in the form

a(x) :=∞

n=0

1

αn

− 1

αn

cos ρnx +

1

αn

cos ρnx − cos ρnx

=∞

n=0

αn − αn

αnαncos ρnx +

2

αnsin

(ρn − ρn)x

2sin

(ρn + ρn)x

2

.

This series converges absolutely and uniformly, a(x) is a continuous function, and

|a(x)| ≤ C ∞

n=0

ξ n.

Differentiating (1.5.5) we calculate analogously

a(x) := −∞

n=0

ρn sin ρnx

αn

− ρn sin ρnx

αn

= −

∞n=0

ρn

αn

− ρn

αn

sin ρnx+

ρn

αn

sin ρnx−sin ρnx

=∞

n=0

αnρn − αnρn

αnαnsin ρnx +

2ρn

αnsin

(ρn − ρn)x

2cos

(ρn + ρn)x

2

=∞

n=0 αnρn − αnρn

αn

αn

sin ρnx + xρn

αn

(ρn − ρn)cos(ρn + ρn)x

2

+ρn

αn

2sin

(ρn − ρn)x

2− x(ρn − ρn)

cos

(ρn + ρn)x

2

= A1(x) + A2(x),

where

A1(x) =∞

n=0

αnρn − αnρn

αnαnsin ρnx + x

∞n=0

ρn

αn

ρn − ρn

cos ρnx, (1.5.6)

A2(x) =∞

n=0

ρn

αn2sin(ρn − ρn)x

2− x(ρn − ρn) cos

(ρn + ρn)x

2



36

+x∞

n=0

ρn

αn(ρn − ρn)

cos

(ρn + ρn)x

2− cos ρnx

. (1.5.7)

The series in (1.5.7) converge absolutely and uniformly, and

|A2(x)| ≤ C

∞n=0 |ρn − ρn|.

The series in (1.5.6) converge in L2(0, 2π), and

A1(x)L2(0,2π) ≤ C Ω.

These estimates imply the assertions of the lemma. 2

1.5.2. Recovery of differential operators from the spectral data. Let us consider

the boundary value problem L = L(q (x), h , H ). Let λn, αnn≥0 be the spectral data of L, ρn =

√ λn. We shall solve the inverse problem of recovering L from the given spectral

data λn, αnn≥0. It was shown in Section 1.1 that the spectral data have the properties:

ρn = n +ω

πn+

κn

n, αn =

π

2+

κn1

n, κn, κn1 ∈ 2, (1.5.8)

αn > 0, λn = λm (n = m). (1.5.9)

More precisely

κn =1

2π

π

0q (t)cos2ntdt + O

1

n

, κn1 = −1

2

π

0(π − t)q (t)sin2ntdt + O

1

n

,

i.e. the main terms depend linearly on the potential.

Consider the function

F (x, t) =∞

n=0 cos ρnx cos ρnt

αn −

cos nx cos nt

α0

n , (1.5.10)

where

α0n =

π

2, n > 0,

π, n = 0.

Since F (x, t) = (a(x+t)+a(x−t))/2, then by virtue of Lemma 1.5.4, F (x, t) is continuous,

andd

dxF (x, x) ∈ L2(0, π).

Theorem 1.5.1. For each fixed x ∈ (0, π], the kernel G(x, t) appearing in representa-tion (1.3.11) satisfies the linear integral equation

G(x, t) + F (x, t) + x

0G(x, s)F (s, t) ds = 0, 0 < t < x. (1.5.11)

This equation is called the Gelfand-Levitan equation.

Thus, Theorem 1.5.1 allows one to reduce our inverse problem to the solution of theGelfand-Levitan equation (1.5.11). We note that (1.5.11) is a Fredholm type integral equa-

tion in which x is a parameter.



37

Proof. One can consider the relation (1.3.11) as a Volterra integral equation with respectto cos ρx. Solving this equation we obtain


0H (x, t)ϕ(t, λ) dt, (1.5.12)

where H (x, t) is a continuous function. Using (1.3.11) and (1.5.12) we calculate

N n=0

ϕ(x, λn)cos ρnt

αn=

N n=0

cos ρnx cos ρnt

αn+

cos ρnt

αn

x

0G(x, s)cos ρns ds

,

N n=0

ϕ(x, λn)cos ρnt

αn=

N n=0

ϕ(x, λn)ϕ(t, λn)

αn+

ϕ(x, λn)

αn

t

0H (t, s)ϕ(s, λn) ds

.

This yields ΦN (x, t) = I N 1(x, t) + I N 2(x, t) + I N 3(x, t) + I N 4(x, t),

where

ΦN (x, t) =N

n=0

ϕ(x, λn)ϕ(t, λn)

αn

− cos nx cos nt

α0n

,

I N 1(x, t) =N

n=0

cos ρnx cos ρnt

αn

− cos nx cos nt

α0n

,

I N 2(x, t) =N

n=0

cos ntα0

n

x

0G(x, s)cos nsds,

I N 3(x, t) =N

n=0

x

0G(x, s)

cos ρnt cos ρns

αn

− cos nt cos ns

α0n

ds,

I N 4(x, t) = −N

n=0

ϕ(x, λn)

αn

t

0H (t, s)ϕ(s, λn) ds.

Let f (x) ∈ AC [0, π]. According to Theorem 1.2.1,

limN →∞

max0≤x≤π

π

0f (t)ΦN (x, t) dt = 0;

furthermore, uniformly with respect to x ∈ [0, π],

limN →∞

π

0f (t)I N 1(x, t) dt =

π

0f (t)F (x, t) dt,

limN →∞

π

0f (t)I N 2(x, t) dt = x

0f (t)G(x, t) dt,

limN →∞

π

0f (t)I N 3(x, t) dt =

π

0f (t)

x

0G(x, s)F (s, t) ds

dt,

limN →∞

π

0f (t)I N 4(x, t) dt = − lim

N →∞

N n=0

ϕ(x, λn)

αn

π

0ϕ(s, λn)

π

sH (t, s)f (t) dt

ds

=

− π

xf (t)H (t, x) dt.



38

Extend G(x, t) = H (x, t) = 0 for x < t. Then, in view of the arbitrariness of f (x), wederive

G(x, t) + F (x, t) + x

0G(x, s)F (s, t) ds − H (t, x) = 0.

For t < x, this yields (1.5.11). 2

The next theorem, which is the main result of Section 1.5, gives us an algorithm for thesolution of the inverse problem as well as necessary and sufficient conditions for its solvability.

Theorem 1.5.2. For real numbers λn, αnn≥0 to be the spectral data for a certain boundary value problem L(q (x), h , H ) with q (x) ∈ L2(0, π), it is necessary and sufficient that the relations (1.5.8)-(1.5.9) hold. Moreover, q (x) ∈ W N

2 iff (1.1.23) holds. The bound-ary value problem L(q (x), h , H ) can be constructed by the following algorithm:

Algorithm 1.5.1. (i) From the given numbers λn, αnn≥0 construct the function

F (x, t) by (1.5.10).(ii) Find the function G(x, t) by solving equation (1.5.11).(iii) Calculate q (x), h and H by the formulae

q (x) = 2d

dxG(x, x), h = G(0, 0), (1.5.13)

H = ω − h − 1

2

π

0q (t) dt. (1.5.14)

The necessity part of Theorem 1.5.2 was proved above, here we prove the sufficiency.Let real numbers λn, αnn≥0 of the form (1.5.8)-(1.5.9) be given. We construct the functionF (x, t) by (1.5.10) and consider equation (1.5.11).

Lemma 1.5.7. For each fixed x ∈ (0, π], equation (1.5.11) has a unique solution G(x, t)in L2(0, x).

Proof. Since (1.5.11) is a Fredholm equation it is sufficient to prove that the homoge-neous equation

g(t) + x

0 F (s, t)g(s) ds = 0 (1.5.15)has only the trivial solution g(t) = 0.

Let g(t) be a solution of (1.5.15). Then x

0g2(t) dt +

x

0

x

0F (s, t)g(s)g(t) dsdt = 0

or

x

0

g2(t) dt +∞

n=0

1

αn x

0

g(t)cos ρnt dt2

−

∞

n=0

1

α0

n x

0

g(t)cos ntdt2

= 0.

Using Parseval‘s equality

x

0g2(t) dt =

∞n=0

1

α0n

x

0g(t)cos ntdt

2,

for the function g(t), extended by zero for t > x, we obtain

∞

n=0

1

αn x

0

g(t)cos ρnt dt2

= 0.



39

Since αn > 0, then x

0g(t)cos ρnt dt = 0, n ≥ 0.

The system of functions cos ρntn≥0 is complete in L2(0, π) (see Levinson‘s theorem [you1,p.118] or Proposition 1.8.6 in Section 1.8). This yields g(t) = 0. 2

Let us return to the proof of Theorem 1.5.2. Let G(x, t) be the solution of (1.5.11). Thesubstitution t → tx, s → sx in (1.5.11) yields

F (x,xt) + G(x,xt) + x 1

0G(x,xs)F (xt, xs) ds = 0, 0 ≤ t ≤ 1. (1.5.16)

It follows from (1.5.11), (1.5.16) and Lemma 1.5.2 that the function G(x, t) is continuous,

and has the same smoothness as F (x, t). In particular,d

dxG(x, x) ∈ L2(0, π).

We construct the function ϕ(x, λ) by (1.3.11) and the function q (x) and the numberh by (1.5.13).

Lemma 1.5.8. The following relations hold

−ϕ(x, λ) + q (x)ϕ(x, λ) = λϕ(x, λ), (1.5.17)

ϕ(0, λ) = 1, ϕ(0, λ) = h. (1.5.18)

Proof. 1) First we assume that a(x) ∈ W

2

2 (0, 2π), where a(x) is defined by (1.5.3).Differentiating the identity

J (x, t) := F (x, t) + G(x, t) + x

0G(x, s)F (s, t) ds ≡ 0, (1.5.19)

we calculateJ t(x, t) = F t(x, t) + Gt(x, t) +

x

0G(x, s)F t(s, t) ds = 0, (1.5.20)

J tt(x, t) = F tt(x, t) + Gtt(x, t) + x

0

G(x, s)F tt(s, t) ds = 0, (1.5.21)

J xx(x, t) = F xx(x, t) + Gxx(x, t) +dG(x, x)

dxF (x, t) + G(x, x)F x(x, t)

+∂G(x, t)

∂xt=x

F (x, t) + x

0Gxx(x, s)F (s, t) ds = 0. (1.5.22)

According to (1.5.10), F tt(s, t) = F ss(s, t) and F t(x, t)|t=0 = 0. Then, (1.5.20) for t = 0gives

∂G(x, t)

∂tt=0

= 0. (1.5.23)

Moreover, integration by parts in (1.5.21) yields

J tt(x, t) = F tt(x, t) + Gtt(x, t) + G(x, x)∂F (s, t)

∂ss=x

−∂G(x, s)

∂s s=x

F (x, t) + x

0

Gss(x, s)F (s, t) ds = 0. (1.5.24)



40

It follows from (1.5.19), (1.5.22), (1.5.24) and the equality

J xx(x, t) − J tt(x, t) − q (x)J (x, t) ≡ 0,

that

(Gxx(x, t) − Gtt(x, t) − q (x)G(x, t)) + x

0(Gxx(x, s) − Gss(x, s) − q (x)G(x, s))F (s, t) ds ≡ 0.

According to Lemma 1.5.7, this homogeneous equation has only the trivial solution, i.e.

Gxx(x, t) − Gtt(x, t) − q (x)G(x, t) = 0 0 < t < x. (1.5.25)

Differentiating (1.3.11) twice, we get

ϕ(x, λ) = −ρ sin ρx + G(x, x)cos ρx + x

0Gx(x, t)cos ρtdt, (1.5.26)

ϕ(x, λ) = −ρ2 cos ρx − G(x, x)ρ sin ρx

+dG(x, x)

dx+

∂G(x, t)

∂xt=x

cos ρx +

x

0Gxx(x, t)cos ρtdt. (1.5.27)

On the other hand, integrating by parts twice, we obtain

λϕ(x, λ) = ρ2 cos ρx + ρ2 x

0G(x, t)cos ρtdt = ρ2 cos ρx + G(x, x)ρ sin ρx

+∂G(x, t)

∂tt=x

cos ρx − ∂G(x, t)

∂tt=0

− x

0Gtt(x, t)cos ρt dt.

Together with (1.3.11) and (1.5.27) this gives

ϕ(x, λ) + λϕ(x, λ)

−q (x)ϕ(x, λ) = 2

dG(x, x)

dx −q (x) cos ρx

−∂G(x, t)

∂t t=0

+ x

0(Gxx(x, t) − Gtt(x, t) − q (x)G(x, t))cos ρt dt.

Taking (1.5.13),(1.5.23) and (1.5.25) into account, we arrive at (1.5.17). The relations(1.5.18) follow from (1.3.11) and (1.5.26) for x = 0.

2) Let us now consider the general case when (1.5.8)-(1.5.9) hold. Then according toLemma 1.5.4, a(x)

∈W 12 (0, 2π). Denote by ϕ(x, λ) the solution of equation (1.1.1) under

the conditions ϕ(0, λ) = 1, ϕ(0, λ) = h. Our goal is to prove that ϕ(x, λ) ≡ ϕ(x, λ).Choose numbers ρn,( j), αn,( j)n≥0, j ≥ 1 of the form

ρn,( j) = n +ω

πn+

κn,( j)

n2, αn,( j) =

π

2+

ωn,( j)

n2+

κn1,( j)

n2, κn,( j), κn1,( j) ∈ 2,

such that for j → ∞,

Ω j := ∞

n=0 |(n + 1)ξ n,( j)

|2

1/2

→0, ξ n,( j) :=

|ρn,( j)

−ρn

|+

|αn,( j)

−αn

|.



41

Denote

a j(x) :=∞

n=0

cos ρn,( j)x

αn,( j)

− cos nx

α0n

j ≥ 1.

By virtue of Lemma 1.5.5, a j (x) ∈ W 22 (0, 2π). Let G j(x, t) be the solution of the Gelfand-Levitan equation

G j(x, t) + F j(x, t) + x

0G j(x, s)F j(s, t) ds = 0, 0 < t < x,

where F j(x, t) = (a j(x + t) + a j(x − t))/2. Take

q j(x) := 2d

dxG j(x, x), h j := G j(0, 0),

ϕ j(x, λ) = cos ρx + x

0 G j(x, t)cos ρtdt. (1.5.28)

Since a j(x) ∈ W 22 (0, 2π), it follows from the first part of the proof of Lemma 1.5.8 that

−ϕ j (x, λ) + q j(x)ϕ j(x, λ) = λϕ j(x, λ), ϕ j(0, λ) = 1, ϕ j(0, λ) = h j .

Further, by virtue of Lemma 1.5.6,

lim j

→∞ a j (x)

−a(x)

W 12

= 0.

From this, taking Lemma 1.5.1 into account, we get

lim j→∞ max

0≤t≤x≤π|G j(x, t) − G(x, t)| = 0, (1.5.29)

lim j→∞

q j − q L2 = 0, lim j→∞h j = h. (1.5.30)


lim j→∞ max


|ϕ j(x, λ) − ϕ(x, λ)| = 0.

On the other hand, according to Lemma 1.5.3 and (1.5.30),

lim j→∞ max


|ϕ j(x, λ) − ϕ(x, λ)| = 0.

Consequently, ϕ(x, λ) ≡ ϕ(x, λ), and Lemma 1.5.8 is proved. 2

Lemma 1.5.9 The following relation holds

H (x, t) = F (x, t) + t

0G(t, u)F (x, u) du, 0 ≤ t ≤ x, (1.5.31)

where H (x, t) is defined in (1.5.12).

Proof. 1) First we assume that a(x) ∈ W 22 (0, 2π). Differentiating (1.5.12) twice, wecalculate

−ρ sin ρx = ϕ(x, λ) + H (x, x)ϕ(x, λ) + x

0 H t(x, t)ϕ(t, λ) dt, (1.5.32)



42

−ρ2 cos ρx = ϕ(x, λ) + H (x, x)ϕ(x, λ)

+dH (x, x)

dx+

∂H (x, t)

∂xt=x

ϕ(x, λ) +

x

0H xx(x, t)ϕ(t, λ) dt. (1.5.33)

On the other hand, it follows from (1.5.12) and (1.5.17) that

−ρ2 cos ρx = ϕ(x, λ) − q (x)ϕ(x, λ) + x

0H (x, t)(ϕ(t, λ) − q (t)ϕ(t, λ)) dt.

Integrating by parts twice and using (1.5.18), we infer

−ρ2 cos ρx = ϕ(x, λ) + H (x, x)ϕ(x, λ) −∂H (x, t)

∂t

t=x+ q (x)

ϕ(x, λ)

+∂H (x, t)

∂tt=0

− hH (x, 0)

+ x

0(H tt(x, t) − q (t)H (x, t))ϕ(t, λ)) dt.

Together with (1.5.33) and

dH (x, x)

dx=∂H (x, t)

∂x+

∂H (x, t)

∂t

t=x

this yields

b0(x) + b1(x)ϕ(x, λ) + x

0b(x, t)ϕ(t, λ) dt = 0, (1.5.34)

where

b0(x) := −∂H (x, t)

∂tt=0

− hH (x, 0)

, b1(x) := 2dH (x, x)

dx+ q (x),

b(x, t) := H xx(x, t) − H tt(x, t) + q (t)H (x, t).

(1.5.35)

Substituting (1.3.11) into (1.5.34) we obtain

b0(x) + b1(x)cos ρx + x

0B(x, t)cos ρtdt = 0, (1.5.36)

whereB(x, t) = b(x, t) + b1(x)G(x, t) +

x

tb(x, s)G(s, t) ds. (1.5.37)

It follows from (1.5.36) for ρ =

n +

1

2π

xthat

b0(x) + x

0B(x, t)cos

n +

1

2

πt

xdt = 0.

By the Riemann-Lebesgue lemma, the integral here tends to 0 as n → ∞; consequently

b0(x) = 0. Further, taking in (1.5.36) ρ =2nπ

x, we get

b1(x) +

x

0B(x, t)cos

2nπt

xdt = 0,



43

therefore, similarly, b1(x) = 0. Hence, (1.5.36) takes the form x

0B(x, t)cos ρtdt = 0, ρ ∈ C,

and consequently B(x, t) = 0. Therefore, (1.5.37) implies

b(x, t) + x

tb(x, s)G(s, t) ds = 0.

From this it follows that b(x, t) = 0. Taking in (1.5.32) x = 0, we get

H (0, 0) = −h. (1.5.38)

Since b0(x) = b1(x) = b(x, t) = 0, we conclude from (1.5.35) and (1.5.38) that the function

H (x, t) solves the boundary value problemH xx(x, t) − H tt(x, t) + q (t)H (x, t) = 0, 0 ≤ t ≤ x,

H (x, x) = −h − 1

2

x

0q (t) dt,

∂H (x, t)

∂tt=0

− hH (x, 0) = 0.

(1.5.39)

The inverse assertion is also valid, namely: If a function H (x, t) satisfies (1.5.39) then(1.5.12) holds.

Indeed, denote

γ (x, λ) := ϕ(x, λ) + x

0H (x, t)ϕ(t, λ) dt.

By similar arguments as above one can calculate

γ (x, λ) + λγ (x, λ) =

2dH (x, x)

dx+ q (x)

ϕ(x, λ) −

∂H (x, t)

∂t

t=0− hH (x, 0)

+ x

0 (H xx(x, t) − H tt(x, t) + q (t)H (x, t))ϕ(t, λ) dt.

In view of (1.5.39) we get γ (x, λ) + λγ (x, λ) = 0. Clearly, γ (0, λ) = 1, γ (0, λ) = 0. Henceγ (x, λ) = cos ρx, i.e. (1.5.12) holds.

Denote

H (x, t) := F (x, t) + t

0G(t, u)F (x, u) du. (1.5.40)

Let us show that the function H (x, t) satisfies (1.5.39).

(i) Differentiating (1.5.40) with respect to t and then taking t = 0, we get

∂ H (x, t)

∂tt=0

= G(0, 0)F (x, 0) = hF (x, 0).

Since H (x, 0) = F (x, 0), this yields

∂ H (x, t)

∂t t=0 −hH (x, 0) = 0.



44

(ii) It follows from (1.3.11) and (1.5.40) that

H (x, x) = F (x, x) + x

0G(x, u)F (x, u) du = −G(x, x),

i.e. according to (1.5.13)

H (x, x) = −h − 12

x

0q (t) dt.

(iii) Using (1.5.40) again, we calculate

H tt(x, t) = F tt(x, t) +dG(t, t)

dtF (x, t) + G(t, t)F t(x, t)

+∂G(t, u)

∂t u=t

F (x, t) + t

0

Gtt(t, u)F (x, u) du,

H xx(x, t) = F xx(x, t) + t

0G(t, u)F xx(x, u) du

= F xx(x, t) + t

0G(t, u)F uu(x, u) du = F xx(x, t) + G(t, t)F t(x, t)

−∂G(t, u)

∂u

u=t

F (x, t) + t

0Guu(t, u)F (x, u) du.

Consequently

H xx(x, t) − H tt(x, t) + q (t)H (x, t) =

q (t) − 2dG(t, t)

dt

F (x, t)

− t

0(Gtt(t, u) − Guu(t, u) − q (t)G(t, u)) dt.

In view of (1.5.13) and (1.5.25) this yields

H xx(x, t) − H tt(x, t) + q (t)H (x, t) = 0.

Since H (x, t) satisfies (1.5.39), then, as was shown above,


0H (x, t)ϕ(t, λ) dt.

Comparing this relation with (1.5.12) we conclude that

x

0(H (x, t) − H (x, t))ϕ(t, λ) dt = 0 for all λ,

i.e. H (x, t) = H (x, t).

2) Let us now consider the general case when (1.5.8)-(1.5.9) hold. Then a(x) ∈ W 12 (0, 2π).Repeating the arguments of the proof of Lemma 1.5.8, we construct numbersρn,( j), αn,( j)n≥0, j ≥ 1, and functions a j (x) ∈ W 22 (0, 2π), j ≥ 1, such that

lim j→∞

a j

(x)−

a(x)W

1

2

= 0.



45

Thenlim

j→∞ max0≤t≤x≤π

|F j(x, t) − F (x, t)| = 0,

and (1.5.29) is valid. Similarly,

lim j→∞ max0≤t≤x≤π |H j(x, t) − H (x, t)| = 0.

It was proved above that

H j(x, t) = F j(x, t) + t

0G j(t, u)F j(x, u) du.

As j → ∞, we arrive at (1.5.31), and Lemma 1.5.9 is proved. 2

Lemma 1.5.10. For each function g(x) ∈ L2(0, π), π

0g2(x) dx =

∞n=0

1

αn

π

0g(t)ϕ(t, λn) dt

2. (1.5.41)

Proof. Denote

Q(λ) := π

0g(t)ϕ(t, λ) dt.

It follows from (1.3.11) that

Q(λ) = π

0h(t)cos ρt dt,

whereh(t) = g(t) +

π

tG(s, t)g(s) ds. (1.5.42)

Similarly, in view of (1.5.12),

g(t) = h(t) +

π

tH (s, t)h(s) ds. (1.5.43)

Using (1.5.42) we calculate π

0h(t)F (x, t) dt =

π

0

g(t) +

π

tG(u, t)g(u) du

F (x, t) dt

= π

0g(t)

F (x, t) +

t

0G(t, u)F (x, u) du

dt

= x

0g(t)F (x, t) +

t

0G(t, u)F (x, u) du dt +

π

xg(t)F (x, t) +

t

0G(t, u)F (x, u) du dt.

From this, by virtue of (1.5.31) and (1.5.11), we derive π

0h(t)F (x, t) dt =

x

0g(t)H (x, t) dt −

π

xg(t)G(t, x) dt. (1.5.44)

It follows from (1.5.10) and Parseval‘s equality that

π

0 h

2

(t) dt + π

0 π

0 h(x)h(t)F (x, t) dxdt



46

= π

0h2(t) dt +

∞n=0

1

αn

π

0h(t)cos ρnt dt

2 − 1

α0n

π

0h(t)cos ntdt

2

=∞

n=0

Q2(n2)

α0n

+∞

n=0

Q2(λn)

αn

− Q2(n2)

α0n

=

∞n=0

Q2(λn)

αn.

Using (1.5.44) we get

∞n=0

Q2(λn)

αn= π

0h2(t) dt +

π

0h(x)

x

0g(t)H (x, t) dt

dx −

π

0h(x)

π

xg(t)G(t, x) dt

dx

= π

0h2(t) dt +

π

0g(t)

π

tH (x, t)h(x) dx

dt −

π

0h(x)

π

xg(t)G(t, x) dt

dx.

In view of (1.5.42) and (1.5.43) we infer

∞n=0

Q2(λn)

αn= π

0h2(t) dt +

π

0g(t)(g(t) − h(t)) dt −

π

0h(x)(h(x) − g(x)) dx =

π

0g2(t) dt,

i.e. (1.5.41) is valid. 2

Corollary 1.5.1. For arbitrary functions f (x), g(x) ∈ L2(0, π),

π

0f (x)g(x) dx =

∞

n=0

1

αn

π

0f (t)ϕ(t, λn) dt

π

0g(t)ϕ(t, λn) dt. (1.5.45)

Indeed, (1.5.45) follows from (1.5.41) applied to the function f + g.

Lemma 1.5.11 The following relation holds

π

0ϕ(t, λk)ϕ(t, λn) dt =

0, n = k,αn, n = k.

(1.5.46)

Proof. 1) Let f (x) ∈ W 22 [0, π]. Consider the series

f ∗(x) =∞

n=0

cnϕ(x, λn), (1.5.47)

where

cn :=1

αn

π

0f (x)ϕ(x, λn) dx. (1.5.48)

Using Lemma 1.5.8 and integration by parts we calculate

cn =1

αnλn

π

0f (x)

− ϕ(x, λn) + q (x)ϕ(x, λn)

dx

=1

αnλn

hf (0)−f (0)+ ϕ(π, λn)f (π)−ϕ(π, λn)f (π) +

π

0ϕ(x, λn)(−f (x) + q (x)f (x)) dx

.

Applying the asymptotic formulae (1.5.8) and (1.1.15) one can check that for n → ∞,

cn = O 1

n2, ϕ(x, λn) = O(1),



47

uniformly for x ∈ [0, π]. Consequently the series (1.5.47) converges absolutely and uniformlyon [0, π]. According to (1.5.45) and (1.5.48),

π

0f (x)g(x) dx =

∞

n=0

cn

π

0g(t)ϕ(t, λn) dt

= π

0g(t)

∞n=0

cnϕ(t, λn) dt = π

0g(t)f ∗(t) dt.

Since g(x) is arbitrary, we obtain f ∗(x) = f (x), i.e.

f (x) =∞

n=0

cnϕ(x, λn). (1.5.49)

2) Fix k ≥ 0, and take f (x) = ϕ(x, λk). Then, by virtue of (1.5.49)

ϕ(x, λk) =∞

n=0

cnkϕ(x, λn), cnk =1

αn

π

0ϕ(x, λk)ϕ(x, λn) dx.

Further, the system cos ρnxn≥0 is minimal in L2(0, π) (see Proposition 1.8.6 in Section1.8), and consequently, in view of (1.3.11), the system ϕ(x, λn)n≥0 is also minimal inL2(0, π). Hence cnk = δ nk ( δ nk is the Kronecker delta), and we arrive at (1.5.46). 2

Lemma 1.5.12. For all n, m ≥ 0,

ϕ(π, λn)

ϕ(π, λn)=

ϕ(π, λm)

ϕ(π, λm). (1.5.50)

Proof. It follows from (1.1.34) that

(λn

−λm)

π

0

ϕ(x, λn)ϕ(x, λm) dx = ϕ(x, λn)ϕ(x, λm)

−ϕ(x, λn)ϕ(x, λm)

π

0

.

Taking (1.5.46) into account, we get

ϕ(π, λn)ϕ(π, λm) − ϕ(π, λn)ϕ(π, λm) = 0. (1.5.51)

Clearly, ϕ(π, λn) = 0 for all n ≥ 0. Indeed, if we suppose that ϕ(π, λm) = 0 for a certainm, then ϕ(π, λm) = 0, and in view of (1.5.51), ϕ(π, λn) = 0 for all n, which is impossible

since ϕ(π, λn) = (

−1)n + O

1

n.

Since ϕ(π, λn) = 0 for all n ≥ 0, (1.5.50) follows from (1.5.51). 2

Denote

H = −ϕ(π, λn)

ϕ(π, λn).

Notice that (1.5.50) yields that H is independent of n. Hence

ϕ(π, λn) + Hϕ(π, λn) = 0, n

≥0.



48

Together with Lemma 1.5.8 and (1.5.46) this gives that the numbers λn, αnn≥0 are thespectral data for the constructed boundary value problem L(q (x), h, H ). Clearly, H = H where H is defined by (1.5.14). Thus, Theorem 1.5.2 is proved. 2

Example 1.5.1. Let λn = n2 (n ≥ 0), αn =π

2(n ≥ 1), and let α0 > 0 be an

arbitrary positive number. Denote a := 1α0

− 1π

. Let us use Algorithm 1.5.1:

1) By (1.5.10), F (x, t) ≡ a.2) Solving equation (1.5.11) we get easily

G(x, t) = − a

1 + ax.

3) By (1.5.13)-(1.5.14),

q (x) =2a2

(1 + ax)2, h = −a, H =

a

1 + aπ=

aα0

π.

By (1.3.11),

ϕ(x, λ) = cos ρx − a

1 + ax

sin ρx

ρ.

Remark 1.5.1. Analogous results are also valid for other types of separated boundaryconditions, i.e. for the boundary value problems L1, L0 and L0

1

. In particular, the followingtheorem holds.

Theorem 1.5.3. For real numbers µn, αn1n≥0 to be the spectral data for a certain boundary value problem L1(q (x), h) with q (x) ∈ L2(0, π), it is necessary and sufficient that µn = µm (n = m), αn1 > 0, and that (1.1.29)-(1.1.30) hold.

1.5.3. Recovery of differential operators from two spectra. Let λnn≥0 andµnn≥0 be the eigenvalues of L and L1 respectively. Then the asymptotics (1.1.13) and(1.1.29) hold, and we have the representations (1.1.26) and (1.1.28) for the characteristic

functions ∆(λ) and d(λ) respectively.

Theorem 1.5.4. For real numbers λn, µnn≥0 to be the spectra for certain bound-ary value problems L and L1 with q (x) ∈ L2(0, π), it is necessary and sufficient that (1.1.13),(1.1.29) and (1.1.33) hold. The function q (x) and the numbers h and H can be constructed by the following algorithm:

(i) From the given numbers λn, µnn≥0 calculate the numbers αn by (1.1.35), where ∆(λ) and d(λ) are constructed by (1.1.26) and (1.1.28).

(ii) From the numbers

λn, αn

n≥

0 construct q (x), h and H by Algorithm 1.5.1.

The necessity part of Theorem 1.5.4 was proved above, here we prove the sufficiency. Letreal numbers λn, µnn≥0 be given satisfying the conditions of Theorem 1.5.4. We constructthe functions ∆(λ) and d(λ) by (1.1.26) and (1.1.28), and calculate the numbers αn by(1.1.35).

Our plan is to use Theorem 1.5.2. For this purpose we should obtain the asymptotics forthe numbers αn. This seems to be difficult because the functions ∆(λ) and d(λ) are byconstruction infinite products. But fortunately, for calculating the asymptotics of αn one

can also use Theorem 1.5.2, as an auxiliary assertion. Indeed, by virtue of Theorem 1.5.2



49

there exists a boundary value problem L = L(q (x), h, H ) with q (x) ∈ L2(0, π) (not unique)such that λnn≥0 are the eigenvalues of L. Then ∆(λ) is the characteristic function of L, and consequently, according to (1.1.22),

∆(λn) = (

−1)n+1 π

2

+κn

n

,

κn

∈l2.

Moreover, sign ∆(λn) = (−1)n+1. Similarly, using Theorem 1.5.3, one can prove that

d(λn) = (−1)n +κn

n, κn ∈ l2.

Moreover, taking (1.1.33) into account, we get sign d(λn) = (−1)n. Hence by (1.1.35),

αn > 0, αn =π

2

+κn1

n

,

κn1

∈l2.

Then, by Theorem 1.5.2 there exists a boundary value problem L = L(q (x), h , H ) withq (x) ∈ L2(0, π) such that λn, αnn≥0 are spectral data of L. Denote by µnn≥0 theeigenvalues of the boundary value problem L1(q (x), h). It remains to show that µn = µn

for all n ≥ 0.Let d(λ) be the characteristic function of L1. Then, by virtue of (1.1.35),

αn = −∆(λn)d(λn). But, by construction, αn = −∆(λn)d(λn), and we conclude that

d(λn) =˜d(λn), n ≥ 0.

Consequently, the function

Z (λ) :=d(λ) − d(λ)

∆(λ)

is entire in λ (after extending it continuously to all removable singularities). On the otherhand, by virtue of (1.1.9),

|d(λ)

| ≤C exp(

|τ |π),

|d(λ)

| ≤C exp(

|τ |π).

Taking (1.1.18) into account, we get for a fixed δ > 0,

|Z (λ)| ≤ C

|ρ| , λ ∈ Gδ, |ρ| ≥ ρ∗.

Using the maximum principle [con1, p.128] and Liouville‘s theorem [con1, p.77], we inferZ (λ) ≡ 0, i.e. d(λ) ≡ d(λ), and consequently µn = µn for all n ≥ 0. 2

1.6. THE METHOD OF SPECTRAL MAPPINGS

The method of spectral mappings presented in this section is an effective tool for inves-tigating a wide class of inverse problems not only for Sturm-Liouville operators, but alsofor other more complicated classes of operators such as differential operators of arbitraryorders, differential operators with singularities and/or turning points, pencils of operators

and others. In the method of spectral mappings we use ideas of the contour integral method.



50

Moreover, we can consider the method of spectral mappings as a variant of the contour in-tegral method which is adapted specially for solving inverse problems. In this section weapply the method of spectral mappings to the solution of the inverse problem for the Sturm-Liouville operator on a finite interval. The results obtained in Section 1.6 by the contourintegral method are similar to the results of Section 1.5 obtained by the transformation op-

erator method. However, the contour integral method is a more universal and perspectivetool for the solution of various classes of inverse spectral problems.

In the method of spectral mappings we start from Cauchy‘s integral formula for analyticfunctions [con1, p.84]. We apply this theorem in the complex plane of the spectral parame-ter for specially constructed analytic functions having singularities connected with the givenspectral characteristics (see the proof of Lemma 1.6.3). This allows us to reduce the inverseproblem to the so-called main equation which is a linear equation in a corresponding Banachspace of sequences. In Subsection 1.6.1 we give a derivation of the main equation, and prove

its unique solvability. Using the solution of the main equation we provide an algorithm forthe solution of the inverse problem. In Subsection 1.6.2 we give necessary and sufficientconditions for the solvability of the inverse problem by the method of spectral mappings.In Subsection 1.6.3 the inverse problem for the non-selfadjoint Sturm-Liouville operator isstudied.

1.6.1. The main equation of the inverse problem. Consider the boundary valueproblem L = L(q (x), h , H ) of the form (1.1.1)-(1.1.2) with real q (x) ∈ L2(0, π), h and H.

Let λn, αnn≥0 be the spectral data of L, ρn =

√ λn. Then (1.5.8)-(1.5.9) are valid.In this section we shall solve the inverse problem of recovering L from the given spectral

data by using ideas of the contour integral method.We note that if y(x, λ) and z (x, µ) are solutions of the equations y = λy and z = µz

respectively, thend

dxy, z = (λ − µ)yz, y, z := yz − yz. (1.6.1)

Denote

D(x,λ,µ) :=ϕ(x, λ), ϕ(x, µ)

λ − µ=

x

0ϕ(t, λ)ϕ(t, µ) dt. (1.6.2)

The last identity follows from (1.6.1).Let us choose a model boundary value problem L = L(q (x), h, H ) with real q (x) ∈

L2(0, π), h and H such that ω = ω (take, for example, q (x) ≡ 0, h = 0, H = ω, orh = H = 0, q (x) ≡ 2ω/π). Let λn, αnn≥0 be the spectral data of L.

Remark 1.6.1. Without loss of generality one can assume that ω = 0. This can beachieved by the shift of the spectrum λn → λn + C , since if λn is the spectrum of L(q (x), h , H ), then

λn + C

is the spectrum of L(q (x) + C,h,H ). In this case ω = 0,

and one can take q (x) ≡ 0, h = H = 0. However we will consider the general case when ωis arbitrary.

Letξ n := |ρn − ρn| + |αn − αn|.

Since ω = ω, it follows from (1.5.8) and the analogous formulae for ρn and αn that

Ω := ∞

n=0

((n + 1)ξ n)21/2

<

∞, n

ξ n <

∞. (1.6.3)



51

Denote in this section

λn0 = λn, λn1 = λn, αn0 = αn, αn1 = αn, ϕni(x) = ϕ(x, λni), ϕni(x) = ϕ(x, λni),

P ni,kj(x) =1

αkj

D(x, λni, λkj ), P ni,kj(x) =1

αkj

D(x, λni, λkj ), i ,j

∈ 0, 1

, n,k

≥0. (1.6.4)

Then, according to (1.6.2),

P ni,kj(x) =ϕni(x), ϕkj (x)αkj (λni − λkj )

=1

αkj

x

0ϕni(t)ϕkj (t) dt.


=1

αkj

x

0ϕni(t)ϕkj (t) dt.

Clearly,

P ni,kj(x) = 1αkj

ϕni(x)ϕkj (x), P ni,kj(x) = 1αkj

ϕni(x)ϕkj (x). (1.6.5)

Below we shall use the following version of Schwarz‘s lemma (see [con1, p.130]):

Lemma 1.6.1. Let f (ρ) be an analytic function for |ρ − ρ0| < a such that f (ρ0) = 0and |f (ρ)| ≤ A, |ρ − ρ0| < a. Then

|f (ρ)| ≤ A

a|ρ − ρ0| for |ρ − ρ0| < a.

In order to obtain the solution of the inverse problem we need several auxiliary proposi-tions.

Lemma 1.6.2. The following estimates are valid for x ∈ [0, π], n,k ≥ 0, i, j, ν = 0, 1,

|ϕ(ν )ni (x)| ≤ C (n + 1)ν , |ϕ(ν )

n0 (x) − ϕ(ν )n1 (x)| ≤ Cξ n(n + 1)ν , (1.6.6)

|P ni,kj(x)

| ≤

C

|n − k| + 1

,

|P

(ν +1)ni,kj (x)

| ≤C (k + n + 1)ν ,

|P ni,k0(x) − P ni,k1(x)| ≤ Cξ k|n − k| + 1

, |P n0,kj (x) − P n1,kj (x)| ≤ Cξ n|n − k| + 1

,

|P n0,k0(x) − P n1,k0(x) − P n0,k1(x) + P n1,k1(x)| ≤ Cξ nξ k|n − k| + 1

.

(1.6.7)

The analogous estimates are also valid for ϕni(x), P ni,kj(x).

Proof. It follows from (1.1.9) and (1.5.8) that

|ϕ(ν )(x, λni)| ≤ C (n + 1)ν .

Moreover, for a fixed a > 0,

|ϕ(ν )(x, λ)| ≤ C (n + 1)ν , |ρ − ρn1| ≤ a.

Applying Schwarz‘s lemma in the ρ - plane to the function f (ρ) := ϕ(ν )(x, λ) − ϕ(ν )(x, λn1)with fixed ν,n,x and a, we get

|ϕ(ν )

(x, λ) − ϕ(ν )

(x, λn1)| ≤ C (n + 1)ν

|ρ − ρn1|, |ρ − ρn1| ≤ a.



52

Consequently,|ϕ(ν )

n0 (x) − ϕ(ν )n1 (x)| ≤ C (n + 1)ν |ρn0 − ρn1|,

and (1.6.6) is proved.Let us show that

|D(x,λ,λkj )| ≤ C exp(|τ |x)|ρ ∓ k| + 1

, λ = ρ2, ±Re ρ ≥ 0, τ := Im ρ, k ≥ 0. (1.6.8)

For definiteness, let σ := Re ρ ≥ 0. Take a fixed δ 0 > 0. For |ρ − ρkj | ≥ δ 0, we have byvirtue of (1.6.2), (1.1.9) and (1.5.8),

|D(x,λ,λkj )| =|ϕ(x, λ), ϕ(x, λkj )|

|λ − λkj | ≤ C exp(|τ |x)|ρ| + |ρkj ||ρ2 − ρ2

kj |.

Since |ρ| + |ρkj ||ρ + ρkj | ≤

√ σ2 + τ 2 + |ρkj | σ2 + τ 2 + ρ2

kj

≤√

2

(it is used here that (a + b)2 ≤ 2(a2 + b2) for all real a, b ), we get

|D(x,λ,λkj )| ≤ C exp(|τ |x)

|ρ − ρkj | .

For |ρ − ρkj | ≥ δ 0, |ρ − k| + 1

|ρ − ρkj | ≤ 1 +|ρkj − k| + 1

|ρ − ρkj | ≤ C,

and consequently1

|ρ − ρkj | ≤ C

|ρ − k| + 1.

This yields (1.6.8) for |ρ − ρkj | ≥ δ 0. For |ρ − ρkj | ≤ δ 0,

|D(x,λ,λkj )| = x

0 ϕ(t, λ)ϕ(t, λkj ) dt ≤ C exp(|τ |x),

i.e. (1.6.8) is also valid for |ρ − ρkj | ≤ δ 0. Similarly one can show that

|D(x,λ,µ)| ≤ C exp(|τ |x)

|ρ ∓ θ| + 1, µ = θ2, |Im θ| ≤ C 0, ±ReρReθ ≥ 0.

Using Schwarz‘s lemma we obtain

|D(x,λ,λk1) − D(x,λ,λk0)| ≤Cξ k exp(

|τ |x)

|ρ ∓ k| + 1 , ±Re ρ ≥ 0, k ≥ 0. (1.6.9)

In particular, this yields

|D(x, λni, λk1) − D(x, λni, λk0)| ≤ Cξ k|n − k| + 1

.

Symmetrically,

|D(x, λn1, λkj )

−D(x, λn0, λkj )

| ≤Cξ n

|n − k| + 1.



53

Applying Schwarz‘s lemma to the function

Qk(x, λ) := D(x,λ,λk1) − D(x,λ,λk0)

for fixed k and x, we get

|D(x, λn0, λk0) − D(x, λn1, λk0) − D(x, λn0, λk1) + D(x, λn1, λk1)| ≤ Cξ nξ k|n − k| + 1

.

These estimates together with (1.6.4), (1.6.5) and (1.5.8) imply (1.6.7). 2


ϕ(x, λ) = ϕ(x, λ) +∞

k=0 ϕ(x, λ), ϕk0(x)

αk0(λ

−λk0)

ϕk0(x) − ϕ(x, λ), ϕk1(x)αk1(λ

−λk1)

ϕk1(x)

, (1.6.10)

ϕ(x, λ), ϕ(x, µ)λ − µ

− ϕ(x, λ), ϕ(x, µ)λ − µ

+∞

k=0

ϕ(x, λ), ϕk0(x)αk0(λ − λk0)

ϕk0(x), ϕ(x, µ)(λk0 − µ)

−ϕ(x, λ), ϕk1(x)αk1(λ − λk1)

ϕk1(x), ϕ(x, µ)(λk1 − µ)

= 0. (1.6.11)

Both series converge absolutely and uniformly with respect to x ∈ [0, π] and λ, µ on com-pact sets.

Proof. 1) Denote λ = minni

λni and take a fixed δ > 0. In the λ - plane we consider

closed contours γ N (with counterclockwise circuit) of the form: γ N = γ +N ∪ γ −N ∪ γ ∪ ΓN ,where

γ ±N = λ : ±Im λ = δ, Reλ ≥ λ, |λ| ≤

N +1

2

2,

γ = λ : λ − λ = δ exp(iα), α ∈π

2,

3π

2

,

ΓN = ΓN ∩ λ : |Im λ| ≤ δ, Re λ > 0, ΓN = λ : |λ| = N +

1

22

.

-

6

'

E

λ

γ N

(N + 1

2)2

6

-λ

(N + 1

2)2

γ 0N

'

E

!

fig. 1.6.1 fig. 1.6.2

Denote γ 0N = γ +N ∪ γ −N ∪ γ ∪ (ΓN \ ΓN ) (with clockwise circuit). Let P (x, λ) =[P jk (x, λ)] j,k=1,2 be the matrix defined by (1.4.15). It follows from (1.4.16) and (1.4.9)



54

that for each fixed x, the functions P jk (x, λ) are meromorphic in λ with simple polesλn and λn. By Cauchy‘s integral formula [con1, p.84],

P 1k(x, λ) − δ 1k =1

2πi

γ 0N

P 1k(x, ξ ) − δ 1k

λ

−ξ

dξ, k = 1, 2,

where λ ∈ intγ 0N and δ jk is the Kronecker delta. Hence

P 1k(x, λ) − δ 1k =1

2πi

γ N

P 1k(x, ξ )

λ − ξ dξ − 1

2πi

ΓN

P 1k(x, ξ ) − δ 1k

λ − ξ dξ,

where ΓN is used with counterclockwise circuit. Substituting into (1.4.17) we obtain

ϕ(x, λ) = ϕ(x, λ) +1

2πi γ N

ϕ(x, λ)P 11(x, ξ ) + ϕ(x, λ)P 12(x, ξ )

λ − ξ

dξ + εN (x, λ),

where

εN (x, λ) = − 1

2πi

ΓN

ϕ(x, λ)(P 11(x, ξ ) − 1) + ϕ(x, λ)P 12(x, ξ )

λ − ξ dξ.

By virtue of (1.4.18),lim

N →∞εN (x, λ) = 0 (1.6.12)

uniformly with respect to x ∈ [0, π] and λ on compact sets. Taking (1.4.16) into account

we calculate

ϕ(x, λ) = ϕ(x, λ) +1

2πi

γ N

ϕ(x, λ)(ϕ(x, ξ )Φ(x, ξ ) − Φ(x, ξ )ϕ(x, ξ ))+

ϕ(x, λ)(Φ(x, ξ )ϕ(x, ξ ) − ϕ(x, ξ )Φ(x, ξ ) dξ

λ − ξ + εN (x, λ).

In view of (1.4.9), this yields

ϕ(x, λ) = ϕ(x, λ) + 12πi

γ N

ϕ(x, λ), ϕ(x, ξ )λ − ξ

M (ξ )ϕ(x, ξ ) dξ + εN (x, λ), (1.6.13)

where M (λ) = M (λ) − M (λ), since the terms with S (x, ξ ) vanish by Cauchy‘s theorem.It follows from (1.4.14) that

Resξ=λkj

ϕ(x, λ), ϕ(x, ξ )λ − ξ

M (ξ )ϕ(x, ξ ) =ϕ(x, λ), ϕkj (x)

αkj (λ − λkj )ϕkj (x).

Calculating the integral in (1.6.13) by the residue theorem [con1, p.112] and using (1.6.12)we arrive at (1.6.10).

2) Since1

λ − µ

1

λ − ξ − 1

µ − ξ

=

1

(λ − ξ )(ξ − µ),

we have by Cauchy‘s integral formula

P jk (x, λ) − P jk (x, µ)

λ − µ=

1

2πi γ 0N

P jk (x, ξ )

(λ − ξ )(ξ − µ)dξ, k, j = 1, 2; λ, µ

∈intγ 0N .



55

Acting in the same way as above and using (1.4.18)-(1.4.19) we obtain


λ − µ=

1

2πi

γ N

P jk (x, ξ )

(λ − ξ )(ξ − µ)dξ + εN jk (x,λ,µ), (1.6.14)

where limN →∞ εN jk (x,λ,µ) = 0, j, k = 1, n.From (1.4.16) and (1.4.11) it follows that

P 11(x, λ)ϕ(x, λ) − P 21(x, λ)ϕ(x, λ) = ϕ(x, λ),

P 22(x, λ)ϕ(x, λ) − P 12(x, λ)ϕ(x, λ) = ϕ(x, λ),

(1.6.15)

P (x, λ)

y(x)y(x)

= y(x), Φ(x, λ)

ϕ(x, λ)ϕ(x, λ)

− y(x), ϕ(x, λ)

Φ(x, λ)Φ(x, λ)

(1.6.16)

for any y(x) ∈ C 1[0, 1]. Taking (1.6.14) and (1.6.16) into account, we calculate

P (x, λ) − P (x, µ)

λ − µ

y(x)y(x)

=

1

2πi

γ N

y(x), Φ(x, ξ )

ϕ(x, ξ )ϕ(x, ξ )

−

y(x), ϕ(x, ξ )

Φ(x, ξ )Φ(x, ξ )

dξ

(λ − ξ )(ξ − µ)+ ε0

N (x,λ,µ), limN →∞

ε0N (x,λ,µ) = 0. (1.6.17)

According to (1.4.15),

P (x, λ)

ϕ(x, λ)ϕ(x, λ)

=

ϕ(x, λ)ϕ(x, λ)

.

Therefore,

det

P (x, λ)

ϕ(x, λ)ϕ(x, λ)

,

ϕ(x, µ)ϕ(x, µ)

= ϕ(x, λ), ϕ(x, µ).

Furthermore, using (1.6.15) we get

det P (x, µ) ϕ(x, λ)ϕ(x, λ)

, ϕ(x, µ)ϕ(x, µ)

= ϕ(x, λ)

P 11(x, µ)ϕ(x, µ)−P 21(x, µ)ϕ(x, µ)−ϕ(x, λ)

P 22(x, µ)ϕ(x, µ)−P 12(x, µ)ϕ(x, µ)

= ϕ(x, λ), ϕ(x, µ).

Thus,

det (P (x, λ)

−P (x, µ))

ϕ(x, λ)˜ϕ(x, λ) ,

ϕ(x, µ)

ϕ(x, µ) =

ϕ(x, λ), ϕ(x, µ) − ϕ(x, λ), ϕ(x, µ).

Consequently, (1.6.17) for y(x) = ϕ(x, λ) yields

ϕ(x, λ), ϕ(x, µ)λ − µ

− ϕ(x, λ), ϕ(x, µ)λ − µ

=

1

2πi γ N ϕ(x, λ), Φ(x, ξ )ϕ(x, ξ ), ϕ(x, µ)

(λ − ξ )(ξ − µ) −ϕ(x, λ), ϕ(x, ξ )Φ(x, ξ ), ϕ(x, µ)

(λ − ξ )(ξ − µ) dξ



56

+ε1N (x,λ,µ), lim

N →∞ε1

N (x,λ,µ) = 0.

By virtue of (1.4.9), (1.4.14) and the residue theorem, we arrive at (1.6.11). 2

Analogously one can obtain the following relation

Φ(x, λ) = Φ(x, λ) +∞

k=0

Φ(x, λ), ϕk0(x)αk0(λ − λk0)

ϕk0(x) − Φ(x, λ), ϕk1(x)αk1(λ − λk1)

ϕk1(x)

. (1.6.18)

It follows from the definition of P ni,kj(x), P ni,kj(x) and (1.6.10)-(1.6.11) that

ϕni(x) = ϕni(x) +∞

k=0

(P ni,k0(x)ϕk0(x) − P ni,k1(x)ϕk1(x)), (1.6.19)

P ni,j(x) − P ni,j(x) +∞

k=0

(P ni,k0(x)P k0,j(x) − P ni,k1(x)P k1,j(x)) = 0. (1.6.20)

Denote

ε0(x) =∞

k=0

1

αk0ϕk0(x)ϕk0(x) − 1

αk1ϕk1(x)ϕk1(x)

, ε(x) = −2ε0(x). (1.6.21)

Lemma 1.6.4. The series in (1.6.21) converges absolutely and uniformly on [0, π]. The

function ε0(x) is absolutely continuous, and ε(x) ∈ L2(0, π).

Proof. We rewrite ε0(x) to the form

ε0(x) = A1(x) + A2(x), (1.6.22)

where

A1(x) =∞

k=0 1

αk0 −1

αk1ϕk0(x)ϕk0(x),

A2(x) =∞

k=0

1

αk1

(ϕk0(x) − ϕk1(x))ϕk0(x) + ϕk1(x)(ϕk0(x) − ϕk1(x))

.

(1.6.23)

It follows from (1.5.8), (1.6.3) and (1.6.6) that the series in (1.6.23) converge absolutely anduniformly on [0, π], and

|A j(x)| ≤ C ∞

k=0

ξ k ≤ C Ω, j = 1, 2. (1.6.24)

Furthermore, using the asymptotic formulae (1.1.9), (1.5.8) and (1.6.3) we calculate

A1(x) =

∞k=0

1

αk0

− 1

αk1

d

dx

ϕk0(x)ϕk0(x)

=

∞k=0

γ k

sin2kx +ηk(x)

k + 1

,

where γ k ∈ l2, and max0≤x≤π

|ηk(x)| ≤ C for k ≥ 0. Hence A1(x) ∈ W 12 (0, π). Similarly,

we get A2(x)

∈W 12 (0, π), and consequently ε0(x)

∈W 12 (0, π). 2



57


q (x) = q (x) + ε(x), (1.6.25)

h = h − ε0(0), H = H + ε0(π), (1.6.26)

where ε(x) and ε0(x) are defined by (1.6.21).Proof. Differentiating (1.6.10) twice with respect to x and using (1.6.1) and (1.6.21) we

getϕ(x, λ) − ε0(x)ϕ(x, λ) = ϕ(x, λ)

+∞

k=0


ϕk0(x) − ϕ(x, λ), ϕk1(x)αk1(λ − λk1)

ϕk1(x)

, (1.6.27)

ϕ(x, λ) = ϕ(x, λ) +∞

k=0


ϕk0

(x)−


ϕk1

(x)+2ϕ(x, λ)

∞k=0

1


αk1ϕk1(x)ϕk1(x)

+

∞k=0

1

αk0(ϕ(x, λ)ϕk0(x))ϕk0(x) − 1

αk1(ϕ(x, λ)ϕk1(x))ϕk1(x)

.

We replace here the second derivatives, using equation (1.1.1), and then replace ϕ(x, λ),using (1.6.10). This yields

q (x)ϕ(x, λ) = q (x)ϕ(x, λ) +∞

k=0

1

αk0

ϕ(x, λ), ϕk0(x)ϕk0(x) − 1

αk1

ϕ(x, λ), ϕk1(x)ϕk1(x)

+2ϕ(x, λ)∞

k=0

1


αk1ϕk1(x)ϕk1(x)

+∞

k=0 1

αk0(ϕ(x, λ)ϕk0(x))ϕk0(x) − 1


.

After canceling terms with ϕ(x, λ) we arrive at (1.6.25).Denote h0 = −h, hπ = H, U 0 = U, U π = V. In (1.6.10) and (1.6.27) we put x = 0 and

x = π. Thenϕ(a, λ) + (ha − ε0(a))ϕ(a, λ) = U a(ϕ(x, λ))

+∞

k=0

ϕ(x, λ), ϕk0(x)|x=a

αk0(λ − λk0)U a(ϕk0) − ϕ(x, λ), ϕk1(x)|x=a

αk1(λ − λk1)U a(ϕk1)

, a = 0, π. (1.6.28)

Let a = 0. Since U 0(ϕ(x, λ)) = 0, ϕ(0, λ) = 1, ϕ(0, λ) =−

h0, we get h0

−h0

−ε0(0) =

0, i.e. h = h − ε0(0). Let a = π. Since

U π(ϕ(x, λ)) = ∆(λ), U π(ϕk0) = 0, U π(ϕk1) = ∆(λk1),

ϕ(x, λ), ϕ(x, µ) = ϕ(π, λ)∆(µ) − ϕ(π, µ)∆(λ),

it follows from (1.6.28) that

ϕ(π, λ) + (hπ

−ε0(π))ϕ(π, λ) = ∆(λ) +

∞

k=0

ϕk1(π)∆(λ)

αk1(λ − λk1)

∆(λk1).



58

For λ = λn1 this yields

ϕn1(π) + (hπ − ε0(π))ϕn1(π) = ∆(λn1)

1 +1

αn1ϕn1(π)∆1(λn1)

,

where ∆1(λ) :=

d

dλ∆(λ). By virtue of (1.1.6) and (1.1.8),

ϕn1(π)β n = 1, β nαn = −∆1(λn1),

i.e.1

αn1ϕn1(π)∆1(λn1) = −1,

Thenϕn1(π) + (hπ

−ε0(π))ϕn1(π) = 0.

On the other hand,ϕn1(π) + hπϕn1(π) = ∆(λn1) = 0.

Thus, (hπ − ε0(π) − hπ)ϕn1(π) = 0, i.e. hπ − hπ = ε0(π). 2

Remark 1.6.2. For each fixed x ∈ [0, π], the relation (1.6.19) can be considered as asystem of linear equations with respect to ϕni(x), n ≥ 0, i = 0, 1. But the series in (1.6.19)converges only ”with brackets”. Therefore, it is not convenient to use (1.6.19) as a main

equation of the inverse problem. Below we will transfer (1.6.19) to a linear equation in acorresponding Banach space of sequences (see (1.6.33) or (1.6.34)).

Let V be a set of indices u = (n, i), n ≥ 0, i = 0, 1. For each fixed x ∈ [0, π], wedefine the vector

ψ(x) = [ψu(x)]u∈V =

ψn0(x)ψn1(x)

n≥0

= [ψ00, ψ01, ψ10, ψ11, . . .]T

by the formulaeψn0(x)ψn1(x)

=

χn −χn

0 1

ϕn0(x)ϕn1(x)

=

χn(ϕn0(x) − ϕn1(x))

ϕn1(x)

,

χn =

ξ −1

n , ξ n = 0,0, ξ n = 0.

We also define the block matrix

H (x) = [H u,v(x)]u,v∈V = H n0,k0(x) H n0,k1(x)H n1,k0(x) H n1,k1(x)

n,k≥0

, u = (n, i), v = (k, j)

by the formulaeH n0,k0(x) H n0,k1(x)H n1,k0(x) H n1,k1(x)

=

χn −χn

0 1

P n0,k0(x) P n0,k1(x)P n1,k0(x) P n1,k1(x)

ξ k 10 −1

= ξ kχn(P n0,k0(x) − P n1,k0(x)) χn(P n0,k0(x) − P n0,k1(x) − P n1,k0(x) + P n1,k1(x))

ξ kP n1,k0(x) P n1,k0(x) − P n1,k1(x) .



59

Analogously we define ψ(x), H (x) by replacing in the previous definitions ϕni(x) b y ϕni(x)and P ni,kj(x) by P ni,kj(x). It follows from (1.6.6) and (1.6.7) that

|ψ(ν )ni (x)| ≤ C (n + 1)ν , |H ni,kj(x)| ≤ Cξ k

(|n − k| + 1), |H

(ν +1)ni,kj (x)| ≤ C (n + k + 1)ν ξ k. (1.6.29)

Similarly,

|ψ(ν )ni (x)| ≤ C (n + 1)ν , |H ni,kj(x)| ≤ Cξ k

(|n − k| + 1), |H

(ν +1)ni,kj (x)| ≤ C (n + k + 1)ν ξ k. (1.6.30)

and also|H ni,kj(x) − H ni,kj(x0)| ≤ C |x − x0|ξ k, x, x0 ∈ [0, π], (1.6.31)

where C does not depend on x, x0, n, i , j and k.

Let us consider the Banach space m of bounded sequences α = [αu]u∈V with the normαm = supu∈V

|αu|. It follows from (1.6.29) and (1.6.30) that for each fixed x ∈ [0, π], the

operators E + H (x) and E − H (x) (here E is the identity operator), acting from m tom , are linear bounded operators, and

H (x), H (x) ≤ C supn

k

ξ k|n − k| + 1

< ∞. (1.6.32)

Theorem 1.6.1. For each fixed x ∈ [0, π], the vector ψ(x) ∈ m satisfies the equation

ψ(x) = (E + H (x))ψ(x) (1.6.33)

in the Banach space m. Moreover, the operator E + H (x) has a bounded inverse operator,i.e. equation (1.6.33) is uniquely solvable.

Proof. We rewrite (1.6.19) in the form

ϕn0(x) − ϕn1(x) = ϕn0(x) − ϕn1(x) +

∞k=0

(P n0,k0(x) − P n1,k0(x))(ϕk0(x) − ϕk1(x))

+(P n0,k0(x) − P n1,k0(x) − P n0,k1(x) + P n1,k1(x))ϕk1(x)

,

ϕn1(x) = ϕn1(x) +∞

k=0

P n1,k0(x)(ϕk0(x) − ϕk1(x)) + (P n1,k0(x) − P n1,k1(x))ϕk1(x)

.

Taking into account our notations, we obtain

ψni(x) = ψni(x) + k,j

H ni,kj(x)ψkj (x), (n, i), (k, j) ∈ V, (1.6.34)

which is equivalent to (1.6.33). The series in (1.6.34) converges absolutely and uniformly forx ∈ [0, π]. Similarly, (1.6.20) takes the form

H ni,kj(x) − H ni,kj(x) +,s

(H ni,s(x)H s,kj (x) = 0, (n, i), (k, j), (, s) ∈ V,

or

(E + H (x))(E − H (x)) = E.



60

Interchanging places for L and L, we obtain analogously

ψ(x) = (E − H (x))ψ(x), (E − H (x))(E + H (x)) = E.

Hence the operator (E + H (x))−1 exists, and it is a linear bounded operator. 2

Equation (1.6.33) is called the main equation of the inverse problem. Solving (1.6.33)we find the vector ψ(x) , and consequently, the functions ϕni(x) . Since ϕni(x) = ϕ(x, λni)are the solutions of (1.1.1), we can construct the function q (x) and the coefficients h andH. Thus, we obtain the following algorithm for the solution of Inverse Problem 1.4.1.

Algorithm 1.6.1. Given the numbers λn, αnn≥0.(i) Choose L such that ω = ω, and construct ψ(x) and H (x).(ii) Find ψ(x) by solving equation (1.6.33).(iii) Calculate q (x), h and H by (1.6.21), (1.6.25)-(1.6.26).

1.6.2. Necessary and sufficient conditions. In this item we provide necessary andsufficient conditions for the solvability of Inverse Problem 1.4.1.

Theorem 1.6.2. For real numbers λn, αnn≥0 to be the spectral data for a cer-tain L(q (x), h , H ) with q (x) ∈ L2(0, π), it is necessary and sufficient that the relations (1.5.8)-(1.5.9) hold. Moreover, q (x) ∈ W N

2 iff (1.1.23) holds. The boundary value problem L(q (x), h , H ) can be constructed by Algorithm 1.6.1.

The necessity part of Theorem 1.6.2 was proved above. We prove here the sufficiency. Letnumbers λn, αnn≥0 of the form (1.5.8)-(1.5.9) be given. Choose L, construct ψ(x), H (x)and consider the equation (1.6.33).

Lemma 1.6.6. For each fixed x ∈ [0, π], the operator E + H (x), acting from m tom, has a bounded inverse operator, and the main equation (1.6.33) has a unique solution ψ(x) ∈ m.

Proof. It is sufficient to prove that the homogeneous equation

(E + H (x))β (x) = 0, (1.6.35)

where β (x) = [β u(x)]u∈V , has only the zero solution. Let β (x) ∈ m be a solution of (1.6.35), i.e.

β ni(x) +k,j

H ni,kj(x)β kj (x) = 0, (n, i), (k, j) ∈ V. (1.6.36)

Denote γ n1(x) = β n1(x), γ n0(x) = β n0(x)ξ n + β n1(x). Then (1.6.36) takes the form

γ ni(x) +∞

k=0

(P ni,k0(x)γ k0(x) − P ni,k1(x)γ k1(x)) = 0, (1.6.37)

and|γ ni(x)| ≤ C (x), |γ n0(x) − γ n1(x)| ≤ C (x)ξ n. (1.6.38)

Construct the functions γ (x, λ), Γ(x, λ) and B(x, λ) by the formulas

γ (x, λ) =

−

∞

k=0 ϕ(x, λ), ϕk0(x)

αk0(λ − λk0)

γ k0(x)

−ϕ(x, λ), ϕk1(x)

αk1(λ − λk1)

γ k1(x), (1.6.39)



61

Γ(x, λ) = −∞

k=0


γ k0(x) − Φ(x, λ), ϕk1(x)αk1(λ − λk1)

γ k1(x)

, (1.6.40)

B(x, λ) = γ (x, λ)Γ(x, λ). (1.6.41)

The idea to consider B(x, λ) comes from the construction of Green‘s function for t = x :

G(x,x,λ) = ϕ(x, λ)Φ(x, λ).In view of (1.6.2) the function γ (x, λ) is entire in λ for each fixed x. The functions

Γ(x, λ) and B(x, λ) are meromorphic in λ with simple poles λni. According to (1.6.2)and (1.6.37), γ (x, λni) = γ ni(x). Let us show that

Resλ=λn0

B(x, λ) =1

αn0

|γ n0(x)|2, Resλ=λn1

B(x, λ) = 0. (1.6.42)

Indeed, by virtue of (1.6.39), (1.6.40) and (1.4.9) we get

Γ(x, λ) = M (λ)γ (x, λ) −∞

k=0

S (x, λ), ϕk0(x)αk0(λ − λk0)

γ k0(x) − S (x, λ), ϕk1(x)αk1(λ − λk1)

γ k1(x)

. (1.6.43)

Since S (x, λ), ϕ(x, µ)|x=0 = −1, it follows from (1.6.1) that

S (x, λ), ϕ(x, µ)λ

−µ

= − 1

λ

−µ

+

x

0S (t, λ)ϕ(t, µ) dt.

Therefore, according to (1.6.43),

Resλ=λn0

Γ(x, λ) =1

αn0γ n0(x), Res

λ=λn1Γ(x, λ) =

1

αn1γ n1(x) − 1

αn1γ n1(x) = 0.

Together with (1.6.41) this yields (1.6.42).Further, consider the integral

I 0N (x) = 12πi

ΓN B(x, λ) dλ, (1.6.44)

where ΓN = λ : |λ| = (N + 1/2)2 . Let us show that

limN →∞

I 0N (x) = 0. (1.6.45)

Indeed, it follows from (1.6.2) and (1.6.39) that

−γ (x, λ) = ∞k=0

1αk0

D(x,λ,λk0)(γ k0(x) − γ k1(x))

+∞

k=0

1

αk0

− 1

αk1

D(x,λ,λk0)γ k1(x) +

∞k=0

1

αk1

D(x,λ,λk0) − D(x,λ,λk1)

γ k1(x).

For definiteness, let σ := Re ρ ≥ 0. By virtue of (1.5.8), (1.6.38), (1.6.8) and (1.6.9), we get

|γ (x, λ)

|=

|γ (x, λ)

| ≤C (x) exp(

|τ

|x)

∞

k=0

ξ k

|ρ − k| + 1

, σ

≥0, τ = Imρ. (1.6.46)



62

Similarly, using (1.6.40) we obtain for sufficiently large ρ∗ > 0 :

|Γ(x, λ)| ≤ C (x)

|ρ| exp(−|τ |x)∞

k=0

ξ k|ρ − k| + 1

, σ ≥ 0, ρ ∈ Gδ, |ρ| ≥ ρ∗.

Then |B(x, λ)| ≤ C (x)

|ρ| ∞

k=0

ξ k|ρ − k| + 1

2

≤ C (x)

|ρ|∞

k=0

((k + 1)ξ k)2 ·∞

k=0

1

(|ρ − k| + 1)2(k + 1)2.

In view of (1.6.3), this implies

|B(x, λ)| ≤C (x)

|ρ|∞k=0

1

|ρ − k|2(k + 1)2 , |ρ| = N +

1

2 , Re ρ ≥ 0. (1.6.47)

Since ρ = (N + 1/2) exp(iθ), θ ∈ [−π/2, π/2], we calculate

|ρ − k| =

N +1

2

2+ k2 − 2

N +

1

2

k cos θ ≥

N +

1

2− k

2. (1.6.48)

Furthermore,

∞k=0

1(k + 1)2(N + 1

2− k)2

≤N

k=0

1(k + 1)2(N + 1

2− k)2

+ 1(N + 2)2

∞k=1

1(k − 1

2)2

=N −1k=1

1

k2(N − k)2+ O

1

N 2

.

Since1

k2(N − k)2=

2

kN 3+

1

k2N 2+

2

N 3(N − k)+

1

N 2(N − k)2,

we getN −1k=1

1

k2(N − k)2=

4

N 3

N −1k=1

1

k+

2

N 2

N −1k=1

1

k2.

Hence ∞k=0

1

(k + 1)2(N + 12− k)2

= O 1

N 2

.

Together with (1.6.47) and (1.6.48) this yields

|B(x, λ)| ≤ C (x)

|ρ|3, λ ∈ ΓN .

Substituting this estimate into (1.6.44) we arrive at (1.6.45).On the other hand, calculating the integral in (1.6.44) by the residue theorem and taking

(1.6.42) into account, we arrive at

limN →∞

N

n=0

1

αn0 |γ n0(x)

|2 = 0.



63

Since αn0 > 0, we get γ n0(x) = 0.Construct the functions

∆(λ) := π(λ0 − λ)∞

n=1

λn − λ

n2, λn = ρ2

n,

∆(λ) := −πλ∞

n=1

n2 − λ

n2= −ρ sin ρπ, λ = ρ2,

f (x, λ) :=γ (x, λ)

∆(λ).

It follows from the relation γ (x, λn0) = γ n0(x) = 0 that for each fixed x ∈ [0, π] thefunction f (x, λ) is entire in λ. On the other hand, we have

∆(λ)∆(λ)

=∞

n=0

1 + λn − n2

λ − λn

.

Fix δ > 0. By virtue of (1.1.17) and (1.5.8), the following estimates are valid in the sectorarg λ ∈ [δ, 2π − δ ] :

|∆(λ)| ≥ C |ρ| exp(|τ |π),λn − n2

λ − λn

≤ C

n2, τ := Imρ.

Consequently|∆(λ)| ≥ C |ρ| exp(|τ |π), arg λ ∈ [δ, 2π − δ ].

Taking (1.6.46) into account we obtain

|f (x, λ)| ≤ C (x)

|ρ| , arg λ ∈ [δ, 2π − δ ].

From this, using the Phragmen-Lindelof theorem [you1, p.80] and Liouville‘s theorem [con1,

p.77] we conclude that f (x, λ) ≡ 0, i.e. γ (x, λ) ≡ 0 and γ n1(x) = γ (x, λn1) = 0. Henceβ ni(x) = 0 for n ≥ 0, i = 0, 1, and Lemma 1.6.6 is proved. 2

Let ψ(x) = [ψu(x)]u∈V be the solution of the main equation (1.6.33).

Lemma 1.6.7. For n ≥ 0, i = 0, 1, the following relations hold

ψni(x) ∈ C 1[0, π], |ψ(ν )ni (x)| ≤ C (n + 1)ν , ν = 0, 1, x ∈ [0, π], (1.6.49)

|ψni(x)

−ψni(x)

| ≤C Ωηn, x

∈[0, π], (1.6.50)

|ψni(x) − ψ

ni(x)| ≤ C Ω, x ∈ [0, π], (1.6.51)

where Ω is defined by (1.6.3), and

ηn := ∞

k=0

1

(k + 1)2(|n − k| + 1)2

1/2.

Here and below, one and the same symbol C denotes various positive constants which

depend here only on L and C 0, where C 0 > 0 is such that Ω ≤ C 0.



64

Proof. 1) Denote R(x) = (E +H (x))−1. Fix x0 ∈ [0, π] and consider the main equation(1.6.33) for x = x0 :

ψ(x0) = (E + H (x0))ψ(x0).


H (x) − H (x0) ≤ C |x − x0|k

ξ k ≤ C 1|x − x0|, x, x0 ∈ [0, π], C 1 > 0.

Take

ω(x0) :=1

2C 1R(x0) .

Then, for |x − x0| ≤ ω(x0),

H (x)

−H (x0)

≤1

2R(x0).

Using Lemma 1.5.1 we get for |x − x0| ≤ ω(x0),

R(x) − R(x0) =∞

k=1

H (x0) − H (x)

kR(x0)

k+1,

R(x) − R(x0) ≤ 2C 1R(x0)2|x − x0|.Consequently, R(x) is continuous with respect to x

∈[0, π], and

R(x) ≤ C, x ∈ [0, π].

We represent R(x) in the form R(x) = E − H (x). Then,

H (x) ≤ C, x ∈ [0, π], (1.6.52)

and(E

−H (x))(E + H (x)) = (E + H (x))(E

−H (x)) = E. (1.6.53)

In coordinates (1.6.53) takes the form

H ni,kj(x) = H ni,kj(x) −,s

H ni,s(x)H s,kj(x), (n, i), (k, j), (, s) ∈ V, (1.6.54)

H ni,kj(x) = H ni,kj(x) −,s

H ni,s(x)H s,kj(x), (n, i), (k, j), (, s) ∈ V. (1.6.55)

The functions H ni,kj(x) are continuous for x ∈ [0, π], and by virtue of (1.6.52), (1.6.54)

and (1.6.30) we have |H ni,kj(x)| ≤ Cξ k, x ∈ [0, π]. (1.6.56)

Substituting this estimate into the right-hand side of (1.6.54) and (1.6.55) and using (1.6.30)we obtain more precisely,

|H ni,kj(x)| ≤ Cξ k 1

|n − k| + 1+ Ωηk

, x ∈ [0, π], (n, i), (k, j) ∈ V, (1.6.57)

|H ni,kj(x)

| ≤Cξ k 1

|n − k| + 1+ Ωηn, x

∈[0, π], (n, i), (k, j)

∈V. (1.6.58)



65

We note that since

∞k=0

|ηk|2 =∞

n=0

∞k=0

1

(k + 1)2(|n − k| + 1)2=

∞k=0

∞n=k

1

(k + 1)2(n − k + 1)2

+

∞n=0

∞k=n+1

1

(k + 1)2(k − n + 1)2 ≤ 2∞

k=1

1

k22

,

it follows that ηk ∈ l2.Solving the main equation (1.6.33) we infer

ψni(x) = ψni(x) −k,j

H ni,kj(x)ψkj (x), x ∈ [0, π], (n, i), (k, j) ∈ V. (1.6.59)

According to (1.6.30) and (1.6.58), the series in (1.6.59) converges absolutely and uniformly

for x ∈ [0, π]; the functions ψni(x) are continuous for x ∈ [0, π],

|ψni(x)| ≤ C, x ∈ [0, π], (n, i) ∈ V,

and (1.6.50) is valid.

2) It follows from (1.6.53) that

H (x) = (E − H (x))H (x)(E − H (x)), (1.6.60)

the functions H ni,kj(x) are continuously differentiable, and

|H ni,kj(x)| ≤ Cξ k. (1.6.61)

Differentiating (1.6.59) we calculate

ψni(x) = ψ

ni(x) −k,j

H ni,kj(x)ψkj (x) −

k,j

H ni,kj(x)ψkj (x). (1.6.62)

By virtue of (1.6.30), (1.6.57) and (1.6.61), the series in (1.6.62) converge absolutely and

uniformly for x ∈ [0, π]; ψni(x) ∈ C 1[0, π],

|ψni(x)| ≤ C (n + 1), x ∈ [0, π], (n, i) ∈ V,

and (1.6.51) is valid. 2

Remark 1.6.3. The estimates for ψni(x) can be also obtained in the following way.

Differentiating (1.6.34) formally we get

˜ψni(x) −

k,j

˜H ni,kj(x)ψkj (x) = ψni(x) +

k,j

˜H ni,kj(x)ψkj (x). (1.6.63)

Define

z (x) = [z u(x)]u∈V , z (x) = [z u(x)]u∈V , A(x) = [Au,v(x)]u,v∈V , u = (n, i), v = (k, j),

by the formulas

z ni(x) =1

n + 1

ψni(x), z ni(x) =

1

n + 1ψ

ni(x)

− k,j

H ni,kj(x)ψkj (x),



66

Ani,kj(x) =k + 1

n + 1H ni,kj(x).

Then (1.6.63) takes the formz (x) = (E + A(x))z (x)

or z ni(x) = z ni(x) +k,j

Ani,kj(x)z kj (x). (1.6.64)


k,j

|Ani,kj(x)| ≤ C ∞

k=0

(k + 1)ξ k(n + 1)(|n − k| + 1)

≤ C Ω

n + 1

∞k=0

1

(|n − k| + 1)2

1/2.

Since

∞k=0

1

(|n − k| + 1)2=

nk=0

1

(n − k + 1)2+

∞k=n+1

1

(k − n + 1)2≤ 2

∞k=1

1

k2, (1.6.65)

we infer k,j

|Ani,kj(x)| ≤ C Ω

n + 1.

For each fixed x

∈[0, π], the operator A(x) is a linear bounded operator, acting from m

to m, and A(x) ≤ C Ω.

The homogeneous equation

uni(x) +k,j

Ani,kj(x)ukj (x) = 0, [uni(x)] ∈ m, (1.6.66)

has only the trivial solution. Indeed, let [uni(x)] ∈ m be a solution of (1.6.66). Then the

functions β ni(x) := (n + 1)uni(x) satisfy (1.6.36). Moreover, (1.6.36) gives

|β ni(x)| ≤ C (x)k,j

|H ni,kj(x)|(k + 1) ≤ C (x)∞

k=0

(k + 1)ξ k|n − k| + 1

≤ C (x)Ω ∞

k=0

1

(|n − k| + 1)2

1/2.

Hence, in view of (1.6.65), |β ni(x)| ≤ C (x), i.e. [β ni(x)] ∈ m. By Lemma 1.6.6, β ni(x) = 0,i.e. uni(x) = 0.

Using (1.6.64), by the same arguments as above, one can show that ψni(x) ∈ C 1[0, π]and

|ψ

ni(x)

| ≤C (n + 1), i.e. (1.6.49) is proved. Hence, it follows from (1.6.63) that

|ψni(x) − ψ

ni(x)| ≤ C ∞

k=0

(k + 1)ξ k|n − k| + 1

≤ C Ω ∞

k=0

1

(|n − k| + 1)2

1/2.

Taking (1.6.65) into account we arrive at (1.6.51). 2

We define the functions ϕni(x) by the formulae

ϕn1(x) = ψn1(x), ϕn0(x) = ψn0(x)ξ n + ψn1(x). (1.6.67)



67

Then (1.6.19) and (1.6.6) are valid. By virtue of (1.6.67) and Lemma 1.6.7, we have

|ϕ(ν )ni (x)| ≤ C (n + 1)ν , ν = 0, 1, (1.6.68)

|ϕni(x) − ϕni(x)| ≤ C Ωηn, |ϕni(x) − ϕni(x)| ≤ C Ω. (1.6.69)

Furthermore, we construct the functions ϕ(x, λ) and Φ(x, λ) via (1.6.10), (1.6.18), and theboundary value problem L(q (x), h , H ) via (1.6.21), (1.6.25)-(1.6.26).Clearly, ϕ(x, λni) = ϕni(x).

Lemma 1.6.8. The following relation holds:

q (x) ∈ L2(0, π).

Proof. Here we follow the proof of Lemma 1.6.4 with necessary modifications. Accordingto (1.6.22), ε0(x) = A1(x) + A2(x), where the functions A j(x) are defined by (1.6.23). Itfollows from (1.5.8), (1.6.3) and (1.6.6) that the series in (1.6.23) converge absolutely anduniformly on [0, π], and (1.6.24) holds. Furthermore,

A1(x) =

∞k=0

1

αk0

− 1

αk1

d

dx

ϕk0(x)ϕk0(x)

= 2

∞

k=0

1

αk0 −1

αk1ϕk0(x)ϕk0(x) + A(x) =

∞

k=0 γ k sin2kx +

ηk(x)

k + 1 + A(x),

where

A(x) =∞

k=0

1

αk0

− 1

αk1

ϕk0(x)(ϕk0(x) − ϕk0(x)) + ϕk0(x)(ϕk0(x) − ϕk0(x))

, (1.6.70)

γ k ∈ l2, ∞

k=0

|γ k|21/2 ≤ C Ω, max

0≤x≤π|ηk(x)| ≤ C.

By virtue of (1.6.68), (1.6.69), (1.6.6) and (1.5.8), the series in (1.6.70) converges absolutelyand uniformly on [0, π], and

|A(x)| ≤ C Ω ∞

k=0

ξ k +∞

k=0

(k + 1)ξ kηk

≤ C Ω2

1 +

∞k=0

|ηk|21/2 ≤ C Ω2,

since ηk ∈ l2. Hence A1(x) ∈ W 12 [0, π]. Similarly we get A2(x) ∈ W 12 [0, π]. Consequently,ε0(x) ∈ W 12 (0, π), ε(x) ∈ L2(0, π), i.e. q (x) ∈ L2(0, π). 2

Lemma 1.6.9. The following relations hold:

ϕni(x) = λniϕni(x), ϕ(x, λ) = λϕ(x, λ), Φ(x, λ) = λΦ(x, λ). (1.6.71)

ϕ(0, λ) = 1, ϕ(0, λ) = h, U (Φ) = 1, V (Φ) = 0. (1.6.72)

Proof. 1) Using (1.6.10), (1.6.18) and acting in the same way as in the proof of Lemma1.6.5, we get

U a

(ϕ(x, λ)) = U a

(ϕ(x, λ))



68

+∞

k=0

ϕ(x, λ), ϕk0(x)|x=a

αk0(λ − λk0)U a(ϕk0) − ϕ(x, λ), ϕk1(x)|x=a

αk1(λ − λk1)U a(ϕk1)

, a = 0, π, (1.6.73)

U a(Φ(x, λ)) = U a(Φ(x, λ))

+

∞k=0

Φ(x, λ), ϕk0(x)

|x=a

αk0(λ − λk0) U a(ϕk0) − Φ(x, λ), ϕk1(x)

|x=a

αk1(λ − λk1) U a(ϕk1), a = 0, π, (1.6.74)

where U 0 = U, U π = V.Since ϕ(x, λ), ϕkj (x)|x=0 = 0, it follows from (1.6.10) and (1.6.73) for x = 0 that

ϕ(0, λ) = 1, U 0(ϕ) = 0, and consequently ϕ(0, λ) = h. In (1.6.74) we put a = 0. SinceU 0(ϕ) = 0 we get U 0(Φ) = 1.

2) In order to prove (1.6.71) we first assume that

Ω1 := ∞k=0

((k + 1)2ξ k)21/2 < ∞. (1.6.75)

In this case, one can obtain from (1.6.60) by differentiation that

H (x) = (E − H (x))H (x)(E − H (x))

−H (x)H (x)(E − H (x)) − (E − H (x))H (x)H (x). (1.6.76)

It follows from (1.6.30), (1.6.56) and (1.6.61) that the series in (1.6.76) converge absolutely

and uniformly for x ∈ [0, π], H ni,kj(x) ∈ C 2[0, π], and

|H ni,kj(x)| ≤ C (n + k + 1)ξ k. (1.6.77)

Furthermore, using (1.6.59) we calculate

ψni(x) = ψni(x)+k,j

H ni,kj(x)ψkj (x)+2k,j

H ni,kj(x)ψkj (x)+

k,j

H ni,kj(x)ψkj (x). (1.6.78)

Since ϕni(x) = λniϕni(x) it follows that

ψn0(x) = λn0ψn0(x) + χn(λn0 − λn1)ϕn1(x), ψn1(x) = λn1ψn1(x),

henceψni(x) ∈ C [0, π], |ψni(x)| ≤ C (n + 1)2, (n, i) ∈ V.

From this and from (1.6.57), (1.6.61), (1.6.65), (1.6.77) and (1.6.30) we deduce that theseries in (1.6.78) converge absolutely and uniformly for x ∈ [0, π], and

ψni(x) ∈ C [0, π], |ψni(x)| ≤ C (n + 1)2, (n, i) ∈ V.

On the other hand, it follows from the proof of Lemma 1.6.8 and from (1.6.75) that in ourcase q (x) − q (x) ∈ C [0, π]; hence

ψni(x) ∈ C [0, π], |ψni(x)| ≤ C (n + 1)2, (n, i) ∈ V.

Together with (1.6.67) this implies that

ϕni(x) ∈ C [0, π], |ϕni(x)| ≤ C (n + 1)2

, |ϕn0(x) − ϕn1(x)| ≤ Cξ n(n + 1)2

, (n, i) ∈ V.



69

Using (1.6.19) we calculate

−ϕni(x) + q (x)ϕni(x) = ϕni(x) +∞

k=0

P ni,k0(x)ϕk0(x) − P ni,k1(x)ϕk1(x)

−2ϕni(x) ∞k=0

1αk0

ϕk0(x)ϕk0(x) − 1αk1

ϕk1(x)ϕk1(x)

−∞

k=0

1

αk0(ϕni(x)ϕk0(x))ϕk0(x) − 1

αk1(ϕni(x)ϕk1(x))ϕk1(x)

.

Taking (1.6.21) and (1.6.25) into account we derive


k=0 P ni,k0(x)ϕk0(x)

−P ni,k1(x)ϕk1(x)+

+∞

k=0

1

αk0

ϕni(x), ϕk0(x)ϕk0(x) − 1

αk1

ϕni(x), ϕk1(x)ϕk1(x)

. (1.6.79)

Using (1.6.10) and (1.6.18) we calculate similarly

ϕ(x, λ) = ϕ(x, λ) +∞

k=0


ϕk0(x) − ϕ(x, λ), ϕk1(x)αk1(λ − λk1)

ϕk1(x)

+

∞k=0

1

αk0

ϕ(x, λ), ϕk0(x)ϕk0(x) − 1

αk1

ϕ(x, λ), ϕk1(x)ϕk1(x)

, (1.6.80)

Φ(x, λ) = Φ(x, λ) +∞

k=0


ϕk0(x) − Φ(x, λ), ϕk1(x)αk1(λ − λk1)

ϕk1(x)

+

∞k=0

1

αk0

Φ(x, λ), ϕk0(x)ϕk0(x) − 1

αk1

Φ(x, λ), ϕk1(x)ϕk1(x)

. (1.6.81)


λniϕni(x) = ϕni(x) +∞

k=0

P ni,k0(x)ϕk0(x) − P ni,k1(x)ϕk1(x)

+

∞k=0

(λni − λk0)P ni,k0(x)ϕk0(x) − (λni − λk1)P ni,k1(x)ϕk1(x)

,

and consequently we arrive at (1.6.37) where

γ ni(x) := ϕni(x) − λniϕni(x).

Then (1.6.36) is valid, where

β n1(x) = γ n1(x), β n0(x) = χn(γ n0(x) − γ n1(x)), χn :=

ξ −1

n , ξ n = 0,0, ξ n = 0.

Moreover,

|γ ni(x)| ≤ C (n + 1)2

, |γ n0(x) − γ n1(x)| ≤ Cξ n(n + 1)2

,



70

and hence|β ni(x)| ≤ C (n + 1)2. (1.6.82)

It follows from (1.6.36), (1.6.30), (1.6.82) and (1.6.75) that

|β ni(x)| ≤ k,j |H ni,kj(x)β kj (x)| ≤ C

∞

k=0

(k + 1)2ξ k

|n − k| + 1 .

Sincek + 1

(n + 1)(|n − k| + 1)≤ 1,

we obtain

|β ni(x)| ≤ C (n + 1)∞

k=0

(k + 1)ξ k ≤ C (n + 1). (1.6.83)

Using (1.6.83) instead of (1.6.82) and repeating the arguments, we infer

|β ni(x)| ≤ k,j

|H ni,kj(x)β kj (x)| ≤ C ∞

k=0

(k + 1)ξ k|n − k| + 1

≤ C.

Then, by virtue of Lemma 1.6.6, β ni(x) = 0, and consequently γ ni(x) = 0. Thus, ϕni(x) =λniϕni(x).

Furthermore, it follows from (1.6.80) that

λϕ(x, λ) = ϕ(x, λ) + ∞k=0

ϕ(x, λ), ϕk0(x)αk0(λ − λk0) λk0ϕk0(x) − ϕ(x, λ), ϕk1(x)αk1(λ − λk1) λk1ϕk1(x)+

∞k=0


(λ − λk0)ϕk0(x) − ϕ(x, λ), ϕk1(x)αk1(λ − λk1)

(λ − λk1)ϕk1(x)

.

From this, by virtue of (1.6.10), it follows that ϕ(x, λ) = λϕ(x, λ). Analogously, using(1.6.81), we obtain Φ(x, λ) = λΦ(x, λ). Thus, (1.6.71) is proved for the case when (1.6.75)is fulfilled.

3) Let us now consider the general case when (1.6.3) holds. Choose numbersρn,( j), αn,( j)n≥0, j ≥ 1, such that

Ω1,( j) := ∞

k=0

(k + 1)2ξ k,( j))21/2

< ∞,

Ω0,( j) := ∞

k=0

((k + 1)ηk,( j))21/2 → 0 as j → ∞,

where ξ k,( j) := |ρk,( j) − ρk| + |αk,( j) − αk|, ηk,( j) := |ρk,( j) − ρk| + |αk,( j) − αk|.For each fixed j ≥ 1, we solve the corresponding main equation:

ψ( j)(x) = (E + H ( j)(x))ψ( j)(x),

and construct the functions ϕ( j)(x, λ) and the boundary value problem L(q ( j)(x), h( j), H ( j)).Using Lemma 1.5.1 one can show that

lim j→∞

q ( j) −

q L2

= 0, lim j→∞

h( j)

= h,



71

lim j→∞ max

0≤x≤π|ϕ( j)(x, λ) − ϕ(x, λ)| = 0. (1.6.84)

Denote by ϕ0(x, λ) the solution of equation (1.1.1) under the conditions ϕ0(0, λ) = 1 ,ϕ0(0, λ) = h. According to Lemma 1.5.3,

lim j→∞ max0≤x≤π |ϕ( j)(x, λ) − ϕ0(x, λ)| = 0.

Comparing this relation with (1.6.84) we conclude that ϕ0(x, λ) = ϕ(x, λ), i.e. ϕ(x, λ) =λϕ(x, λ). Similarly we get Φ(x, λ) = λΦ(x, λ).

4) Denote ∆(λ) := V (ϕ). It follows from (1.6.73) and (1.6.74) for a = π that

∆(λ) = ∆(λ) +∞

k=0

ϕ(x, λ), ϕk0(x)|x=π

αk0(λ

−λk0)

∆(λk0) − ϕ(x, λ), ϕk1(x)|x=π

αk1(λ

−λk1)

∆(λk1)

, (1.6.85)

0 = V (Φ) +∞

k=0

Φ(x, λ), ϕk0(x)|x=π

αk0(λ − λk0)∆(λk0) − Φ(x, λ), ϕk1(x)|x=π

αk1(λ − λk1)∆(λk1)

. (1.6.86)

In (1.6.85) we set λ = λn1 :

0 = ∆(λn1) +∞

k=0

P n1,k0(π)∆(λk0) − P n1,k1(π)∆(λk1)

= 0.

Since P n1,k1(π) = δ nk we get∞

k=0

P n1,k0(π)∆(λk0) = 0.

Then, by virtue of Lemma 1.6.6, ∆(λk0) = 0, k ≥ 0.Substituting this into (1.6.86) and using the relation Φ(x, λ), ϕk1(x)|x=π = 0, we obtain

V (Φ) = 0, i.e. (1.6.72) is valid. Notice that we additionally proved that ∆(λn) = 0, i.e.the numbers λnn≥0 are eigenvalues of L. 2

It follows from (1.6.18) for x = 0 that

M (λ) = M (λ) +∞

k=0

1

αk0(λ − λk0)− 1

αk1(λ − λk1)

.

According to Theorem 1.4.6,

M (λ) =∞

k=0

1

αk1(λ − λk1)

,

hence

M (λ) =∞

k=0

1

αk(λ − λk).

Thus, the numbers λn, αnn≥0 are the spectral data for the constructed boundary valueproblem L. Notice that if (1.1.23) is valid we can choose L = L(q (x), h, H ) with q (x) ∈ W N

2

such that nN +1ξ n ∈ l2, and obtain that q (x) ∈ W N 2 . This completes the proof of Theorem

1.6.2. 2



72

Example 1.6.1. Let λn = n2 (n ≥ 0), αn =π

2(n ≥ 1), and let α0 > 0 be

an arbitrary positive number. We choose L such that q (x) = 0, h = H = 0. Then

λn = 0 (n ≥ 0), αn =π

2(n ≥ 1), and α0 = π. Denote a :=

1

α0

− 1

π. Clearly, a > − 1

π.

Then, (1.6.19) and (1.6.21) yield

1 = (1 + ax)ϕ00(x), ε0(x) = aϕ00(x),

and consequently,

ϕ00(x) =1

1 + ax, ε0(x) =

a

1 + ax.


q (x) = 2a

2

(1 + ax)2 , h = −a, H = a1 + aπ = aα0π . (1.6.87)

By (1.6.10),

ϕ(x, λ) = cos ρx − a

1 + ax

sin ρx

ρ.

1.6.3. The nonselfadjoint case. In the nonselfadjoint case Theorem 1.6.1 remainsvalid, i.e. by necessity the main equation (1.6.33) is uniquely solvable. Hence, the results of Subsection 1.6.1 are valid, and Algorithm 1.6.1 can be applied for the solution of the inverseproblem also in the nonselfadjoint case, but the sufficiency part in Theorem 1.6.2 must becorrected, since Lemma 1.6.6 is not valid in the general case. In the nonselfadjoint case wemust require that the main equation is uniquely solvable (see Condition S in Theorem 1.6.3).As it is shown in Example 1.6.2, Condition S is essential and cannot be omitted. On theother hand, we provide here classes of operators for which the unique solvability of the mainequation can be proved.

Let us consider a boundary value problem L of the form (1.1.1)-(1.1.2) where q (x) ∈L2(0, π) is a complex-valued function, and h, H are complex numbers. For simplicity, weconfine ourselves to the case when all zeros of the characteristic function ∆(λ) are simple.In this case we shall write L ∈ V. If L ∈ V we have

αn = 0, λn = λk (n = k), (1.6.88)

and (1.5.8) holds.

Theorem 1.6.3 For complex numbers λn, αnn≥0 to be the spectral data for a certain L(q (x), h , H ) ∈ V with q (x) ∈ L2(0, π), it is necessary and sufficient that

1) the relations (1.6.88) and (1.5.8) hold;2) (Condition S) for each fixed x ∈ [0, π], the linear bounded operator E + H (x), acting

from m to m, has a unique inverse operator. Here L ∈ V is taken such that ω = ω.The boundary value problem L(q (x), h , H ) can be constructed by Algorithm 1.6.1.

The proof of Theorem 1.6.3 is essentially the same as the proof of Theorem 1.6.2, but

only Lemma 1.6.6 must be omitted.



73

Example 1.6.2. Consider Example 1.6.1, but let α0 now be an arbitrary complexnumber. Then the main equation of the inverse problem takes the form

1 = (1 + ax)ϕ00(x),

and Condition S means that 1 + ax = 0 for all x ∈ [0, π], i.e. a /∈ (−∞, −1/π]. Hence,Condition S is fulfilled if and only if α0 /∈ (−∞, 0]. Thus, the numbers n2, αnn≥0; αn =π

2(n ≥ 1), are the spectral data if and only if α0 /∈ (−∞, 0]. The boundary value problem

can be constructed by (1.6.87). For the selfadjoint case we have α0 > 0, and Condition Sis always fulfilled.

In Theorem 1.6.3 one of the conditions under which arbitrary complex numbersλn, αnn≥0 are the spectral data for a boundary value problem L of the form (1.1.1)-(1.1.2)

is the requirement that the main equation is uniquely solvable (Condition S). This conditionis difficult to check in the general case. In this connection it is important to point out classesof operators for which the unique solvability of the main equation can be checked. Here wesingle out three such classes which often appear in applications.

(i) The selfadjoint case. In this case Condition S is fulfilled automatically (see Theorem1.6.2 above).

(ii) Finite-dimensional perturbations. Let a model boundary value problem L with thespectral data

λn, αn

n

≥0 be given. We change a finite subset of these numbers. In other

words, we consider numbers λn, αnn≥0 such that λn = λn and αn = αn for n > N, andarbitrary in the rest. Then, according to (1.6.19), the main equation becomes the followinglinear algebraic system:

ϕni(x) = ϕni(x) +N

k=0

(P ni,k0(x)ϕk0(x) − P ni,k1(x)ϕk1(x)), n = 0, N , i = 0, 1, x ∈ [0, π],

and Condition S is equivalent to the condition that the determinant of this system differs

from zero for each x ∈ [0, π]. Such perturbations are very popular in applications. We notethat for the selfadjoint case the determinant is always nonzero.

(iii) Local solvability of the main equation. For small perturbations of the spectral dataCondition S is fulfilled automatically. More precisely, the following theorem is valid.

Theorem 1.6.4 Let L = L(q (x), h, H ) ∈ V be given. There exists a δ > 0 (which depends on L ) such that if complex numbers λn, αnn≥0 satisfy the condition Ω < δ,then there exists a unique boundary value problem L(q (x), h , H ) ∈ V with q (x) ∈ L2(0, π)

for which the numbers λn, αnn≥0 are the spectral data, and

q − q L2(0,π) < C Ω, |h − h| < C Ω, |H − H | < C Ω, (1.6.89)

where C depends on L only.

Proof. In this proof C denotes various constants which depend on L only. SinceΩ < ∞, the asymptotical formulae (1.5.8) are fulfilled. Choose δ 0 > 0 such that if Ω < δ 0then (1.6.88) is valid.



74


H (x) ≤ C supn

k

ξ n|n − k| + 1

< C Ω.

Choose δ ≤ δ 0 such that if Ω < δ, then H (x) ≤ 1/2 for x ∈ [0, π]. In this case thereexists (E + H (x))−1, and

(E + H (x))−1 ≤ 2.

Thus, all conditions of Theorem 1.6.3 are fulfilled. By Theorem 1.6.3, there exists a uniqueboundary value problem L(q (x), h , H ) ∈ V for which the numbers λn, αnn≥0 are thespectral data. Moreover, (1.6.68) and (1.6.69) are valid. Repeating now the arguments inthe proof of Lemma 1.6.8 we get

max0≤x≤π |ε0(x)| ≤ C Ω, ε(x)L2(0,π) ≤ C Ω.

Together with (1.6.25)-(1.6.26) this yields (1.6.89). 2

Similarly we also can obtain the stability of the solution of the inverse problem in the uniform norm ; more precisely we get:

Theorem 1.6.5. Let L = L(q (x), h, H ) ∈ V be given. There exists δ > 0 (which depends on L ) such that if complex numbers λn, αnn≥0 satisfy the condition Ω1 < δ,

then there exists a unique boundary value problem L(q (x), h , H ) ∈ V for which the numbers λn, αnn≥0 are the spectral data. Moreover, the function q (x) − q (x) is continuous on [0, π], and

max0≤x≤π

|q − q | < C Ω1, |h − h| < C Ω1, |H − H | < C Ω1,

where C depends on L only.

Remark 1.6.4. Using the method of spectral mappings we can solve the inverseproblem not only in L2(0, π) and W N

2 (0, π) but also for other classes of potentials. As

example we formulate here the following theorem for the non-selfadjoint case.

Theorem 1.6.6. For complex numbers λn, αnn≥0 to be the spectral data for a certain L(q (x), h , H ) ∈ V with q (x) ∈ D ⊂ L(0, π), it is necessary and sufficient that 1) (1.6.88) holds,2) (Asymptotics) there exists L = L(q (x), h, H ) ∈ V with q (x) ∈ D such that nξ n ∈ l2,3) (Condition S) for each fixed x ∈ [0, π], the linear bounded operator E + H (x), acting

from m to m, has a unique inverse operator,4) ε(x)

∈D, where the function ε(x) is defined by (1.6.21).

The boundary value problem L(q (x), h , H ) can be constructed by Algorithm 1.6.1.

1.7. THE METHOD OF STANDARD MODELS

In the method of standard models we construct a sequence of model operators whichapproximate in an adequate sense the unknown operator and allow us to construct the

potential step by step. The method of standard models provides an effective algorithm for



75

the solution of inverse problems. This method has a wide area for applications. It works formany important classes of inverse problems when other methods turn out to be unsuitable.For example, in [yur9] the so-called incomplete inverse problems for higher-order differentialoperators were studied when only some part of the spectral information is accessible tomeasurement and when there is a priori information about the operator or its spectrum.

In [yur27] the method of standard models was applied for solving the inverse problem forsystems of differential equations with nonlinear dependence on the spectral parameter. Theinverse problem for integro-differential operators was treated in [yur13]. This method wasalso used for studying an inverse problem in elasticity theory [yur12].

However the method of standard models requires rather strong restrictions on the op-erator. For example, for the Sturm-Liouville operator the method works in the classes of analytic or piecewise analytic potentials on the interval [0, π]. We can also deal with moregeneral classes, for example, the class of piecewise operator-analytic functions (see [fag2]) or

other classes of functions which can be expanded in series which generalize the Taylor series.In this section we present the method of standard models using the Sturm-Liouville

operator as a model. In order to show ideas we confine ourselves to the case when thepotential q (x) in the boundary value problem (1.1.1)-(1.1.2) is an analytic function on[0, π]. Other, more complicated cases can be treated similarly.

Let Φ(x, λ) be the Weyl solution of (1.1.1) which satisfies the conditions U (Φ) =1, V (Φ) = 0, and let M (λ) := Φ(0, λ) be the Weyl function for the boundary value prob-lem L (see Subsection 1.4.4). From several equivalent formulations of the inverse problems

which were studied in Section 1.4 we will deal with here Inverse Problem 1.4.4 of recoveringL from the Weyl function.Let the Weyl function M (λ) of the boundary value problem L be given. Our goal is to

construct the potential q (x) which is an analytic function on [0, π] (to simplify calculationswe assume that the coefficients h and H are known). We take a model boundary valueproblem L = L(q (x), h , H ) with an analytic potential q (x). Since

−Φ(x, λ) + q (x)Φ(x, λ) = λΦ(x, λ), −Φ(x, λ) + q (x)Φ(x, λ) = λΦ(x, λ),

we get

(q (x) − q (x))Φ(x, λ)Φ(x, λ) = (Φ(x, λ)Φ(x, λ) − Φ(x, λ)Φ(x, λ)),and consequently π

0q (x)Φ(x, λ)Φ(x, λ) dx = M (λ), (1.7.1)

where q (x) = q (x) − q (x), M (λ) = M (λ) − M (λ).We are interested in the asymptotics of the terms in (1.7.1). First we prove an auxiliary

assertion. Denote Q = ρ : arg ρ ∈ [δ 0, π − δ 0], δ 0 > 0. Then there exists ε0 > 0 suchthat

|Im ρ| ≥ ε0|ρ| for ρ ∈ Q. (1.7.2)Lemma 1.7.1. Let

r(x) =xk

k!

γ + p(x)

, H (x, ρ) = exp(2iρx)

1 +

ξ (x, ρ)

ρ

, x ∈ [0, a],

where p(x) ∈ C [0, a], p(0) = 0, and where the function ξ (x, ρ) is continuous and bounded for x ∈ [0, a], ρ ∈ Q, |ρ| ≥ ρ∗. Then for ρ → ∞, ρ ∈ Q,

a

0

r(x)H (x, ρ) dx =(−1)k

(2iρ)k+1

(γ + o(1)).



76

Proof. We calculate

(2iρ)k+1(−1)k a

0r(x)H (x, ρ) dx = I 1(ρ) + I 2(ρ) + I 3(ρ),

where

I 1(ρ) = γ (2iρ)k+1(−1)k a

0

xk

k!exp(2iρx) dx,

I 2(ρ) = (2iρ)k+1(−1)k a

0

xk

k! p(x) exp(2iρx) dx,

I 3(ρ) = (2i)k+1(−ρ)k a

0r(x) exp(2iρx)ξ (x, ρ) dx.

Since

∞0

xk

k! exp(2iρx) dx =

(

−1)k

(2iρ)k+1 , ρ ∈ Q,

it follows that I 1(ρ) − γ → 0 as |ρ| → ∞, ρ ∈ Q.

Furthermore, take ε > 0 and choose δ = δ (ε) such that for x ∈ [0, δ ], | p(x)| <ε

2εk+1

0 ,

where ε0 is defined in (1.7.2). Then, using (1.7.2) we infer

|I 2(ρ)| <ε

2(2|ρ|ε0)k+1

δ

0

xk

k!exp(−2ε0|ρ|x) dx + (2|ρ|)k+1

a

δ

xk

k!| p(x)| exp(−2ε0|ρ|x) dx

<ε

2+ (2|ρ|)k+1 exp(−2ε0|ρ|δ )

a−δ

0

(x + δ )k

k!| p(x + δ )| exp(−2ε0|ρ|x) dx.

By arbitrariness of ε we obtain that I 2(ρ) → 0 for |ρ| → ∞, ρ ∈ Q.Since |(γ + p(x))ξ (x, ρ)| < C, then for |ρ| → ∞, ρ ∈ Q,

|I 3(ρ)| < C |ρ|k a

0

xk

k!exp(−2ε0|ρ|x) dx ≤ C

|ρ|εk+10

→ 0.

2

Suppose that for a certain fixed k ≥ 0 the Taylor coefficients q j := q ( j)(0), j = 0, k − 1,have been already found. Let us choose a model boundary value problem L such that thefirst k Taylor coefficients of q and q coincide, i.e. q j = q j, j = 0, k − 1. Then, using(1.7.1) and Lemma 1.7.1 we can calculate the next Taylor coefficient q k = q (k)(0). Namely,the following assertion is valid.

Lemma 1.7.2. Fix k. Let the functions q (x) and q (x) be analytic for x ∈ [0, a], a >0, with q j := q j

−q j = 0 for j = 0, k

−1. Then

q k =1

4(−1)k lim

|ρ|→∞ρ∈Q

(2iρ)k+3M (λ). (1.7.3)

Proof. It follows from (1.1.10), (1.1.16) and (1.4.9) that

Φ(x, λ) =1

iρexp(iρx)

1 + O

1

ρ

.



77

Hence, taking (1.7.1) into account, we have

M (λ) = π

0q (x)Φ(x, λ)Φ(x, λ) dx =

a

0

xk

k!q (k)(0) + p(x) 1

(iρ)2 exp(2iρx)1 +ξ (x, ρ)

ρ dx + π

a q (x)Φ(x, λ)Φ(x, λ) dx.

Using Lemma 1.7.1 we obtain for |ρ| → ∞, ρ ∈ Q :

M (λ) =4(−1)k

(2iρ)k+3

q (k)(0) + o(1)

.

2

Thus, we arrive at the following algorithm for the solution of Inverse Problem 1.4.4 inthe class of analytic potentials.

Algorithm 1.7.1. Let the Weyl function M (λ) of the boundary value problem (1.1.1)-(1.1.2) with an analytic potential be given . Then:

(i) We calculate q k = q (k)(0), k ≥ 0. For this purpose we successively perform the following operations for k = 0, 1, 2, . . . : We construct a model boundary value problem Lwith an analytic potential q (x) such that q j = q j, j = 0, k − 1 and arbitrary in the rest,and we calculate q k = q (k)(0) by (1.7.3).

(ii) We construct the function q (x) by the formula

q (x) =∞

k=0

q kxk

k!, 0 < x < R,

where

R =

limk→∞

|q k|k!

1/k−1.

If R < π then for R < x < π the function q (x) (which is analytic on [0, π] ) is constructed

by analytic continuation or by the step method which is described below in Remark 1.7.1.

Remark 1.7.1. The method of standard models also works when q (x) is a piecewiseanalytic function. Indeed, the functions

Φa(x, λ) =Φ(x, λ)

Φ(a, λ), M a(λ) =

Φ(a, λ)

Φ(a, λ)=

S (a, λ) + M (λ)ϕ(a, λ)

S (a, λ) + M (λ)ϕ(a, λ)

are the Weyl solution and the Weyl function for the interval x∈

[a, π] respectively. Supposethat q (x) has been already constructed on [0, a]. Then we can calculate M a(λ) and go onto the interval [a, π].

1.8. LOCAL SOLUTION OF THE INVERSE PROBLEM

In this section we present a method for the local solution of Inverse Problem 1.4.2 of recovering Sturm-Liouville operators from two spectra. Without loss of generality we will

consider the case when H = 0. In this method, which is due to Borg [Bor1], the inverse



78

problem is reduced to a nonlinear integral equation (see (1.8.13)) which can be solved locally.An important role in Borg‘s method is played by products of eigenfunctions of the boundaryvalue problems considered. In order to derive and study Borg‘s nonlinear equation one mustprove that such products are complete or form a Riesz basis in L2(0, π). For uniquenesstheorems it is sufficient to prove the completeness of the products (see Subsection 1.4.3).

For the local solution of the inverse problem and for studying stability of the solution suchproducts must be a Riesz basis. For convenience of the reader we provide in Subsection 1.8.5necessary information about Riesz bases in a Hilbert space.

We note that for Sturm-Liouville operators Borg‘s method is weaker than the Gelfand-Levitan method and the method of spectral mappings. However, Borg‘s method turns outto be useful when other methods do not work (see, for example, [kha1],[dub1],[yur13] andthe references therein).

1.8.1. Derivation of the Borg equation. Let λni = ρ2ni, n ≥ 0, i = 1, 2, be the

eigenvalues of the boundary value problems Li :

y := −y + q (x)y = λy, (1.8.1)

y(0) − hy(0) = y(i−1)(π) = 0, (1.8.2)

where q (x) ∈ L2(0, π) is a real function, and h is a real number. Then (see Section 1.1)

ρn1 = n +

1

2 +

a

n +

κn1

n , ρn2 = n +

a

n +

κn2

n , (1.8.3)

where

κni ∈ l2, a =1

π

h +

1

2

π

0q (t) dt

.

Let Li and Li, i = 1, 2, be such that a = a, then

Λ := ∞

n=0 |λn1

−λn1

|2 +

|λn2

−λn2

|2

1/2

<

∞.

Denote

yni(x) = ϕ(x, λni), yni(x) = ϕ(x, λni), sni(x) = S (x, λni), sni(x) = S (x, λni),

Gni(x, t) =

S (x, λni)C (t, λni) − S (t, λni)C (x, λni), 0 ≤ t ≤ x ≤ π,0, 0 ≤ x < t ≤ π,

where ϕ(x, λ), C (x, λ) and S (x, λ) are solutions of (1.8.1) under the conditions C (0, λ) =

ϕ(0, λ) = S (0, λ) = 1, C (0, λ) = S (0, λ) = 0, ϕ(0, λ) = h.Since

yni(x) = λniyni(x), yni(x) = λniyni(x),

we get π

0r(x)yni(x)yni(x) dx =

π

0(yni(x)yni(x) − yni(x)yni(x)) dx

=

yni(x)yni(x) − yni(x)yni(x)

π

0, (1.8.4)



79

where r := q − q. Let

p = −1

2

π

0r(x) dx. (1.8.5)

From the equality a = a, it follows that p = h − h; hence we infer from the boundaryconditions that (1.8.4) takes the form π

0r(x)yni(x)yni(x) dx = yni(π)yni(π) − yni(π)yni(π) − p, n ≥ 0, i = 1, 2. (1.8.6)

Since yni(x) are eigenfunctions for the boundary value problems Li, we have

yn1(π) = 0, yn2(π) = 0, n ≥ 0. (1.8.7)

Furthermore, one can easily verify by differentiation that the function yni(x) satisfies

the following Volterra integral equation

yni(x) = yni(x) + psni(x) + π

0Gni(x, t)r(t)yni(t) dt, n ≥ 0, i = 1, 2. (1.8.8)

Solving (1.8.8) by the method of successive approximations we deduce

yni(x) = yni(x) + psni(x) + ϕni(x), n ≥ 0, i = 1, 2, (1.8.9)

whereϕni(x) =

∞ j=1

π

0. . .

π

0

j

Gni(x, t1)Gni(t1, t2) . . . Gni(t j−1, t j )×

r(t1) . . . r(t j)(yni(t j) + psni(t j)) dt1 . . . d t j. (1.8.10)

Substituting (1.8.9)-(1.8.10) into (1.8.6) and using (1.8.5) and (1.8.7), we get

π

0r(x)ξ ni(x) dx = ωni, n

≥0, i = 1, 2, (1.8.11)

where

ξ ni(x) = y2ni(x) − 1

2,

ωn1 = yn1(π)(yn1(π) + psn1(π) + ϕn1(π))− π

0 pr(x)sn1(x)yn1(x) dx−

π

0r(x)yn1(x)ϕn1(x) dx,

ωn2 = −yn2(π)(yn2(π)+ psn2(π)+ϕn2(π))− π

0 pr(x)sn2(x)yn2(x) dx−

π

0r(x)yn2(x)ϕn2(x) dx.


ξ ni(x) =1

2

cos2ρnix +

x

0V 1(x, t)cos2ρnit dt

, (1.8.12)

where V 1(x, t) is a continuous function. We introduce the functions ηn(x)n≥0 and thenumbers z nn≥0 by the formulae

η2n(x) = ξ n2(x), η2n+1(x) = ξ n1(x), z 2n = 2ρn2, z 2n+1 = 2ρn1.



80

Then, according to (1.8.3) and (1.8.12),

z n = n +a

n+

κn

n, κn ∈ l2,

ηn(x) =

1

2 cos z nx + x

0 V 1(x, t)cos z nt dt.

Hence, by virtue of Propositions 1.8.6 and 1.8.2, the set of functions ηn(x)n≥0 formsa Riesz basis in L2(0, π). Denote by χn(x)n≥0 the corresponding biorthonormal basis.Then from (1.8.11) we infer

r(x) =∞

n=0

ωn2χ2n(x) + ωn1χ2n+1(x)

,

and consequently, in view of (1.8.5),

r(x) = f (x) +∞

j=1

π

0. . .

π

0 j

H j (x, t1, . . . t j)r(t1) . . . r(t j) dt1 . . . d t j, (1.8.13)

where

f (x) =∞

n=0

(−yn2(π)yn2(π)χ2n(x) + yn1(π)yn1(π)χ2n+1(x)), (1.8.14)

H 1(x, t1) =∞

n=0

− yn2(π)

Gn2(π, t1)yn2(t1) − 1

2sn2(π)

χ2n(x)

+yn1(π)∂Gn1(x, t1)

∂xx=π

yn1(t1) − 1

2sn1(π)

χ2n+1(x)

, (1.8.15)

H j (x, t1, . . . , t j) =∞

n=0

−

yn2(t1) + yn2(π)Gn2(π, t1)

Gn2(t1, t2) . . . Gn2(t j−2, t j−1)×

Gn2(t j−1, t j)yn2(t j ) − 1

2sn2(t j−1)

χ2n(x) −

yn1(t1) − yn1(π)∂Gn1(x, t1)∂x

x=π

×Gn1(t1, t2) . . . Gn1(t j−2, t j−1)

Gn1(t j−1, t j)yn1(t j) − 1

2sn1(t j−1)

χ2n+1(x)

, j ≥ 2. (1.8.16)

Equation (1.8.13) is called the Borg equation.

Remark 1.8.1. In order to prove that the system of functions ηn(x)n≥0 forms aRiesz basis in L2(0, π) we used above the transformation operator. However in many casesof practical interest the Borg method can be applied, although the transformation operatordoes not work. In Subsection 1.8.3 we describe Borg‘s original idea which allows us to provethat the products of eigenfunctions form a Riesz basis in L2(0, π).

1.8.2. The main theorem. Using the Borg equation we can prove the followingtheorem on the local solution of the inverse problem and on the stability of the solution.

Theorem 1.8.1. For the boundary value problems Li of the form (1.8.1)-(1.8.2), there exists δ > 0 (which depends on L

i) such that if real numbers

λ

nin≥0, i = 1, 2, satisfy



81

the condition Λ < δ, then there exists a unique real pair q (x) ∈ L2(0, π) and h, for which the numbers λnin≥0, i = 1, 2, are the eigenvalues of Li. Moreover,

q − q L2 < C Λ, |h − h| < C Λ. (1.8.17)

Here and below C denotes various positive constants which depend on Li.Proof. One can choose δ 1 > 0 such that if real numbers λnin≥0, i = 1, 2, satisfy the

condition Λ < δ 1, then λni = λkj for (n, i) = (k, j), and

yn2(π) = 0, yn1(π) = 0 for all n ≥ 0. (1.8.18)

Indeed, according to Theorem 1.1.1, ϕ(x, λn2) is an eigenfunction of the boundary valueproblem L2, and ϕ(π, λn2) = 0, ϕ(π, λn2) = 0 for all n ≥ 0. Moreover, by virtue of (1.1.9) and (1.8.3),

ϕ(π, λn2) = cos nπ + O 1

n

= (−1)n + O

1

n

.

Thus,|ϕ(π, λn2)| ≥ C > 0.

On the other hand, using (1.3.11) we calculate

ϕ(π, λn2) − ϕ(π, λn2) = (cos ρn2π − cos ρn2π) + π

0 G(π, t)(cos ρn2t − cos ρn2t) dt,

and consequently|ϕ(π, λn2) − ϕ(π, λn2)| < C |ρn2 − ρn2|.

Then for sufficiently small Λ we have

yn2(π) := ϕ(π, λn2) = 0 for all n ≥ 0.

Similarly, one can derive the second inequality in (1.8.18).It is easy to verify that the following estimates are valid for n ≥ 0, i = 0, 1, 0 ≤ x, t ≤ π,

|yni(x)| < C, |yn1(π)| < C (n + 1), |∂Gni(x, t)

∂x| < C, |Gni(x, t)| <

C

n + 1,

|yn2(π)| < C |λn2 − λn2|, |yn1(π)| <C

n + 1|λn1 − λn1|, |sn1(π)| <

C

n + 1.

Indeed, the estimate|yni(x)

|< C follows from (1.1.9) and (1.8.3). Since ϕ(π, λn2) = 0,

we getyn2(π) = ϕ(π, λn2) − ϕ(π, λn2).


ϕ(x, λ) = −ρ sin ρx + G(x, x)cos ρx + x

0

∂G(x, t)

∂xcos ρtdt.

Hence

ϕ(π, λn2) − ϕ(π, λn2) = −ρn2 sin ρn2π + ρn2 sin ρn2π



82

+G(π, t)(cos ρn2π − cos ρn2π) + π

0

∂G(x, t)

∂xx=π

(cos ρn2t − cos ρn2t) dt,

and consequently|yn2(π)| < C |λn2 − λn2|.

The other estimates are proved similarly.

Using (1.8.14)-(1.8.16) we obtain

f < C Λ, H 1 < C Λ, H j < C j, ( j ≥ 2), (1.8.19)

where . is the norm in L2 with respect to all of the arguments.Indeed, according to (1.8.14),

f (x) =

∞n=0

f nχn(x),

wheref 2n = −yn2(π)yn2(π), f 2n+1 = yn1(π)yn1(π).

Then, using the preceding estimates for yni(x), we get

|f 2n| ≤ C |λn2 − λn2|, |f 2n+1| ≤ C |λn1 − λn1|.

By virtue of (1.8.58) this yields f < C Λ. Other estimates in (1.8.19) can be verifiedsimilarly.

Consider in L2(0, π) the nonlinear integral equation (1.8.13). We rewrite (1.8.13) in theform

r = f + Ar,

where

Ar =∞

j=1

A jr,

(A jr)(x) = π

0. . .

π

0 j

H j(x, t1, . . . t j)r(t1) . . . r(t j) dt1 . . . d t j.


A1r ≤ C Λr, A1r − A1r ≤ C Λr − r,

A j r ≤ (C r)

j

, A jr − A j r ≤ r − rC max(r, r) j−1

, j ≥ 2,f < C Λ.

(1.8.20)

Fix C ≥ 1/2 such that (1.8.20) is valid. If

r ≤ 1

2C , r ≤ 1

2C ,

then (1.8.20) implies

Ar ≤ C Λr + 2C 2

r2

,



83

Ar − Ar ≤ C r − r

Λ + 2 max(r, r)

.

Moreover, if

Λ ≤ 1

4C , r ≤ 1

8C 2, r ≤ 1

8C 2,

then Ar ≤ 12r, Ar − Ar ≤ 1

2r − r. (1.8.21)

Therefore, there exists sufficiently small δ 2 > 0 such that if Λ < δ 2, then equation (1.8.13)can be solved by the method of successive approximations:

r0 = f, rk+1 = f + Ark, k ≥ 0,

r = r0 +∞

k=0

(rk+1

−rk). (1.8.22)

Indeed, put δ 2 = 116C 3

. If Λ < δ 2, then f ≤ 116C 2

, Λ ≤ 14C

. Using (1.8.21), by inductionwe get

rk ≤ 2f , rk+1 − rk ≤ 1

2k+1f , k ≥ 0.

Consequently, the series (1.8.22) converges in L2(0, π) to the solution of equation (1.8.13).Moreover, for this solution the estimate r ≤ 2f is valid.

Let δ = min(δ 1, δ

2), and let r(x) be the above constructed solution of (1.8.13). Define

p via (1.8.5) and q (x), h by the formulas q = q + r, h = h + p. Clearly, (1.8.17) is valid.Thus, we have constructed the boundary value problems Li.

It remains to be shown that the numbers λnin≥0 are the eigenvalues of the constructedboundary value problems Li. For this purpose we consider the functions yni(x), which aresolutions of equation (1.8.8). Then (1.8.9)-(1.8.10) hold. Clearly,

−yni(x) + q (x)yni(x) = λniyni(x), yni(0) = 1, yni(0) = h,

and consequently (1.8.6) is valid.Furthermore, multiplying (1.8.13) by ηn(x), integrating with respect to x and taking

(1.8.9)-(1.8.10) into account, we obtain π

0r(x)yn1(x)yn1(x) dx = yn1(π)yn1(π) − p,

π

0r(x)yn2(x)yn2(x) dx = −yn2(π)yn2(π) − p.

Comparing these relations with (1.8.6) and using (1.8.18), we conclude that (1.8.7) holds,i.e. yni(x) are eigenfunctions, and λnin≥0 are eigenvalues for Li. The uniqueness followsfrom Borg‘s theorem (see Theorem 1.4.4). 2

1.8.3. The case of Dirichlet boundary conditions. Similar results are also valid forDirichlet boundary conditions; in this subsection we prove the corresponding theorem (seeTheorem 1.8.2). The most interesting part of this subsection is Borg‘s original method forproving that products of eigenfunctions form a Riesz basis in L2(0, π) (see Remark 1.8.1).



84

Let λ0ni = (ρ0

ni)2, n ≥ 1, i = 1, 2, be the eigenvalues of the boundary value problemsL0

i of the form−y + q (x)y = λy, q (x) ∈ L2(0, π), (1.8.23)

y(0) = y(i−1)(π) = 0 (1.8.24)

with real q. Then (see Section 1.1)

ρ0n1 = n +

a0

n+

κ0n1

n, ρ0

n2 =

n − 1

2

+

a0

n+

κ0n2

n,

κ0

ni

∈ l2, a0 =

1

2π

π

0q (t) dt.

Let L0i and L0

i , i = 1, 2 be such that a0 = a0, then

Λ0 := ∞

n=1

|λ0

n1 − λ0n1|2 + |λ0

n2 − λ0n2|2

1/2< ∞.

Hence π

0r(t) dt = 0, (1.8.25)

where r = q − q.Acting in the same way as in Subsection 1.8.1 and using the same notations we get π

0r(x)sni(x)sni(x) dx = sni(π)sni(π) − sni(π)sni(π), n ≥ 1, i = 1, 2. (1.8.26)

sn1(π) = 0, sn2(π) = 0, n ≥ 1. (1.8.27)

sni(x) = sni(x) + π

0Gni(x, t)r(t)sni(t) dt, n ≥ 1, i = 1, 2. (1.8.28)

Solving (1.8.28) by the method of successive approximations we obtain

sni(x) = sni(x) + ψni(x), (1.8.29)

where

ψni(x) =

∞ j=1

π

0. . . π

0

j

Gni(x, t1)Gni(t1, t2) . . . Gni(t j−1, t j)×

r(t1) . . . r(t j)sni(t j) dt1 . . . d t j. (1.8.30)

Substituting (1.8.29)-(1.8.30) into (1.8.26) and using (1.8.27), we get π

0r(x)uni(x) dx = ω0

ni, n ≥ 1, i = 1, 2, (1.8.31)

whereu0

ni(x) = 2n2s2ni(x),

ω0n1 = 2n2

sn1(π)(sn1(π) + ψ

n1(π)) − π

0r(x)sn1(x)ψn1(x) dx

,

ω0n2 = 2n2

− sn2(π)(sn2(π) + ψn2(π)) −

π

0r(x)sn2(x)ψn2(x) dx

.

(1.8.32)

We introduce the functions un(x)n≥1 and the numbers ωnn≥0 by the formulae

u2n(x) := u0n1(x), u2n−1(x) := u

0n2(x), ω2n := ω

0n1, ω2n−1 := ω

0n2, (n ≥ 1), ω0 := 0.



85

Then (1.8.31) takes the form π

0r(x)un(x) dx = ωn, n ≥ 1. (1.8.33)

Denote

vn(x) := 1n

un(x), wn(x) := 1 − un(x) (n ≥ 1), w0(x) := 1.

Using the results of Section 1.1, one can easily calculate

vn(x) = sin nx + O1

n

, wn(x) = cos nx + O

1

n

, n → ∞. (1.8.34)

By virtue of (1.8.25) and (1.8.33) we have

π

0 r(x)wn(x) dx = −ωn, n ≥ 0. (1.8.35)

Lemma 1.8.1. The sets of the functions vn(x)n≥1 and wn(x)n≥0 are both Riesz bases in L2(0, π).

Proof. In order to prove the lemma we could use the transformation operator as above,but we provide here another method which is due to G. Borg [Bor1].

Let q (x) ∈ W 12 , and let y be a solution of (1.8.23) satisfying the condition y(0) = 0.Similarly, let y be such that

−y + q (x)y = λy, y(0) = 0, where q (x)

∈W 12 . Denote

u = yy.

Thenu = yy + yy, u = −2λu + (q + q )u + 2yy,

u + 4λu − (q + q )u − (q + q )u = 2(qyy + qy y).

From this, taking into account the relations

2(qyy + qy y) = (q + q )u + (q − q )(yy − yy), (yy − yy)(x) = x

0(q (s) − q (s))u(s) ds,

we derive

u + 4λu − 2(q (x) + q (x))u − (q (x) + q (x))u = (q (x) − q (x)) x

0(q (s) − q (s))u(s) ds.

Denote v = u. Since u(0) = u(0) = 0, we have v(0) = 0, u(x) = x

0v(s) ds, and

consequently

−v + 2(q (x) + q (x))v + x

0N (x, s)v(s) ds = 4λv, (1.8.36)

whereN (x, s) = q (x) + q (x) + (q (x) − q (x))

x

s(q (ξ ) − q (ξ )) dξ.

In particular, for q = q we obtain that the functions vn(x)n≥1 are the eigenfunctions forthe boundary value problem

−v + p(x)v + x

0 M (x, s)v(s) ds = 4λv, v(0) = v(π) = 0,



86

where p(x) = 4q (x), M (x, t) = 2q (x). Then for the sequence vn(x)n≥1, there exists thebiorthonormal sequence v∗n(x)n≥1 in L2(0, π) ( v∗n(x) are eigenfunctions of the adjointboundary value problem

−v∗ + p(x)v∗ +

π

x

M (s, x)v∗(s) ds = 4λv∗, v∗(0) = v∗(π) = 0.

By virtue of (1.8.34) and Proposition 1.8.4 this gives that vn(x)n≥1 is a Riesz basis inL2(0, π).

Furthermore, let us show that the system of the functions wn(x)n≥0 is complete inL2(0, π). Indeed, suppose that π

0f (x)wn(x) dx = 0, n ≥ 0, f (x) ∈ L2(0, π).

This yields π

0f (x) dx = 0,

and consequently π

0f (x)un(x) dx = 0, n ≥ 1.

Integrating by parts we infer, using vn(0) = vn(π) = 0, that

π

0 vn(x) dx π

x f (t) dt = 0, n ≥ 1.

The system vn(x)n≥1 is complete in L2(0, π). Hence π

xf (t) dt = 0 for x ∈ [0, π], i.e.

f (x) = 0 a.e. on (0, π). Therefore wn(x)n≥0 is complete in L2(0, π). Since wn(x)n≥0

is, in view of (1.8.34), quadratically close to the Riesz basis cos nxn≥0, it follows fromProposition 1.8.5 that wn(x)n≥0 is a Riesz basis in L2(0, π). 2

Remark 1.8.2. With the help of equation (1.8.36) one can also prove the completeness

of the products of eigenfunctions sni(x)sni(x), and hence provide another method of provingBorg‘s uniqueness theorem (see Section 1.4) without using the transformation operator.

Denote by v∗n(x)n≥1 and w∗n(x)n≥0 the biorthonormal bases to vn(x)n≥1 and

wn(x)n≥0 respectively. Then from (1.8.35) we infer (like in Subsection 1.8.1)

r(x) = f 0(x) +∞

j=1

π

0. . .

π

0

j

H 0 j (x, t1, . . . , t j )r(t1) . . . r(t j) dt1 . . . d t j, (1.8.37)

where

f 0(x) = −∞

n=1

2n2(sn1(π)sn1(π)v∗2n(x) − sn2(π)sn2(π)v∗2n−1(x)),

H 01 (x, t1) = −∞

n=1

2n2

sn1(π)∂Gn1(x, t1)

∂xx=π

sn1(t1)v∗2n(x) − sn2(π)Gn2(π, t1)sn2(t1)v∗2n−1(x)

,

H 0 j (x, t1, . . . , t j) = −∞

n=1

2n2

sn1(π)

∂Gn1(x, t1)

∂x x=π− sn1(t1)

Gn1(t1, t2) . . . Gn1(t j−1, t j )



87

sn1(t j)v∗2n(x)−

sn2(π)Gn2(π, t1)+sn2(t1)

Gn2(t1, t2) . . . Gn2(t j−1, t j)sn2(t j )v∗2n−1(x)

, j ≥ 2.

Using the nonlinear equation (1.8.37), by the same arguments as in Subsection 1.8.2, wearrive at the following theorem.

Theorem 1.8.2. For the boundary value problems L0

iof the form (1.8.23)-(1.8.24),

there exists δ > 0 such that if real numbers λ0nin≥1, i = 1, 2, satisfy the condition

Λ0 < δ, then there exists a unique real function q (x) ∈ L2(0, π), for which the numbers λ0

nin≥1, i = 1, 2, are the eigenvalues of L0i . Moreover,

q − q L2 < C Λ0.

Remark 1.8.3. Using the Riesz basis vn(x)n≥1 we can also derive another nonlinearequation. Indeed, denote

z (x) := π

xr(t) dt.

After integration by parts, (1.8.33) takes the form π

0z (x)vn(x) dx =

ωn

n, n ≥ 1.

Since vn(x)n≥1 and v∗n(x)n≥1 are biorthonormal bases, this yields

z (x) = ∞n=1

ω

0

nn v∗n(x)

or π

xr(t) dt =

∞n=1

ωn

nv∗n(x), r(x) = −

∞n=1

ωn

n

d

dxv∗n(x).

From this, in view of (1.8.30) and (1.8.32), we get analogously to Subsection 1.8.1

r(x) = f 1(x) +∞

j=1

π

0

. . . π

0 j

H j1(x, t1, . . . , t j)r(t1) . . . r(t j) dt1 . . . d t j,

where

f 1(x) = −∞

n=1

2n(sn1(π)sn1(π)d

dxv∗2n(x) − sn2(π)sn2(π)

d

dxv∗2n−1(x)),

H 11(x, t1) = −∞

n=1

2n


∂x

x=π

sn1(t1)d

dxv∗2n(x)

−sn2(π)Gn2(π, t1)sn2(t1)d

dxv∗2n−1(x)

,

H j1(x, t1, . . . , t j) = −∞

n=1

2n


∂xx=π

−sn1(t1)

Gn1(t1, t2) . . . Gn1(t j−1, t j)sn1(t j)d

dxv∗2n(x)

−sn2(π)Gn2(π, t1) + sn2(t1)Gn2(t1, t2) . . . Gn2(t j−1, t j)sn2(t j)

d

dxv∗2n−1(x), j ≥ 2.



88

1.8.4. Stability of the solution of the inverse problem in the uniform norm. Letλni = ρ2

ni, n ≥ 0, i = 1, 2, be the eigenvalues of the boundary value problems Li of the form(1.8.1)-(1.8.2), where q is a real continuous function, and h is a real number. The eigenval-ues λni coincide with zeros of the characteristic functions ∆i(λ) := ϕ(i−1)(π, λ), i = 1, 2,where ϕ(x, λ) is the solution of (1.8.1) under the conditions ϕ(0, λ) = 1, ϕ(0, λ) = h.

Without loss of generality we shall assume in the sequel that in (1.8.3) a = 0. In this case,by virtue of (1.8.3), we have

∞n=0

|ρn1 − (n + 1/2)| + |ρn2 − n|

< ∞. (1.8.38)

Let the boundary value problems Li be chosen such that

Λ1 :=

∞n=0

|λn1 − λn1| + |λn2 − λn2| < ∞. (1.8.39)

The quantity Λ1 describes the distance of the spectra.

Theorem 1.8.3. There exists δ > 0 (which depends on Li ) such that if Λ1 < δ then

max0≤x≤π

|q (x) − q (x)| < C Λ1, |h − h| < C Λ1.

Here and below, C denotes various positive constants which depend on Li.

We first prove some auxiliary propositions. Let

αn := π

0ϕ2(x, λn2) dx

be the weight numbers for L2.

Lemma 1.8.2. There exists δ 1 > 0 such that if Λ1 < δ 1 then

∞n=0

|αn − αn| < C Λ1. (1.8.40)

Proof. According to (1.1.38),

αn = −∆2(λn2)∆1(λn2), (1.8.41)

where ∆2(λ) =d

dλ∆2(λ).

The functions ∆i(λ) are entire in λ of order 1/2, and consequently by Hadamard‘sfactorization theorem [con1, p.289], we have

∆i(λ) = Bi

∞k=0

1 − λ

λki

(1.8.42)

(the case when λ = 0 is the eigenvalue of Li requires minor modifications). Then

∆i(λ)

∆i(λ)

=Bi

Bi

∞

k=0

λki

λki

∞

k=0 1 +λki − λki

λki − λ .



89

Since

limλ→−∞

∆i(λ)

∆i(λ)= 1, lim

λ→−∞

∞k=0

1 +

λki − λki

λki − λ

= 1,

it follows thatBi

Bi

∞k=0

λki

λki= 1, (1.8.43)

and consequently∆i(λ)

∆i(λ)=

∞k=0

λki − λ

λki − λ.


˙∆2(λn2) = −

B2

λn2

∞

k=0k=n

1 −λn2

λk2.

Therefore, taking (1.8.43) into account we calculate

˙∆2(λn2)

∆2(λn2)=

∞k=0k=n

λk2 − λn2

λk2 − λn2.

Then, by virtue of (1.8.41),

αn

αn=

∞k=0

λk1 − λn2

λk1 − λn2

∞k=0k=n

λk2 − λn2

λk2 − λn2

orαn

αn=

∞k=0

(1 − θ(1)kn )

∞k=0k=n

(1 − θ(2)kn ), (1.8.44)

where

θ(i)kn =

λki − λki

λki − λn2+

λn2 − λn2

λki − λn2.

Denote

θn =∞

k=0

|θ(1)kn | +

∞k=0k=n

|θ(2)kn |.

Then

∞n=0

θn ≤ ∞n=0

∞k=0

|λk1 −˜λk1||λk1 − λn2| + |

˜λn2 − λn2||λk1 − λn2|

+∞

n=0

∞k=0k=n

|λk2 − λk2||λk2 − λn2| +

|λn2 − λn2||λk2 − λn2|

≤∞

n=0

|λn2 − λn2|

2∞k=0k=n

1

|λk2 − λn2| +∞

k=0

1

|λk1 − λn2|



90

+∞

n=0

|λn1 − λn1|∞

k=0

1

|λn1 − λk2| . (1.8.45)

Using the asymptotics of the eigenvalues (see Section 1.1), we obtain

1

|λk2 − λn2| <

C

|k2 − n2| , k = n.

Since ∞k=1k=n

1

|k2 − n2| < 1, n ≥ 1,

we have ∞

k=1k=n

1

|λk2

−λn2

|

< C. (1.8.46)

Similarly,∞

k=0

1

|λk1 − λn2| +1

|λn1 − λk2|

< C. (1.8.47)

It follows from (1.8.45)-(1.8.47) that

∞n=0

θn < C Λ1. (1.8.48)

Choose δ 1 > 0 such that if Λ1 < δ 1 then θn < 1/4. Since for |ξ | < 1/2,

| ln(1 − ξ )| ≤∞

k=1

|ξ |k

k≤

∞k=1

|ξ |k ≤ 2|ξ |,


lnαn

αn ≤∞

k=0 |ln(1

−θ

(1)kn )

|+

∞

k=0k=n |

ln(1

−θ

(2)kn )

|< 2θn.

Using the properties of ln, we get αn

αn

− 1 < 4θn or |αn − αn| < Cθn. Using (1.8.48) we

arrive at (1.8.40). 2

Lemma 1.8.3. There exists δ 2 > 0 such that if Λ1 < δ 2 then

|G(x, t) − G(x, t)| < C Λ1, 0 ≤ x ≤ t ≤ π, (1.8.49)

|ϕ(π, λn1)ϕ(π, λn1)| < |λn1 − λn1|, (1.8.50)

|ϕ(π, λn2)ϕ(π, λn2)| < C |λn2 − λn2|. (1.8.51)

Proof. The function G(x, t) is the solution of the Gelfand-Levitan equation (1.5.11).By virtue of Lemma 1.5.6, |F (x, t)| < C Λ1. Hence, from the unique solvability of theGelfand-Levitan equation and from Lemma 1.5.1 we obtain (1.8.49).

Furthermore, it follows from (1.1.9) and (1.8.3) that

|ϕ(π, λn1)| < C (n + 1).



91


ϕ(π, λn1) = ϕ(π, λn1) − ϕ(π, λn1) = (cos ρn1π − cos ρn1π) + π

0G(π, t)(cos ρn1t − cos ρn1t) dt.

Hence, for Λ1 < δ 2 we have in view of (1.8.49) that

|ϕ(π, λn1)| < C |ρn1 − ρn1|,and we arrive at (1.8.50). Similarly we obtain (1.8.51). 2

Lemma 1.8.4. Let g(x) be a continuous function on [0, π], and let z nn≥0 be num-bers such that ∞

n=0

|z n − n| < ∞.

Suppose that

Θ :=∞

n=0

|εn| < ∞, εn := π

0g(x)cos z nxdx.

Then |g(x)| < M Θ, where the constant M depends only on the set z nn≥0.

Proof. Since the system of the functions cos z nxn≥0 is complete in L2(0, π), it followsthat εn uniquely determine the function g(x). From the equality

π

0 g(x)cos nxdx = εn + π

0 g(x)(cos nx − cos z nx) dx

we obtain

g(x) =∞

n=0

εn

α0n

cos nx +∞

n=0

1

α0n

cos nx π

0(cos nt − cos z nt)g(t) dt.

Hence, the function g(x) is the solution of the integral equation

g(x) = ε(x) + π

0 H (x, t)g(t) dt, (1.8.52)

where

ε(x) =∞

n=0

εn

α0n

cos nx, H (x, t) =∞

n=0

1

α0n

cos nx(cos nt − cos z nt).

The series converge absolutely and uniformly for 0 ≤ x, t ≤ π, and

|ε(x)

|<

2

π

Θ,

|H (x, t)

|< C

∞

n=0 |z n

−n

|.

Let us show that the homogeneous equation

y(x) = π

0H (x, t)y(t) dt, y(x) ∈ C [0, π] (1.8.53)

has only the trivial solution. Indeed, it follows from (1.8.53) that

y(x) =∞

n=0

1

α0n

cos nx π

0

(cos nt

−cos z nt)y(t) dt.



92

Hence π

0y(t)cos ntdt =

π

0(cos nt − cos z nt)y(t) dt

or π

0y(t)cos z nt dt = 0.

From this, in view of the completeness of the system cos z nxn≥0, follows that y(x) = 0.Thus, the integral equation (1.8.52) is uniquely solvable, and hence |g(x)| < M Θ. 2

Remark 1.8.4. It is easy to see from the proof of Lemma 1.8.4 that if we considernumbers z n such that

∞n=0

|z n − z n| < δ

for sufficiently small δ, then the constant M will not depend on z n.

Proof of Theorem 1.8.3. Using

−ϕ(x, λ) + q (x)ϕ(x, λ) = λϕ(x, λ), −ϕ(x, λ) + q (x)ϕ(x, λ) = λϕ(x, λ),

we get π

0q (x)ϕ(x, λ)ϕ(x, λ) dx = ϕ(π, λ)ϕ(π, λ) − ϕ(π, λ)ϕ(π, λ) + h − h.

Since

h +

1

2 π

0 q (x) dx = 0,

we get π

0q (x)

ϕ(x, λ)ϕ(x, λ) − 1

2

dx = ϕ(π, λ)ϕ(π, λ) − ϕ(π, λ)ϕ(π, λ). (1.8.54)

Substituting (1.4.7) into (1.8.54) we obtain for λ = λn1 and λ = λn2,

π

0

g(x)cos2ρn1x dx = ϕ(π, λn1)ϕ(π, λn1),

π

0g(x)cos2ρn2x dx = ϕ(π, λn2)ϕ(π, λn2),

whereg(x) = 2

q (x) +

π

xV (x, t)q (t) dt

. (1.8.55)

Taking (1.4.8) and (1.8.49) into account, we get for Λ1 < δ 2,

|V (x, t)| < C. (1.8.56)

We use Lemma 1.8.4 for z 2n+1 = 2ρn,1, z 2n = 2ρn,2, ε2n+1 = ϕ(π, λn1)ϕ(π, λn1), ε2n =ϕ(π, λn2)ϕ(π, λn2). It follows from (1.8.35), (1.8.36), (1.8.50) and (1.8.51) that for Λ1 <δ 2, |g(x)| < C Λ1. Since q (x) is the solution of the Volterra integral equation (1.8.55), itfollows from (1.8.56) that |q (x)| < C Λ1, and hence |h| < C Λ1. 2

1.8.5. Riesz bases. 1) For convenience of the reader we provide here some facts aboutRiesz bases in a Hilbert space. For further discussion and proofs see [goh1], [you1] and

[Bar1]. Let B be a Hilbert space with the scalar product (., .).



93

Definition 1.8.1. A sequence f j j≥1 of vectors of a Hilbert space B is called a basis of this space if every vector f ∈ B can be expanded in a unique way in a series

f =∞

j=1

c jf j, (1.8.57)

which converges in the norm of the space B.

Clearly, if f j j≥1 is a basis then f j j≥1 is complete and minimal in B. We remindthat ” f j j≥1 is complete” means that the closed linear span of f j j≥1 coincides with B,and ” f j j≥1 is minimal” means that each element of the sequence lies outside the closedlinear span of the others.

Definition 1.8.2. Two sequences f j j≥1 and χ j j≥1 of B are called biorthonormal if (f j, χk) = δ jk ( δ jk is the Kronecker delta).

If f j j≥1 is a basis then the biorthonormal sequence χ j j≥1 exists and it is defineduniquely. Moreover, χ j j≥1 is also a basis of B. In the expansion (1.8.57) the coefficientsc j have the form

c j = (f, χ j). (1.8.58)

Definition 1.8.3. A sequence f j j≥1 is called almost normalized if

inf j

f j > 0 and sup j

f j < ∞.

If a basis f j j≥1 is almost normalized then the biorthonormal basis χ j j≥1 is alsoalmost normalized.

Definition 1.8.4. A sequence e j j≥1 is called orthogonal if (e j, ek) = 0 for j = k. Asequence e j j≥1 is called orthonormal if (e j, ek) = δ jk .

If e j j≥1 is orthonormal and complete in B then e j j≥1 is a basis of B. For anorthonormal basis e j j≥1 the relations (1.8.57)-(1.8.58) take the form

f = ∞ j=1

(f, e j)e j, (1.8.59)

and for each vector f ∈ B the following Parseval‘s equality holds

f 2 =∞

j=1

|(f, e j )|2.

2) Let now e j j≥1 be an orthonormal basis of the Hilbert space B, and let A : B → Bbe a bounded linear invertible operator. Then A−1 is bounded, and according to (1.8.59)we have for any vector f ∈ B,

A−1f =∞

j=1

(A−1f, e j)e j =∞

j=1

(f, A∗−1e j)e j.

Consequently

f =∞

j=1

(f, χ j)f j,



94

wheref j = Ae j, χ j = A∗−1e j. (1.8.60)

Obviously (f j, χk) = δ jk ( j, k ≥ 1), i.e. the sequences f j j≥1 and χ j j≥1 are biorthonor-mal. Therefore if (1.8.57) is valid then c j = (f, χ j), i.e. the expansion (1.8.57) is unique.

Thus, every bounded linear invertible operator transforms any orthonormal basis into someother basis of the space B.

Definition 1.8.5. A basis f j j≥1 of B is called a Riesz basis if it is obtained froman orthonormal basis by means of a bounded linear invertible operator.

According to (1.8.60), a basis which is biorthonormal to a Riesz basis is a Riesz basisitself. Using (1.8.60) we calculate

inf j

f j ≥ A−1 and sup j

f j ≤ A,

i.e. every Riesz basis is almost normalized.For the Riesz basis f j j≥1 (f j = Ae j ) the following inequality is valid for all vectors

f ∈ B :

C 1∞

j=1

|(f, χ j)|2 ≤ f 2 ≤ C 2∞

j=1

|(f, χ j)|2, (1.8.61)

where χ j j≥1 is the corresponding biorthonormal basis ( χ j = A∗−1e j ), and the constantsC 1, C 2 depend only on the operator A.

Indeed, using Parseval‘s equality and (1.8.60) we calculate

A−1f 2 =∞

j=1

|(A−1f, e j)|2 =∞

j=1

|(f, χ j)|2.

Sincef = AA−1f ≤ A · A−1f , A−1f ≤ A−1 · f ,

we getC 1

A−1f

2

≤ f

2

≤C 2

A−1f

2,

whereC 1 = A−1−2, C 2 = A2,

and consequently (1.8.61) is valid.

The following assertions are obvious.

Proposition 1.8.1. If f j j≥1 is a Riesz basis of B and γ j j≥1 are complex numbers such that 0 < C 1 ≤ |γ j| ≤ C 2 < ∞, then γ j f j j≥1 is a Riesz basis of B.

Proposition 1.8.2. If f j j≥1 is a Riesz basis of B and A is a bounded linear invertible operator, then Af j j≥1 is a Riesz basis of B.

3) Now we present several theorems which give us sufficient conditions on a sequencef j j≥1 to be a Riesz basis of B. First we formulate the definitions:

Definition 1.8.6. Two sequences of vectors f j j≥1 and g j j≥1 from B are calledquadratically close if

∞

j=1 g j

−f j

2 <

∞.



95

Definition 1.8.7. A sequence of vectors g j j≥1 is called ω - linearly independent if the equality

∞ j=1

c jg j = 0

is possible only for c j = 0 ( j ≥ 1).Assumption 1.8.1. Let f j = Ae j, j ≥ 1 be a Riesz basis of B, where e j j≥1 is

an orthonormal basis of B, and A is a bounded linear invertible operator. Let g j j≥1 besuch that

Ω := ∞

j=1

g j − f j21/2

< ∞,

i.e. g j j≥1 is quadratically close to f j j≥1.

Proposition 1.8.3 (stability of bases). Let Assumption 1.8.1 hold. If

Ω <1

A−1 ,

then g j j≥1 is a Riesz basis of B.

Proof. Consider the operator T :

T ∞

j=1c jf j =

∞ j=1

c j(f j − g j), c j ∈ l2. (1.8.62)

In other words, T e j = f j − g j. Clearly, T is a bounded linear operator and T ≤ Ω.

Moreover,∞

j=1

T e j2 < ∞. Since

A − T = (E − T A−1)A, and T A−1 < 1,

then A − T is a linear bounded invertible operator. On the other hand,

(A − T )e j = g j,

and consequently g j j≥1 is a Riesz basis of B. 2

Proposition 1.8.4 (Bari [Bar1]). Let Assumption 1.8.1 hold. If g j j≥1 is ω−linearly independent then g j j≥1 is a Riesz basis of B.

Proof. Define the operator T by (1.8.62). The equation (A

−T )f = 0 has only the

trivial solution. Indeed, if (A − T )f = 0 then from

(A − T )f =∞

j=1

(f, e j)f j −∞

j=1

(f, e j)(f j − g j) =∞

j=1

(f, e j)g j

it follows that ∞ j=1

(f, e j)g j = 0.



96

Hence (f, e j) = 0, j ≥ 1, by the ω - linearly independence of g j j≥1. From this weget f = 0. Thus, the operator A − T is a linear bounded invertible operator. Since(A − T )e j = g j, the sequence g j j≥1 is a Riesz basis of B. 2

Proposition 1.8.5. Let Assumption 1.8.1 hold. If g j j≥1 is complete in B then

g j j≥1 is a Riesz basis of B.Proof. Choose N such that

∞ j=N +1

g j − f j21/2

<1

A−1 ,

and consider the sequence ψ j j≥1 :

ψ j = f j, j = 1, N ,

g j, j > N.

By virtue of Proposition 1.8.3, the sequence ψ j j≥1 is a Riesz basis of B. Let ψ∗ j j≥1

be the biorthonormal basis to ψ j j≥1. Denote

D := det[(g j, ψ∗n)] j,n=1,N .

Let us show that D = 0. Suppose, on the contrary, that D = 0. Then the linear algebraicsystem

N n=1

β n(g j, ψ∗n) = 0, j = 1, N ,

has a non-trivial solution β nn=1,N . Consider the vector

f :=N

n=1

β nψ∗n.

Since

ψ∗nn≥1

is a Riesz basis we get f = 0. On the other hand, we calculate:

(i) for j = 1, N : (g j, f ) =N

n=1

β n(g j, ψ∗n) = 0;

(ii) for j > N : (g j, f ) = (ψ j, f ) =N

n=1

β n(ψ j, ψ∗n) = 0.

Thus, (g j, f ) = 0 for all j ≥ 1. Since g j j≥1 is complete we get f = 0. This contradiction

shows that D = 0.Let us show that g j j≥1 is ω− linearly independent. Indeed, let c j j≥1 be complexnumbers such that ∞

j=1

c jg j = 0. (1.8.63)

Since g j = ψ j for j > N we get

N

j=1

c j(g j, ψ∗n) = 0, n = 1, N .



97

The determinant of this linear system is equal to D = 0. Consequently, c j = 0, j = 1, N .Then (1.8.63) takes the form

∞ j=N +1

c jψ j = 0.

Since ψ j j≥1 is a Riesz basis we have c j = 0, j > N. Thus, the sequence g j j≥1 is ω−linearly independent, and by Proposition 1.8.4 g j j≥1 is a Riesz basis of B. 2

Proposition 1.8.6. Let numbers ρnn≥0, ρn = ρk (n = k) of the form

ρn = n +a

n+

κn

n, κn ∈ l2, a ∈ C (1.8.64)

be given. Then cos ρnxn≥0 is a Riesz basis in L2(0, π).

Proof. First we show that

cos ρnx

n≥

0 is complete in L2(0, π). Let f (x)∈

L2(0, π)be such that π

0f (x)cos ρnx dx = 0, n ≥ 0. (1.8.65)

Consider the functions

∆(λ) := π(λ0 − λ)∞

n=1

λn − λ

n2, λn = ρ2

n,

F (λ) =

1

∆(λ) π

0 f (x)cos ρx dx, λ = ρ2

, λ = λn.

It follows from (1.8.65) that F (λ) is entire in λ (after removing singularities). On theother hand, it was shown in the proof of Lemma 1.6.6 that

|∆(λ)| ≥ C |ρ| exp(|τ |π), arg λ ∈ [δ, 2π − δ ], δ > 0, τ := Imρ,

and consequently

|F (λ)

| ≤C

|ρ|, arg λ

∈[δ, 2π

−δ ].

From this, using the Phragmen-Lindelof theorem [you1, p.80] and Liouville‘s theorem [con1,p.77] we conclude that F (λ) ≡ 0, i.e. π

0f (x)cos ρxdx = 0 for ρ ∈ C.

Hence f = 0. Thus, cos ρnxn≥0 is complete in L2(0, π). We note that completeness of

cos ρnx

n

≥0 follows independently also from Levinson‘s theorem [you1, p.118]. Further-

more, it follows from (1.8.64) that cos ρnxn≥0 is quadratically close to the orthogonalbasis cos nxn≥0. Then, by virtue of Proposition 1.8.5, cos ρnxn≥0 is a Riesz basis inL2(0, π). 2

1.9. REVIEW OF THE INVERSE PROBLEM THEORY



98

In this section we give a short review of results on inverse problems of spectral analysis forordinary differential equations. We describe briefly only the main directions of this theory,mention the most important monographs and papers, and refer to them for details.

The greatest success in the inverse problem theory was achieved for the Sturm-Liouvilleoperator

y := −y + q (x)y. (1.9.1)

The first result in this direction is due to Ambarzumian [amb1]. He showed that if theeigenvalues of the boundary value problem

−y + q (x)y = λy, y(0) = y(π) = 0

are λk = k2, k ≥ 0, then q = 0. But this result is an exception from the rule, and ingeneral the specification of the spectrum does not uniquely determine the operator (1.9.1).

Afterwards Borg [Bor1] proved that the specification of two spectra of Sturm-Liouville op-erators uniquely determines the potential q. Levinson [Lev1] suggested a different methodto prove Borg‘s results. Tikhonov in [Tik1] obtained the uniqueness theorem for the inverseSturm-Liouville problem on the half-line from the given Weyl function.

An important role in the spectral theory of Sturm-Liouville operators was played by thetransformation operator. Marchenko [mar3]-[mar4] first applied the transformation operatorto the solution of the inverse problem. He proved that a Sturm-Liouville operator on thehalf-line or a finite interval is uniquely determined by specifying the spectral function. For a

finite interval this corresponds to the specification of the spectral data (see Subsection 1.4.2).Transformation operators were also used in the fundamental paper of Gelfand and Levitan[gel1], where they obtained necessary and sufficient conditions along with a method forrecovering a Sturm-Liouville operator from its spectral function. In [gas1] similar results wereestablished for the inverse problem of recovering Sturm-Liouville operators on a finite intervalfrom two spectra. Another approach for the solution of inverse problems was suggested byKrein [kre1], [kre2]. Blokh [blo1] and Rofe-Beketov [rof1] studied the inverse problem on theline with given spectral matrix. The inverse scattering theory on the half-line and on the linewas constructed in [agr1], [cha1], [deg1], [dei1], [fad1], [fad2], [kay1], [kur1] and other works.An interesting approach connected with studying isospectral sets for the Sturm-Liouvilleoperators on a finite interval was described in [car1], [dah1], [isa1], [isa2], [mcl2], [pos1] and[shu1]. The created methods also allow one to study stability of the solution of the inverseproblem for (1.9.1) (see [ale1], [Bor1], [dor1], [hoc1], [hoc2], [mar6], [miz1], [rja1] and [yur3]).

In the last twenty years there appeared many new areas for applications of inverse Sturm-Liouville problems; we briefly mention some of them.

Boundary value problems with discontinuity conditions inside the interval are connectedwith discontinuous material properties. For example, discontinuous inverse problems appearin electronics for constructing parameters of heterogeneous electronic lines with desirabletechnical characteristics ([lit1], [mes1]). Spectral information can be used to reconstruct thepermittivity and conductivity profiles of a one-dimensional discontinuous medium ([kru1],[she1]). Boundary value problems with discontinuities in an interior point also appear ingeophysical models for oscillations of the Earth ([And1], [lap1]). Here the main discontinuityis caused by reflection of the shear waves at the base of the crust. Discontinuous inverseproblems (in various formulations) have been considered in [hal1], [kob1], [kob2], [kru1],[pro1] and [she1].



99

Many further applications connect with the differential equation of the form

−y + q (x)y = λr(x)y (1.9.2)

with turning points when the function r(x) has zeros or (and) changes sign. For example,turning points connect with physical situations in which zeros correspond to the limit of motion of a wave mechanical particle bound by a potential field. Turning points appear alsoin elasticity, optics, geophysics and other branches of natural sciences. Moreover, a wideclass of differential equations with Bessel-type singularities and their perturbations can bereduced to differential equations having turning points. Inverse problems for equations withturning points and singularities help to study blow-up solutions for some nonlinear integrableevolution equations of mathematical physics (see [Con1]). Inverse problems for the equation(1.9.1) with singularities and for the equation (1.9.2) with turning points and singularitieshave been studied in [Bel1], [Bou1], [car2], [elr1], [FY1], [gas2], [pan1], [sta1], [yur28], [yur29]

and other works. Some aspects of the turning point theory and a number of applications aredescribed in [ebe1]-[ebe4], [gol1], [mch1] and [was1].

The inverse problem for the Sturm-Liouville equation in the impedance form

−( p(x)y) = λp(x)y, p(x) > 0, 0 < x < 1, (1.9.3)

when p/p ∈ L2(0, 1) has been studied in [and1], [and2] and [col1]. Equation (1.9.3) canbe transformed to (1.9.1) when p ∈ L2(0, 1) but not when only p/p ∈ L2(0, 1). The lackof the smoothness leads to additional technical difficulties in the investigation of the inverse

problem. In [akt2], [dar1], [gri1] and [yak1] the inverse problem for the equation (1.9.2) witha lack of the smoothness of r is treated.

In [bro1], [hal2], [hal3], [mcl4], [She1], [yan1] and other papers, the so-called nodal in-verse problem is considered when the potential is to be reconstructed from nodes of theeigenfunctions. Many works are devoted to incomplete inverse problems when only a part of the spectral information is available for measurement and (or) there is a priori informationabout the operator or its spectrum (see [akt1], [akt4], [FY2], [ges2], [gre1], [kli1], [mcl3],[run1], [sac2], [sad1], [tik1], [Zho1] and the references therein). Sometimes in incomplete in-

verse problems we have a lack of information which leads to nonuniqueness of the solution of the inverse problem (see, for example, [FY2] and [tik1]). We also mention inverse problemsfor boundary value problems with nonseparated boundary conditions ([Gus1], [mar7], [pla1],[yur2] and [yur20]) and with nonlocal boundary conditions of the form T

0y(x) dσ j(x) = 0

(see [kra1]). Some aspects of numerical solutions of the inverse problems have been studiedin [ahm1], [bAr1], [fab1], [kHa1], [kno1], [low1], [mue1], [neh1], [pai1] and [zhi1].

Numerous applications of the inverse problem theory connect with higher-order differen-tial operators of the form

ly := y(n) +n−2k=0

pk(x)y(k), n > 2. (1.9.4)

In contrast to the case of Sturm-Liouville operators, inverse problems for the operator (1.9.4)turned out to be much more difficult for studying. Fage [fag2] and Leontjev [leo1] deter-

mined that for n > 2 transformation operators have a more complicated structure than for



100

Sturm-Liouville operators which makes it more difficult to use them for solving the inverseproblem. However, in the case of analytic coefficients the transformation operators havethe same ”triangular” form as for Sturm-Liouville operators (see [kha2], [mat1] and [sak1]).Sakhnovich [sak2]-[sak3] and Khachatryan [kha3]-[kha4] used a ”triangular” transformationoperator to investigate the inverse problem of recovering selfadjoint differential operators

on the half-line with analytic coefficients from the spectral function, as well as the inversescattering problem.

A more effective and universal method in the inverse spectral theory is the method of spectral mappings connected with ideas of the contour integral method. Levinson [Lev1] firstapplied ideas of the contour integral method for investigating the inverse problem for theSturm-Liouville operator (1.9.1). Developing Levinson‘s ideas Leibenson in [lei1]-[lei2] inves-tigated the inverse problem for (1.9.4) on a finite interval under the restriction of ”separation”of the spectrum. The spectra and weight numbers of certain specially chosen boundary value

problems for the differential operators (1.9.4) appeared as the spectral data of the inverseproblem. The inverse problem for (1.9.4) on a finite interval with respect to a system of spectra was investigated under various conditions on the operator in [bar1], [mal1], [yur4],[yur8] and [yur10]. Things are more complicated for differential operators on the half-line,especially for the nonselfadjoint case, since the spectrum can have rather ”bad” behavior.Moreover, for the operator (1.9.4) even a formulation of the inverse problem is not obvioussince the spectral function turns out to be unsuitable for this purpose. On a finite intervalthe formulation of the inverse problem is not obvious as well, since waiving the requirement

of ”separation” of the spectra leads to a violation of uniqueness for the solution of the inverseproblem. In [yur1], [yur16] and [yur21] the so-called Weyl matrix was introduced and studiedas the main spectral characteristics for the nonselfadjoint differential operator (1.9.4) on thehalf-line and on a finite interval. This is a generalization of the classical Weyl function forthe Sturm-Liouville operator. The concept of the Weyl matrix and a development of ideasof the contour integral method enable one to construct a general theory of inverse problemsfor nonselfadjoint differential operators (1.9.4) (see [yur1] for details). The inverse scatteringproblem on the line for the operator (1.9.4) has been treated in various settings in [bea1],[bea2], [Cau1], [dei2], [dei3], [kau1], [kaz1], [kaz2], [suk1] and other works. We note that the

use of the Riemann problem in the inverse scattering theory can be considered as a particularcase of the contour integral method (see, for example, [bea1]).

The inverse problem for the operator (1.9.4) with locally integrable coefficients has beenstudied in [yur14]. There the generalized Weyl functions are introduced, and the Riemann-Fage formula [fag1] for the solution of the Cauchy problem for higher-order partial differentialequations is used.

Inverse problems for higher-order differential operators with singularities have been stud-ied in [kud1], [yur17], [yur18] and [yur22]. Incomplete inverse problems for higher-order

differential operators and their applications are considered in [kha1], [bek1], [elc1], [mal1],[mcl1], [pap1], [yur8]-[yur10] and [yur26]. In particular, in [yur9] the inverse problem of recovering a part of the coefficients of the differential operator (1.9.4) from a part of theWeyl matrix has been studied (the rest of the coefficients are known a priori). For suchinverse problems the method of standard models was suggested and a constructive solutionfor a wide class of incomplete inverse problems was given. This method was also applied forthe solution of the inverse problem of elasticity theory when parameters of a beam are to bedetermined from given frequences of its natural oscillations (see [yur12]). This problem can



101

be reduced to the inverse problem of recovering the fourth-order differential operator

(hµ(x)y) = λh(x)y, µ = 1, 2, 3

from the Weyl function.

Many papers are devoted to inverse problems for systems of differential equations (see[alp1], [amo1], [aru1], [bea3], [bea4], [bou1], [Cha1], [Che1], [cox1], [gas3], [gas4], [goh2],[hin1], [lee1], [li1], [mal2], [pal1], [sak5], [sAk1], [sha1], [sha2], [win1], [yam1], [zam1], [zho1],[zho2], [zho3] and the references therein). Some systems can be treated similarly to theSturm-Liouville operator. As example we mention inverse problems for the Dirac system([lev4], [aru1], [gas3], [gas4], [zam1]). But in the general case, the inverse problems forsystems deal with difficulties like those for the operator (1.9.4). Such systems have beenstudied in [bea3], [bea4], [lee1], [zho1], [zho2] and [zho3].

We mention also inverse spectral problems for discrete operators (see [atk1], [ber1], [gla1],[gus1], [gus2], [Kha1], [nik1], [yur23], [yur24] and [yur26]), for differential operators withdelay ([pik1], [pik2]), for nonlinear differential equations ([yur25]), for integro-differentialand integral operators ([ere1], [yur6], [yur7], [yur13]), for pencils of operators ([akt3], [bea5],[bro2], [Chu1], [gas5], [yur5], [yur27], [yur31], [yur32]) and others.

Extensive literature is devoted to one more wide area of the inverse problem theory. In1967 Gardner, Green, Kruskal and Miura [gar1] found a remarkable method for solving someimportant nonlinear equations of mathematical physics (the Korteweg-de Vries equation, thenonlinear Schrodinger equation, the Boussinesq equation and many others) connected with

the use of inverse spectral problems. This method is described in [abl1], [lax1], [tak1], [zak1]and other works (see also Sections 4.1-4.2).

Many contributions are devoted to the inverse problem theory for partial differentialequations. This direction is reflected fairly completely in [ang1], [ani1], [ber1], [buk1], [cha1],[cha3], [Isa1], [kir1], [lav1], [new1], [niz1], [Pri1], [Pri2], [ram1], [rom1] and [rom2]. In Section2.4 we study an inverse problem for a wave equation as a model inverse problem for par-tial differential equations and show connections with inverse spectral problems for ordinarydifferential equations.



102

II. STURM-LIOUVILLE OPERATORS ON THE HALF-LINE

In this chapter we present an introduction to the inverse problem theory for Sturm-Liouville operators on the half-line. First, in Sections 2.1-2.2 nonselfadjoint operators with

integrable complex-valued potentials are considered. We introduce and study the Weyl func-tion as the main spectral characteristic, prove an expansion theorem and solve the inverseproblem of recovering the Sturm-Liouville operator from its Weyl function. For this purposewe use ideas of the contour integral method and develop the method of spectral mappingspresented in Chapter I. Moreover connections with the transformation operator method areestablished. In Section 2.3 the most important particular cases, which often appear in appli-cations, are considered, namely: selfadjoint operators, nonselfadjoint operators with a simplespectrum and perturbations of the discrete spectrum of a model operator. We introduce here

the so-called spectral data which describe the set of singularities of the Weyl function. Thesolution of the inverse problem of recovering the Sturm-Liouville operator from its spectraldata is provided. In Sections 2.4-2.5 locally integrable complex-valued potentials are stud-ied. In Section 2.4 we consider an inverse problem for a wave equation. In Section 2.5 thegeneralized Weyl function is introduced as the main spectral characteristic. We prove anexpansion theorem and solve the inverse problem of recovering the Sturm-Liouville operatorfrom its generalized Weyl function. The inverse problem to construct q from the Weylsequence is considered in Section 2.6.

2.1. PROPERTIES OF THE SPECTRUM. THE WEYL FUNCTION.

We consider the differential equation and the linear form L = L(q (x), h) :

y := −y + q (x)y = λy, x > 0, (2.1.1)

U (y) := y(0)−

hy(0), (2.1.2)

where q (x) ∈ L(0, ∞) is a complex-valued function, and h is a complex number. Letλ = ρ2, ρ = σ + iτ, and let for definiteness τ := Im ρ ≥ 0. Denote by Π the λ -plane with the cut λ ≥ 0, and Π1 = Π \ 0; notice that here Π and Π1 must beconsidered as subsets of the Riemann surface of the squareroot-function. Then, under themap ρ → ρ2 = λ, Π1 corresponds to the domain Ω = ρ : Im ρ ≥ 0, ρ = 0. PutΩδ = ρ : Im ρ ≥ 0, |ρ| ≥ δ . Denote by W N the set of functions f (x), x ≥ 0 suchthat the functions f ( j)(x), j = 0, N − 1 are absolutely continuous on [0, T ] for each fixed

T > 0, and f

( j)

(x) ∈ L(0, ∞) , j = 0, N.2.1.1. Jost and Birkhoff solutions. In this subsection we construct a special funda-

mental system of solutions for equation (2.1.1) in Ω having asymptotic behavior at infinitylike exp(±iρx).

Theorem 2.1.1. Equation (2.1.1) has a unique solution y = e(x, ρ), ρ ∈ Ω, x ≥ 0,satisfying the integral equation

e(x, ρ) = exp(iρx)−

1

2iρ ∞

x

(exp(iρ(x−

t))−

exp(iρ(t−

x))) q (t)e(t, ρ) dt. (2.1.3)



103

The function e(x, ρ) has the following properties:(i1) For x → ∞, ν = 0, 1, and each fixed δ > 0,

e(ν )(x, ρ) = (iρ)ν exp(iρx)(1 + o(1)), (2.1.4)

uniformly in Ωδ. For I m ρ > 0, e(x, ρ) ∈ L2(0, ∞). Moreover, e(x, ρ) is the unique solution of (2.1.1) (up to a multiplicative constant) having this property.(i2) For |ρ| → ∞, ρ ∈ Ω, ν = 0, 1,

e(ν )(x, ρ) = (iρ)ν exp(iρx)

1 +ω(x)

iρ+ o

1

ρ

, ω(x) := −1

2

∞x

q (t) dt, (2.1.5)

uniformly for x ≥ 0.(i3) For each fixed x

≥0, and ν = 0, 1, the functions e(ν )(x, ρ) are analytic for Im ρ > 0,

and are continuous for ρ ∈ Ω.(i4) For real ρ = 0, the functions e(x, ρ) and e(x, −ρ) form a fundamental system of solutions for (2.1.1), and

e(x, ρ), e(x, −ρ) = −2iρ, (2.1.6)

where y, z := yz − yz is the Wronskian.The function e(x, ρ) is called the Jost solution for (2.1.1).

Proof. We transform (2.1.3) by means of the replacement

e(x, ρ) = exp(iρx)z (x, ρ) (2.1.7)

to the equation

z (x, ρ) = 1 − 1

2iρ

∞x

(1 − exp(2iρ(t − x))) q (t)z (t, ρ) dt, x ≥ 0, ρ ∈ Ω. (2.1.8)


z 0(x, ρ) = 1, z k+1(x, ρ) = − 12iρ

∞x

(1 − exp(2iρ(t − x))) q (t)z k(t, ρ) dt, (2.1.9)

z (x, ρ) =∞

k=0

z k(x, ρ). (2.1.10)

Let us show by induction that

|z k(x, ρ)

| ≤(Q0(x))k

|ρ|k

k!

, ρ

∈Ω, x

≥0, (2.1.11)

whereQ0(x) :=

∞x

|q (t)| dt.

Indeed, for k = 0, (2.1.11) is obvious. Suppose that (2.1.11) is valid for a certain fixedk ≥ 0. Since |1 − exp(2iρ(t − x))| ≤ 2, (2.1.9) implies

|z k+1(x, ρ)

| ≤

1

|ρ| ∞

x |q (t)z k(t, ρ)

|dt. (2.1.12)



104

Substituting (2.1.11) into the right-hand side of (2.1.12) we calculate

|z k+1(x, ρ)| ≤ 1

|ρ|k+1k!

∞x

|q (t)|(Q0(t))k dt =(Q0(t))k+1

|ρ|k+1(k + 1)!.

It follows from (2.1.11) that the series (2.1.10) converges absolutely for x ≥ 0, ρ ∈ Ω,and the function z (x, ρ) is the unique solution of the integral equation (2.1.8). Moreover,by virtue of (2.1.10) and (2.1.11),

|z (x, ρ)| ≤ exp(Q0(x)/|ρ|), |z (x, ρ) − 1| ≤ (Q0(x)/|ρ|)exp(Q0(x)/|ρ|). (2.1.13)

In particular, (2.1.13) yields for each fixed δ > 0,

z (x, ρ) = 1 + o(1), x→ ∞

, (2.1.14)

uniformly in Ωδ, and

z (x, ρ) = 1 + O1

ρ

, |ρ| → ∞, ρ ∈ Ω, (2.1.15)

uniformly for x ≥ 0. Substituting (2.1.15) into the right-hand side of (2.1.8) we obtain

z (x, ρ) = 1−

1

2iρ ∞

x(1

−exp(2iρ(t

−x))) q (t) dt + O 1

ρ2,|ρ| → ∞

, (2.1.16)

uniformly for x ≥ 0.

Lemma 2.1.1. Let q (x) ∈ L(0, ∞), and denote

J q(x, ρ) := ∞

xq (t) exp(2iρ(t − x)) dt, ρ ∈ Ω. (2.1.17)

Then

lim|ρ|→∞ supx≥0 |J q(x, ρ)| = 0. (2.1.18)

Proof. 1) First we assume that q (x) ∈ W 1. Then integration by parts in (2.1.17) yields

J q(x, ρ) = −q (x)

2iρ− 1

2iρ

∞x

q (t) exp(2iρ(t − x)) dt,

and consequently

supx≥0 |J q(x, ρ)| ≤C q|ρ| .

2) Let now q (x) ∈ L(0, ∞). Fix ε > 0 and choose q ε(x) ∈ W 1 such that ∞0

|q (t) − q ε(t)| dt <ε

2.

Then

|J q(x, ρ)| ≤ |J qε(x, ρ)| + |J q−qε(x, ρ)| ≤ C qε

|ρ|

+ε

2.



105

Hence, there exists ρ0 > 0 such that supx≥0

|J q(x, ρ)| ≤ ε for |ρ| ≥ ρ0, ρ ∈ Ω. By virtue of

arbitrariness of ε > 0 we arrive at (2.1.18). 2

Let us return to the proof of Theorem 2.1.1. It follows from (2.1.16) and Lemma 2.1.1that

z (x, ρ) = 1 + ω(x)iρ

+ o1ρ, |ρ| → ∞, ρ ∈ Ω, (2.1.19)

uniformly for x ≥ 0. From (2.1.7), (2.1.9)-(2.1.11), (2.1.14) and (2.1.19) we derive (i1)−(i3)for ν = 0. Furthermore, (2.1.3) and (2.1.7) imply

e(x, ρ) = (iρ)exp(iρx)

1 − 1

2iρ

∞x

(1 + exp(2iρ(t − x))) q (t)z (t, ρ) dt

. (2.1.20)

Using (2.1.20) we get (i1) − (i3) for ν = 1. It is easy to verify by differentiation that thefunction e(x, ρ) is a solution of (2.1.1). For real ρ = 0, the functions e(x, ρ) and e(x, −ρ)satisfy (2.1.1), and by virtue of (2.1.4), lim

x→∞ e(x, ρ), e(x, −ρ) = −2iρ. Since the Wronskian

e(x, ρ), e(x, −ρ) does not depend on x, we arrive at (2.1.6). 2

Remark 2.1.1. If q (x) ∈ W N , then there exist functions ωsν (x) such that forρ → ∞, ρ ∈ Ω, ν = 0, 1, 2,

e(ν )(x, ρ) = (iρ)ν exp(iρx)1 +N +1

s=1

ωsν (x)

(iρ)s

+ o1

ρN +1, ω1ν (x) = ω(x). (2.1.21)

Indeed, let q (x) ∈ W 1. Substituting (2.1.19) into the right-hand side of (2.1.8) we get

z (x, ρ) = 1 − 1

2iρ

∞x

(1 − exp(2iρ(t − x))) q (t) dt

− 1

2(iρ)2

∞x

(1 − exp(2iρ(t − x))) q (t)ω(t) dt + o 1

ρ2

, |ρ| → ∞.

Integrating by parts and using Lemma 2.1.1, we obtain

z (x, ρ) = 1 +ω(x)

iρ+

ω20(x)

(iρ)2+ o

1

ρ2

, |ρ| → ∞, ρ ∈ Ω, (2.1.22)

where

ω20(x) = −1

4q (x) +

1

4

∞x

q (t) ∞

tq (s) ds

dt = −1

4q (x) +

1

8

∞x

q (t) dt2

.

By virtue of (2.1.7) and (2.1.22), we arrive at (2.1.21) for N = 1, ν = 0. Using inductionone can prove (2.1.21) for all N.

If we additionally assume that xq (x) ∈ L(0, ∞), then the Jost solution e(x, ρ) existsalso for ρ = 0. More precisely, the following theorem is valid.

Theorem 2.1.2. Let (1 + x)q (x) ∈ L(0, ∞). The functions e(ν )(x, ρ), ν = 0, 1 are continuous for Im ρ ≥ 0, x ≥ 0, and

|e(x, ρ) exp(−iρx)| ≤ exp(Q1(x)), (2.1.23)



106

|e(x, ρ)exp(−iρx) − 1| ≤

Q1(x) − Q1

x +

1

|ρ|

exp(Q1(x)), (2.1.24)

|e(x, ρ)exp(−iρx) − iρ| ≤ Q0(x)exp(Q1(x)), (2.1.25)

where

Q1(x) := ∞x Q0(t) dt = ∞

x (t − x)|q (t)| dt.

First we prove an auxiliary assertion.

Lemma 2.1.2. Assume that c1 ≥ 0, u(x) ≥ 0, v(x) ≥ 0, (a ≤ x ≤ T ≤ ∞), u(x) is bounded, and (x − a)v(x) ∈ L(a, T ). If

u(x) ≤ c1 + T

x(t − x)v(t)u(t) dt, (2.1.26)

then u(x) ≤ c1 exp

T

x(t − x)v(t) dt

. (2.1.27)

Proof. Denote

ξ (x) = c1 + T

x(t − x)v(t)u(t) dt.

Then

ξ (T ) = c1, ξ (x) =−

T

x

v(t)u(t) dt, ξ = v(x)u(x),

and (2.1.26) yields0 ≤ ξ (x) ≤ ξ (x)v(x).

Let c1 > 0. Then ξ (x) > 0, andξ (x)

ξ (x)≤ v(x).

Hence

ξ (x)

ξ (x)

≤ v(x) − ξ (x)

ξ (x) 2

≤ v(x).

Integrating this inequality twice we get

−ξ (x)

ξ (x)≤ T

xv(t) dt, ln

ξ (x)

ξ (T )≤ T

x(t − x)v(t) dt,

and consequently,

ξ (x)

≤c1 exp

T

x

(t

−x)v(t) dt.

According to (2.1.26), u(x) ≤ ξ (x), and we arrive at (2.1.27).If c1 = 0, then ξ (x) = 0. Indeed, suppose on the contrary that ξ (x) = 0. Since

ξ (x) ≥ 0, ξ (x) ≤ 0, there exists T 0 ≤ T such that ξ (x) > 0 for x < T 0, and ξ (x) ≡ 0for x ∈ [T 0, T ]. Repeating these arguments we get for x < T 0 and sufficiently small ε > 0,

lnξ (x)

ξ (T 0 − ε)≤ T 0−ε

x(t − x)v(t) dt ≤

T 0

x(t − x)v(t) dt,

which is impossible. Thus, ξ (x) ≡ 0, and (2.1.27) becomes obvious. 2



107

Proof of Theorem 2.1.2. For Im ρ ≥ 0 we have

sin ρ

ρexp(iρ)

≤ 1. (2.1.28)

Indeed, (2.1.28) is obvious for real ρ and for |ρ| ≥ 1, I m ρ ≥ 0. Then, by the maximumprinciple [con1, p.128], (2.1.28) is also valid for |ρ| ≤ 1, I m ρ ≥ 0.It follows from (2.1.28) that

1 − exp(2iρx)

2iρ

≤ x for Im ρ ≥ 0, x ≥ 0. (2.1.29)

Using (2.1.9) and (2.1.29) we infer

|z k+1(x, ρ)| ≤ ∞x t|q (t)z k(t, ρ)| dt, k ≥ 0, I m ρ ≥ 0, x ≥ 0,

and consequently by induction

|z k(x, ρ)| ≤ 1

k!

∞x

t|q (t)| dtk

, k ≥ 0, I m ρ ≥ 0, x ≥ 0.

Then, the series (2.1.10) converges absolutely and uniformly for Im ρ ≥ 0, x ≥ 0, and thefunction z (x, ρ) is continuous for Im ρ ≥ 0, x ≥ 0. Moreover,

|z (x, ρ)| ≤ exp ∞

xt|q (t)| dt

, Im ρ ≥ 0, x ≥ 0. (2.1.30)

Using (2.1.7) and (2.1.20) we conclude that the functions e(ν )(x, ρ), ν = 0, 1 are continuousfor Im ρ ≥ 0, x ≥ 0.

Furthermore, it follows from (2.1.8) and (2.1.29) that

|z (x, ρ)

| ≤1 +

∞

x(t

−x)

|q (t)z (t, ρ)

|dt, Im ρ

≥0, x

≥0.

By virtue of Lemma 2.1.2, this implies

|z (x, ρ)| ≤ exp(Q1(x)), Im ρ ≥ 0, x ≥ 0, (2.1.31)

i.e. (2.1.23) is valid. We note that (2.1.31) is more precise than (2.1.30).Using (2.1.8), (2.1.29) and (2.1.31) we calculate

|z (x, ρ)

−1| ≤

∞

x(t

−x)

|q (t)

|exp(Q

1(t)) dt

≤exp(Q

1(x))

∞

x(t

−x)

|q (t)

|dt,

and consequently,

|z (x, ρ) − 1| ≤ Q1(x)exp(Q1(x)), Im ρ ≥ 0, x ≥ 0. (2.1.32)

More precisely,

|z (x, ρ)

−1

| ≤ x+ 1

|ρ|

x

(t

−x)

|q (t)

|exp(Q1(t)) dt +

1

|ρ| ∞

x+1

|ρ| |q (t)

|exp(Q1(t)) dt



108

≤ exp(Q1(x)) ∞

x(t − x)|q (t)| dt −

∞x+ 1

|ρ|

t − x − 1

|ρ||q (t)| dt

=

Q1(x) − Q1

x +

1

|ρ|

exp(Q1(x)),

i.e. (2.1.24) is valid. At last, from (2.1.20) and (2.1.31) we obtain

|e(x, ρ) exp(−iρx) − iρ| ≤ ∞

x|q (t)| exp(Q1(t)) dt ≤ exp(Q1(x))

∞x

|q (t)| dt,

and we arrive at (2.1.25). Theorem 2.1.2 is proved. 2

Remark 2.1.2. Consider the function

q (x) =2a2

(1 + ax)2

,

where a is a complex number such that a /∈ (−∞, 0]. Then q (x) ∈ L(0, ∞), but xq (x) /∈L(0, ∞). The Jost solution has in this case the form (see Example 2.3.1)

e(x, ρ) = exp(iρx)

1 − a

iρ(1 + ax)

,

i.e. e(x, ρ) has a singularity at ρ = 0, hence we cannot omit the integrability condition in

Theorem 2.1.2.For the Jost solution e(x, ρ) there exists a transformation operator. More precisely, the

following theorem is valid.

Theorem 2.1.3. Let (1 + x)q (x) ∈ L(0, ∞). Then the Jost solution e(x, ρ) can be represented in the form

e(x, ρ) = exp(iρx) + ∞

xA(x, t)exp(iρt) dt, Im ρ ≥ 0, x ≥ 0, (2.1.33)

where A(x, t) is a continuous function for 0 ≤ x ≤ t < ∞, and

A(x, x) =1

2

∞x

q (t) dt. (2.1.34)

|A(x, t)| ≤ 1

2Q0

x + t

2

exp

Q1(x) − Q1

x + t

2

, (2.1.35)

1 + ∞

x |A(x, t)

|dt

≤exp(Q1(x)),

∞

x |A(x, t)

|dt

≤Q1(x)exp(Q1(x)). (2.1.36)

Moreover, the function A(x1, x2) has first derivatives ∂A∂xi

, i = 1, 2; the functions

∂A(x1, x2)

∂xi+

1

4q x1 + x2

2

are absolutely continuous with respect to x1 and x2, and satisfy the estimates

∂A(x1, x2)

∂xi+

1

4q x1 + x2

2



109

≤ 1

2Q0(x1)Q0

x1 + x2

2

exp

Q1(x1) − Q1

x1 + x2

2

, i = 1, 2. (2.1.37)

Proof. The arguments for proving this theorem are similar to those in Section 1.3 withadequate modifications. This theorem also can be found in [mar1] and [lev2].

According to (2.1.7) and (2.1.10) we have

e(x, ρ) =∞

k=0

εk(x, ρ), εk(x, ρ) = z k(x, ρ)exp(iρx). (2.1.38)

Let us show by induction that the following representation is valid

εk(x, ρ) = ∞

xak(x, t)exp(iρt) dt, k ≥ 1, (2.1.39)

where the functions ak(x, t) do not depend on ρ.First we calculate ε1(x, ρ). By virtue of (2.1.9) and (2.1.38),

ε1(x, ρ) = ∞

x

sin ρ(s − x)

ρexp(iρs)q (s) ds =

1

2

∞x

q (s) 2s−x

xexp(iρt) dt

ds.

Interchanging the order of integration we obtain that (2.1.39) holds for k = 1, where

a1(x, t) =1

2 ∞

(t+x)/2

q (s) ds.

Suppose now that (2.1.39) is valid for a certain k ≥ 1. Then

εk+1(x, ρ) = ∞

x

sin ρ(s − x)

ρq (s)εk(s, ρ) ds

= ∞

x

sin ρ(s − x)

ρq (s)

∞s

ak(s, u)exp(iρu) du

ds

= 12

∞x

q (s) ∞

sak(s, u)

s+u−x

−s+u+xexp(iρt) dt

du

ds.

We extend ak(s, u) by zero for u < s. For s ≥ x this yields

∞s

ak(s, u) s+u−x

−s+u+xexp(iρt) dt

du =

∞x

exp(iρt) t+s−x

t−s+xak(s, u) du

dt.

Therefore

εk+1(x, ρ) = 12

∞x

exp(iρt) ∞

xq (s)

t+s−x

t−s+xak(s, u) du

ds

dt = ∞

xak+1(x, t)exp(iρt) dt,

where

ak+1(x, t) =1

2

∞x

q (s) t+s−x

t−s+xak(s, u) du

ds, t ≥ x.

Changing the variables according to u + s = 2α, u − s = 2β, we obtain

ak+1(x, t) = ∞

(t+x)/2 (t−x)/2

0

q (α−

β )ak(α−

β, α + β ) dβ dα.



110

Taking H k(α, β ) = ak(α − β, α + β ), t + x = 2u, t − x = 2v, we calculate for 0 ≤ v ≤ u,

H 1(u, v) =1

2

∞u

q (s) ds, H k+1(u, v) = ∞

u

v

0q (α − β )H k(α, β ) dβ

dα. (2.1.40)

It can be shown by induction that

|H k+1(u, v)| ≤ 1

2Q0(u)

(Q1(u − v) − Q1(u))k

k!, k ≥ 0, 0 ≤ v ≤ u. (2.1.41)

Indeed, for k = 0, (2.1.41) is obvious. Suppose that (2.1.41) is valid for H k(u, v). Then(2.1.40) implies

|H k+1(u, v)| ≤ 1

2 ∞

uQ0(α)

v

0|q (α − β )|(Q1(α − β ) − Q1(α))k−1

(k−

1)!dβ dα.

Since the functions Q0(x) and Q1(x) are monotonic, we get

|H k+1(u, v)| ≤ 1

2· Q0(u)

(k − 1)!

∞u

(Q1(α − v) − Q1(α))k−1(Q0(α − v) − Q0(α)) dα

=1

2Q0(u)

(Q1(u − v) − Q1(u))k

k!,

i.e. (2.1.41) is proved. Therefore, the series

H (u, v) =∞

k=1

H k(u, v)

converges absolutely and uniformly for 0 ≤ v ≤ u, and

H (u, v) =1

2

∞u

q (s) ds + ∞

u

v

0q (α − β )H (α, β ) dβ

dα, (2.1.42)

|H (u, v)| ≤ 1

2Q0(u)exp

Q1(u − v) − Q1(u)

. (2.1.43)

Put

A(x, t) = H t + x

2,

t − x

2

. (2.1.44)

Then

A(x, t) =∞

k=1

ak(x, t),

the series converges absolutely and uniformly for 0 ≤ x ≤ t , and (2.1.33)-(2.1.35) are valid.Using (2.1.35) we calculate ∞

x|A(x, t)| dt ≤ exp(Q1(x))

∞x

Q0(ξ )exp(−Q1(ξ )) dξ

= exp(Q1(x)) ∞

x

d

dξ

exp(−Q1(ξ ))

dξ = exp(Q1(x)) − 1,

and we arrive at (2.1.36).



111


∂H (u, v)

∂u= −1

2q (u) −

v

0q (u − β )H (u, β ) dβ, (2.1.45)

∂H (u, v)∂v = ∞u

q (α − v)H (α, v) dα. (2.1.46)

It follows from (2.1.45)-(2.1.46) and (2.1.43) that

∂H (u, v)

∂u+

1

2q (u)

≤ 1

2

v

0|q (u − β )|Q0(u) exp(Q1(u − β ) − Q1(u)) dβ,

∂H (u, v)

∂v

≤ 1

2

∞u

|q (α − v)|Q0(α) exp(Q1(α − v) − Q1(α)) dα.

Since v

0|q (u − β )| dβ =

u

u−v|q (s)| ds ≤ Q0(u − v),

Q1(α − v) − Q1(α) = α

α−vQ0(t) dt ≤

u

u−vQ0(t) dt = Q1(u − v) − Q1(u), u ≤ α,

we get

∂H (u, v)

∂u+

1

2q (u) ≤

1

2Q0(u)exp(Q1(u − v) − Q1(u))

v

0|q (u − β )| dβ

≤ 1

2Q0(u − v)Q0(u) exp(Q1(u − v) − Q1(u)), (2.1.47)

∂H (u, v)

∂v

≤ 1

2Q0(u) exp(Q1(u − v) − Q1(u))

∞u

|q (α − v)| dα

≤ 1

2Q0(u − v)Q0(u) exp(Q1(u − v) − Q1(u)). (2.1.48)


∂A(x, t)

∂x=

1

2

∂H (u, v)

∂u− ∂H (u, v)

∂v

,

∂A(x, t)

∂t=

1

2

∂H (u, v)

∂u+

∂H (u, v)

∂v

,

where

u =t + x

2, v =

t − x

2.

Hence,∂A(x, t)

∂x +

1

4 q x + t

2 =

1

2∂H (u, v)

∂u +

1

2 q (u)−1

2

∂H (u, v)

∂v ,∂A(x, t)

∂t+

1

4q x + t

2

=

1

2

∂H (u, v)

∂u+

1

2q (u)

+

1

2

∂H (u, v)

∂v.

Taking (2.1.47) and (2.1.48) into account we arrive at (2.1.37). 2

Let (1 + x)q (x) ∈ L(0, ∞). We introduce the potentials

q r(x) = q (x), x ≤ r,

0, x > r,r

≥0 (2.1.49)



113

According to Lemma 2.1.2,

ur(x, ρ) ≤ exp(Q1(r)) ∞

rt|q (t)| dt exp

r

x(t − x)|q (t)| dt

, x ≤ r,

and consequently

ur(x, ρ) ≤ exp(Q1(x)) ∞

rt|q (t)| dt ≤ exp(Q1(0))

∞r

t|q (t)| dt, x ≤ r.

Together with (2.1.57) this yields (2.1.54).Denote

vr(x, ρ) = |(er(x, ρ) − e(x, ρ)) exp(−iρx)|, Im ρ ≥ 0, x ≥ 0, r ≥ 0.


vr(x, ρ) ≤ ∞

x|q r(t)z r(t, ρ) − q (t)z (t, ρ)| dt, Im ρ ≥ 0, x ≥ 0, r ≥ 0. (2.1.58)

Let x ≥ r. By virtue of (2.1.23), (2.1.49) and (2.1.58),

vr(x, ρ) ≤ ∞

x|q (t)| exp(Q1(t)) dt

≤Q0(x)exp(Q1(x))

≤Q0(r)exp(Q1(0)), Im ρ

≥0, 0

≤r

≤x. (2.1.59)

For x ≤ r, (2.1.58) gives

vr(x, ρ) ≤ ∞

r|q (t)z (t, ρ)| dt +

r

x|q (t)|ur(t, ρ) dt.

Using (2.1.23) and (2.1.54) we infer

vr(x, ρ) ≤ ∞

r|q (t)| exp(Q1(t)) dt + exp(Q1(0))

∞r

s|q (s)| ds r

x|q (t)| dt,

and consequently

vr(x, ρ) ≤

Q0(r) + Q0(0) ∞

rt|q (t)| dt

exp(Q1(0)), 0 ≤ x ≤ r, Imρ ≥ 0.

Together with (2.1.59) this yields (2.1.55). 2

Theorem 2.1.4. For each δ > 0, there exists a = aδ ≥ 0 such that equation (2.1.1)has a unique solution y = E (x, ρ), ρ ∈ Ωδ, satisfying the integral equation

E (x, ρ) = exp(−iρx)+1

2iρ

x

aexp(iρ(x−t))q (t)E (t, ρ) dt+

12iρ

∞x

exp(iρ(t−x))q (t)E (t, ρ) dt.

(2.1.60)The function E (x, ρ), called the Birkhoff solution for (2.1.1), has the following properties:(i1) E (ν )(x, ρ) = (−iρ)ν exp(−iρx)(1 + o(1)), x → ∞ ν = 0, 1, uniformly for |ρ| ≥δ, Imρ ≥ α, for each fixed α > 0;(i2) E (ν )(x, ρ) = (−iρ)ν exp(−iρx)(1 + O(ρ−1)), |ρ| → ∞ ρ ∈ Ω, uniformly for x ≥ a;(i3) for each fixed x ≥ 0, the functions E (ν )(x, ρ) are analytic for Im ρ > 0, |ρ| ≥ δ, and

are continuous for ρ ∈ Ωδ;



114

(i4) the functions e(x, ρ) and E (x, ρ) form a fundamental system of solutions for (2.1.1),and e(x, ρ), E (x, ρ) = −2iρ.(i5) If δ ≥ Q0(0), then one can take above a = 0.

Proof. For fixed δ > 0 choose a = aδ ≥ 0 such that Q0(a) ≤ δ. We transform (2.1.60)

by means of the replacement E (x, ρ) = exp(−iρx)ξ (x, ρ) to the equation

ξ (x, ρ) = 1 +1

2iρ

x

aexp(2iρ(x − t))q (t)ξ (t, ρ) dt +

1

2iρ

∞x

q (t)ξ (t, ρ) dt. (2.1.61)


ξ 0(x, ρ) = 1, ξ k+1(x, ρ) =1

2iρ

x

aexp(2iρ(x − t))q (t)ξ k(t, ρ) dt +

1

2iρ

∞x

q (t)ξ k(t, ρ) dt,

ξ (x, ρ) =

∞k=0

ξ k(x, ρ).

This yields

|ξ k+1(x, ρ)| ≤ 1

2|ρ| ∞

a|q (t)ξ k(t, ρ)| dt,

and hence

|ξ k(x, ρ)| ≤(Q0(a)

2|ρ|k

.

Thus for x ≥ a, |ρ| ≥ Q0(a), we get

|ξ (x, ρ)| ≤ 2, |ξ (x, ρ) − 1| ≤ Q0(a)

|ρ| .


E (x, ρ) = exp(−iρx)−iρ +

1

2

x

aexp(2iρ(x − t))q (t)ξ (t, ρ) dt − 1

2

∞x

q (t)ξ (t, ρ) dt

.

(2.1.62)

Since |ξ (x, ρ)| ≤ 2 for x ≥ a, ρ ∈ Ωδ, it follows from (2.1.61) and (2.1.62) that

|E (ν )(x, ρ)(−iρ)−ν exp(iρx) − 1| ≤ 1

|ρ| x

aexp(−2τ (x − t))|q (t)| dt +

∞x

|q (t)| dt

≤ 1

|ρ|

exp(−τ x) x/2

a|q (t)| dt +

∞x/2

|q (t)| dt

,

and consequently (i1) − (i2) are proved.

The other assertions of Theorem 2.1.4 are obvious.2

2.1.2. Properties of the spectrum. Denote

∆(ρ) = e(0, ρ) − h e(0, ρ). (2.1.63)

By virtue of Theorem 2.1.1, the function ∆(ρ) is analytic for I m ρ > 0, and continuousfor ρ ∈ Ω. It follows from (2.1.5) that for |ρ| → ∞, ρ ∈ Ω,

e(0, ρ) = 1 +ω1

iρ+ o1

ρ, ∆(ρ) = (iρ)1 +ω11

iρ+ o1

ρ, (2.1.64)



115

where ω1 = ω(0), ω11 = ω(0) − h. Using (2.1.7), (2.1.16) and (2.1.20) one can obtain moreprecisely

e(0, ρ) = 1 +ω1

iρ+

1

2iρ

∞0

q (t) exp(2iρt) dt + O 1

ρ2

,

∆(ρ) = (iρ)

1 + ω11

iρ− 1

2iρ

∞0


ρ2

.

(2.1.65)

DenoteΛ = λ = ρ2 : ρ ∈ Ω, ∆(ρ) = 0,

Λ = λ = ρ2 : Im ρ > 0, ∆(ρ) = 0,

Λ = λ = ρ2 : Im ρ = 0, ρ = 0, ∆(ρ) = 0.

Obviously, Λ = Λ ∪ Λ is a bounded set, and Λ is a bounded and at most countable set.Denote

Φ(x, λ) =e(x, ρ)

∆(ρ). (2.1.66)

The function Φ(x, λ) satisfies (2.1.1) and on account of (2.1.63) and Theorem 2.1.1 also theconditions

U (Φ) = 1, (2.1.67)

Φ(x, λ) = O(exp(iρx)), x

→ ∞, ρ

∈Ω, (2.1.68)

where U is defined by (2.1.2). The function Φ(x, λ) is called the Weyl solution for L.Note that (2.1.1), (2.1.67) and (2.1.68) uniquely determine the Weyl solution.

Denote M (λ) := Φ(0, λ). The function M (λ) is called the Weyl function for L. Itfollows from (2.1.66) that

M (λ) =e(0, ρ)

∆(ρ). (2.1.69)

Clearly,

Φ(x, λ) = S (x, λ) + M (λ)ϕ(x, λ), (2.1.70)where the functions ϕ(x, λ) and S (x, λ) are solutions of (2.1.1) under the initial conditions

ϕ(0, λ) = 1, ϕ(0, λ) = h, S (0, λ) = 0, S (0, λ) = 1.

We recall that the Weyl function plays an important role in the spectral theory of Sturm-Liouville operators (see [lev3] for more details).

By virtue of Liouville‘s formula for the Wronskian [cod1, p.83], ϕ(x, λ), Φ(x, λ) doesnot depend on x. Since for x = 0,

ϕ(x, λ), Φ(x, λ)|x=0 = U (Φ) = 1,

we inferϕ(x, λ), Φ(x, λ) ≡ 1. (2.1.71)

Theorem 2.1.5. The Weyl function M (λ) is analytic in Π \ Λ and continuous in Π1 \ Λ. The set of singularities of M (λ) (as an analytic function) coincides with the set

Λ0 := λ : λ ≥ 0 ∪ Λ.



116

Theorem 2.1.5 follows from (2.1.63), (2.1.69) and Theorem 2.1.1. By virtue of (2.1.70),the set of singularities of the Weyl solution Φ(x, λ) coincides with Λ0 for all x ≥ 0, sincethe functions ϕ(x, λ) and S (x, λ) are entire in λ for each fixed x ≥ 0.

Definition 2.1.1. The set of singularities of the Weyl function M (λ) is called the

spectrum of L. The values of the parameter λ, for which equation (2.1.1) has nontrivialsolutions satisfying the conditions U (y) = 0, y(∞) = 0 (i.e. limx→∞ y(x) = 0 ), are calledeigenvalues of L, and the corresponding solutions are called eigenfunctions.

Remark 2.1.3. One can introduce the operator

Lo : D(Lo) → L2(0, ∞), y → −y + q (x)y

with the domain of definition D(Lo) = y : y ∈ L2(I ) ∩ AC loc(I ), y ∈ AC loc(I ), Loy ∈L2(I ), U (y) = 0

, where I := [0,

∞). It is easy to verify that the spectrum of Lo coincides

with Λ0. For the Sturm-Liouville equation there is no difference between working either withthe operator Lo or with the pair L. However, for generalizations for many other classes of inverse problems, from methodical point of view it is more natural to consider the pair L(see, for example, [yur1] ).

Theorem 2.1.6. L has no eigenvalues λ > 0.

Proof. Suppose that λ0 = ρ20 > 0 is an eigenvalue, and let y0(x) be a correspond-

ing eigenfunction. Since the functions e(x, ρ0), e(x, −ρ0) form a fundamental system

of solutions of (2.1.1), we have y0(x) = Ae(x, ρ0) + Be(x, −ρ0). For x → ∞, y0(x) ∼0, e(x, ±ρ0) ∼ exp(±iρ0x). But this is possible only if A = B = 0. 2

Theorem 2.1.7. Let λ0 /∈ [0, ∞). For λ0 to be an eigenvalue, it is necessary andsufficient that ∆(ρ0) = 0. In other words, the set of nonzero eigenvalues coincides withΛ. For each eigenvalue λ0 ∈ Λ there exists only one (up to a multiplicative constant)eigenfunction, namely,

ϕ(x, λ0) = β 0e(x, ρ0), β 0 = 0. (2.1.72)

Proof. Let λ0 ∈ Λ. Then U (e(x, ρ0)) = ∆(ρ0) = 0 and, by virtue of (2.1.4),lim

x→∞ e(x, ρ0) = 0. Thus, e(x, ρ0) is an eigenfunction, and λ0 = ρ20 is an eigenvalue. More-

over, it follows from (2.1.66) and (2.1.71) that ϕ(x, λ), e(x, ρ) = ∆(ρ), and consequently(2.1.72) is valid.

Conversely, let λ0 = ρ20, Im ρ0 > 0 be an eigenvalue, and let y0(x) be a corresponding

eigenfunction. Clearly, y0(0) = 0. Without loss of generality we put y0(0) = 1. Theny0(0) = h, and hence y0(x) ≡ ϕ(x, λ0). Since the functions E (x, ρ0) and e(x, ρ0) form afundamental system of solutions of equation (2.1.1), we get y0(x) = α0E (x, ρ0) + β 0e(x, ρ0).

As x → ∞, we calculate α0 = 0, i.e. y0(x) = β 0e(x, ρ0). This yields (2.1.72). Conse-quently, ∆(ρ0) = U (e(x, ρ0)) = 0, and ϕ(x, λ0) and e(x, ρ0) are eigenfunctions. 2

Thus, the spectrum of L consists of the positive half-line λ : λ ≥ 0, and the discreteset Λ = Λ ∪ Λ. Each element of Λ is an eigenvalue of L. According to Theorem 2.1.6,the points of Λ are not eigenvalues of L, they are called spectral singularities of L.

Example 2.1.1. Let q (x) ≡ 0, h = iθ, where θ is a real number. Then ∆(ρ) = iρ−h,and Λ = ∅, Λ = θ, i.e. L has no eigenvalues, and the point ρ0 = θ is a spectral

singularity for L.



117

It follows from (2.1.5), (2.1.64), (2.1.66) and (2.1.69) that for |ρ| → ∞, ρ ∈ Ω,

M (λ) =1

iρ

1 +

m1

iρ+ o

1

ρ

, (2.1.73)

Φ(ν )(x, λ) = (iρ)ν −1 exp(iρx)

1 + B(x)iρ

+ o1

ρ

, (2.1.74)

uniformly for x ≥ 0; here

m1 = h, B(x) = h +1

2

x

0q (s) ds.

Taking (2.1.65) into account one can derive more precisely

M (λ) = 1iρ

1 + m1

iρ+ 1

iρ

∞0


ρ2

, |ρ| → ∞, ρ ∈ Ω. (2.1.75)

Moreover, if q (x) ∈ W N , then by virtue of (2.1.21) we get

M (λ) =1

iρ

1 +

N +1s=1

ms

(iρ)s+ o

1

ρN +1

, |ρ| → ∞, ρ ∈ Ω, (2.1.76)

where m1 = h, m2 = −1

2q (0) + h2, · · · .Denote

V (λ) =1

2πi

M −(λ) − M +(λ)

, λ > 0, (2.1.77)

whereM ±(λ) = lim

z→0,Rez>0M (λ ± iz ).

It follows from (2.1.73) and (2.1.77) that for ρ > 0, ρ → +∞,

V (λ) = 12πi

− 1

iρ

1 − m1

iρ+ o

1ρ

− 1

iρ

1 + m1

iρ+ o

1ρ

,

and consequently

V (λ) =1

πρ

1 + o

1

ρ

, ρ > 0, ρ → +∞. (2.1.78)

In view of (2.1.75), we calculate more precisely

V (λ) = 1πρ

1 + 1

ρ

∞0

q (t)sin2ρtdt + O 1

ρ2

, ρ > 0, ρ → +∞. (2.1.79)

Moreover, if q (x) ∈ W N +1, then (2.1.76) implies

V (λ) =1

πρ

1 +

N +1s=1

V sρs

+ o

1

ρN +1

, ρ > 0, ρ → +∞, (2.1.80)

where V 2s = (−

1)sm2s, V 2s+1 = 0.



118

Remark 2.1.4. Analogous results are also valid if we replace U (y) = 0 by the bound-ary condition U 0(y) := y(0) = 0. In this case, the Weyl solution Φ0(x, λ) and the Weylfunction M 0(λ) are defined by the conditions Φ0(0, λ) = 1, Φ0(x, λ) = O(exp(iρx)), x →∞, M 0(λ) := Φ

0(0, λ), and

Φ0(x, λ) = e(x, ρ)e(0, ρ)

= C (x, λ) + M 0(λ)S (x, λ), M 0(λ) = e(0, ρ)e(0, ρ)

, (2.1.81)

where C (x, λ) is a solution of (2.1.1) under the conditions C (0, λ) = 1, C (0, λ) = 0.

2.1.3. An expansion theorem. In the λ - plane we consider the contour γ = γ ∪γ

(with counterclockwise circuit), where γ is a bounded closed contour encircling the setΛ ∪ 0, and γ is the two-sided cut along the arc λ : λ > 0, λ /∈ intγ .

6

-'

E

γ

γ

γ

fig. 2.1.1.

Theorem 2.1.8. Let f (x) ∈ W 2. Then, uniformly for x ≥ 0,

f (x) =

1

2πi γ ϕ(x, λ)F (λ)M (λ) dλ, (2.1.82)

whereF (λ) :=

∞0

ϕ(t, λ)f (t) dt.

Proof. We will use the contour integral method. For this purpose we consider thefunction

Y (x, λ) = Φ(x, λ) x

0ϕ(t, λ)f (t) dt + ϕ(x, λ)

∞

xΦ(t, λ)f (t) dt. (2.1.83)

Since the functions ϕ(x, λ) and Φ(x, λ) satisfy (2.1.1) we transform Y (x, λ) as follows

Y (x, λ) =1

λΦ(x, λ)

x

0(−ϕ(t, λ) + q (t)ϕ(t, λ))f (t) dt

+1

λϕ(x, λ)

∞x

(−Φ(t, λ) + q (t)Φ(t, λ))f (t) dt.



119

Two-fold integration by parts of terms with second derivatives yields in view of (2.1.71)

Y (x, λ) =1

λ

f (x) + Z (x, λ)

, (2.1.84)

whereZ (x, λ) = (f (0) − hf (0))Φ(x, λ) + Φ(x, λ)

x

0ϕ(t, λ)f (t) dt

+ϕ(x, λ) ∞

xΦ(t, λ)f (t) dt. (2.1.85)

Similarly,

F (λ) = − 1

λ

f (0) − hf (0)

+

1

λ

∞0

ϕ(t, λ)f (t) dt, λ > 0. (2.1.86)

The function ϕ(x, λ) satisfies the integral equation (1.1.11). Denote

µT (λ) = max0≤x≤T

(|ϕ(x, λ)| exp(−|τ |x)), τ := Imρ.

Then (1.1.11) gives for |ρ| ≥ 1, x ∈ [0, T ],

|ϕ(x, λ)| exp(−|τ |x) ≤ C +µT (λ)

|ρ| T

0|q (t)| dt,

and consequently

µT (λ) ≤ C +µT (λ)

|ρ| T

0|q (t)| dt ≤ C +

µT (λ)

|ρ| ∞

0|q (t)| dt.

From this we get |µT (λ)| ≤ C for |ρ| ≥ ρ∗. Together with (1.1.12) this yields for ν =0, 1, |ρ| ≥ ρ∗,

|ϕ(ν )(x, λ)| ≤ C |ρ|ν exp(|τ |x), (2.1.87)

uniformly for x ≥ 0. Furthermore, it follows from (2.1.74) that for ν = 0, 1, |ρ| ≥ ρ∗,

|Φ(ν )(x, λ)| ≤ C |ρ|ν −1 exp(−|τ |x), (2.1.88)

uniformly for x ≥ 0. By virtue of (2.1.85), (2.1.87) and (2.1.88) we get

Z (x, λ) = O1

ρ

, |λ| → ∞,

uniformly for x ≥ 0. Hence, (2.1.84) implies

limR→∞

supx≥0

f (x) − 1

2πi

|λ|=R

Y (x, λ) dλ = 0, (2.1.89)

where the contour in the integral is used with counterclockwise circuit. Consider the contourγ 0R = (γ ∩ λ : |λ| ≤ R) ∪ λ : |λ| = R (with clockwise circuit).



120

6

-'E

s

~

R

γ 0R

fig. 2.1.2

By Cauchy‘s theorem [con1, p.85]

1

2πi

γ 0R

Y (x, λ)dλ = 0.

Taking (2.1.89) into account we obtain

limR→∞

sup

x≥0 f (x)

−

1

2πi γ R

Y (x, λ) dλ = 0,

where γ R = γ ∩λ : |λ| ≤ R (with counterclockwise circuit). From this, using (2.1.83) and(2.1.70), we arrive at (2.1.82), since the terms with S (x, λ) vanish by Cauchy‘s theorem.We note that according to (2.1.79), (2.1.86) and (2.1.87),

F (λ) = O1

λ

, M (λ) = O

1

ρ

, ϕ(x, λ) = O(1), x ≥ 0, λ > 0, λ → ∞,

and consequently the integral in (2.1.82) converges absolutely and uniformly for x

≥0. 2

Remark 2.1.5. If q (x) and h are real, and (1 + x)q (x) ∈ L(0, ∞), then (see Section2.3) Λ = ∅, Λ ⊂ (−∞, 0) is a finite set of simple eigenvalues, V (λ) > 0 for λ > 0 ( V (λ)is defined by (2.1.77)), and M (λ) = O(ρ−1) as ρ → 0. Then (2.1.82) takes the form

f (x) = ∞

0ϕ(x, λ)F (λ)V (λ)dλ +

λj∈Λ

ϕ(x, λ j)F (λ j)Q j, Q j := Resλ=λj

M (λ).

orf (x) =

∞

−∞ϕ(x, λ)F (λ) dσ(λ),

where σ(λ) is the spectral function of L (see [lev3]). For λ < 0, σ(λ) is a step-function;for λ > 0, σ(λ) is an absolutely continuous function, and σ(λ) = V (λ).

Remark 2.1.6. It follows from the proof that Theorem 2.1.8 remains valid also forf (x) ∈ W 1.

2.2. RECOVERY OF THE DIFFERENTIAL EQUATION FROM THE WEYL

FUNCTION.



121

In this section we study the inverse problem of recovering the pair L = L(q (x), h) of the form (2.1.1)-(2.1.2) from the given Weyl function M (λ). For this purpose we will usethe method of spectral mappings described in Section 1.6 for Sturm-Liouville operators ona finite interval. Since for the half-line the arguments are more or less similar, the proofs inthis section are shorter than in Section 1.6.

First, let us prove the uniqueness theorem for the solution of the inverse problem. Weagree, as in Chapter 1, that together with L we consider here and in the sequel a pairL = L(q (x), h) of the same form but with different coefficients. If a certain symbol γ denotes an object related to L, then the corresponding symbol γ with tilde denotes theanalogous object related to L, and γ := γ − γ.

Theorem 2.2.1 If M (λ) = M (λ), then L = L. Thus, the specification of the Weyl function uniquely determines q (x) and h .

Proof. Let us define the matrix P (x, λ) = [P jk (x, λ)] j,k=1,2 by the formula

P (x, λ)

ϕ(x, λ) Φ(x, λ)

ϕ(x, λ) Φ(x, λ)

=


.

By virtue of (2.1.71), this yields

P j1(x, λ) = ϕ( j−1)(x, λ)Φ(x, λ) − Φ( j−1)(x, λ)ϕ(x, λ)

P j2(x, λ) = Φ( j

−1)

(x, λ)ϕ(x, λ) − ϕ( j

−1)

(x, λ)˜Φ(x, λ)

, (2.2.1)

ϕ(x, λ) = P 11(x, λ)ϕ(x, λ) + P 12(x, λ)ϕ(x, λ)

Φ(x, λ) = P 11(x, λ)Φ(x, λ) + P 12(x, λ)Φ(x, λ)

. (2.2.2)

Using (2.2.1), (2.1.87)-(2.1.88) we get for |λ| → ∞, λ = ρ2,

P jk (x, λ) − δ jk = O(ρ−1), j ≤ k; P 21(x, λ) = O(1). (2.2.3)

If M (λ) ≡ M (λ) , then in view of (2.2.1) and (2.1.70), for each fixed x, the functionsP jk (x, λ) are entire in λ. Together with (2.2.3) this yields P 11(x, λ) ≡ 1, P 12(x, λ) ≡ 0.Substituting into (2.2.2) we get ϕ(x, λ) ≡ ϕ(x, λ), Φ(x, λ) ≡ Φ(x, λ) for all x and λ, andconsequently, L = L. 2

Let us now start to construct the solution of the inverse problem. We shall say thatL ∈ V N if q (x) ∈ W N . We shall subsequently solve the inverse problem in the classes V N .

Let a model pair L = L(q (x), h) be chosen such that

∞ρ∗

ρ4|V (λ)|2 dρ < ∞, V := V − V (2.2.4)

for sufficiently large ρ∗ > 0. The condition (2.2.4) is needed for technical reasons. Inprincipal, one could take any L (for example, with q (x) = h = 0 ) but generally speakingproofs would become more complicated. On the other hand, (2.2.4) is not a very strongrestriction, since by virtue of (2.1.79),

ρ2V (λ) =1

π ∞

0

q (t)sin2ρtdt + O1

ρ.



122

In particular, if q (x) ∈ L2, then (2.2.4) is fulfilled automatically for any q (x) ∈ L2. Hence,for N ≥ 1, the condition (2.2.4) is fulfilled for any model L ∈ V N .


∞

λ∗ |V (λ)

|dλ = 2

∞

ρ∗

ρ

|V (λ)

|dρ <

∞, λ∗ = (ρ∗)2, λ = ρ2. (2.2.5)

Denote

D(x,λ,µ) =ϕ(x, λ), ϕ(x, µ)

λ − µ= x

0ϕ(t, λ)ϕ(t, µ) dt, r(x,λ,µ) = D(x,λ,µ)M (µ),

D(x,λ,µ) =ϕ(x, λ), ϕ(x, µ)

λ − µ= x

0ϕ(t, λ)ϕ(t, µ) dt, r(x,λ,µ) = D(x,λ,µ)M (µ).

(2.2.6)

Lemma 2.2.1. The following estimate holds

|D(x,λ,µ)|, |D(x,λ,µ)| ≤ C x exp(|Im ρ|x)

|ρ ∓ θ| + 1, λ = ρ2, µ = θ2 ≥ 0, ±θReρ ≥ 0. (2.2.7)

Proof. Let ρ = σ + iτ. For definiteness, let θ ≥ 0 and σ ≥ 0. All other cases can betreated in the same way. Take a fixed δ 0 > 0. For |ρ − θ| ≥ δ 0 we have by virtue of (2.2.6)and (2.1.87),

|D(x,λ,µ)| = ϕ(x, λ), ϕ(x, µ)λ − µ ≤ C exp(|τ |x) |ρ| + |θ||ρ2 − θ2| . (2.2.8)

Since|ρ| + |θ||ρ + θ| =

√ σ2 + τ 2 + θ

(σ + θ)2 + τ 2≤

√ σ

2 + τ 2 + θ√ σ2 + τ 2 + θ2

≤√

2

(here we use that (a + b)2 ≤ 2(a2 + b2) for all real a, b ), (2.2.8) implies

|D(x,λ,µ)| ≤C exp(

|τ

|x)

|ρ − θ| . (2.2.9)

For |ρ − θ| ≥ δ 0, we get|ρ − θ| + 1

|ρ − θ| ≤ 1 +1

δ 0,

and consequently1

|ρ − θ| ≤ C 0|ρ − θ| + 1

with C 0 =δ 0 + 1

δ 0.

Substituting this estimate into the right-hand side of (2.2.9) we obtain

|D(x,λ,µ)| ≤ C exp(|τ |x)

|ρ − θ| + 1,

and (2.2.7) is proved for |ρ − θ| ≥ δ 0.For |ρ − θ| ≤ δ 0, we have by virtue of (2.2.6) and (2.1.87),

|D(x,λ,µ)| ≤ x

0 |ϕ(t, λ)ϕ(t, µ)| dt ≤ C x exp(|τ |x),



123

i.e. (2.2.7) is also valid for |ρ − θ| ≤ δ 0. 2

Lemma 2.2.2. The following estimates hold

∞

1

dθ

θ(

|R

−θ

|+ 1)

= O

ln R

R , R → ∞, (2.2.10)

∞1

dθ

θ2(|R − θ| + 1)2= O

1

R2

, R → ∞. (2.2.11)

Proof. Since

1

θ(R − θ + 1)=

1

R + 1

1

θ+

1

R − θ + 1

,

1

θ(θ − R + 1)=

1

R − 1

1

θ − R + 1− 1

θ

,

we calculate for R > 1, ∞1

dθ

θ(|R − θ| + 1)= R

1

dθ

θ(R − θ + 1)+ ∞

R

dθ

θ(θ − R + 1)

=1

R + 1

R

1

1

θ+

1

R − θ + 1

dθ +

1

R − 1

∞R

1

θ − R + 1− 1

θ

dθ

=2 ln R

R + 1+

ln R

R − 1,

i.e. (2.2.10) is valid. Similarly, for R > 1, ∞1

dθ

θ2(|R − θ| + 1)2= R

1

dθ

θ2(R − θ + 1)2+ ∞

R

dθ

θ2(θ − R + 1)2

≤ 1

(R + 1)2

R

1

1

θ+

1

R − θ + 1

2dθ +

1

R2

∞R

dθ

(θ − R + 1)2

=2

(R + 1)2

R

1

dθ

θ2

+ R

1

dθ

θ(R − θ + 1)+

1

R2

∞

1

dθ

θ2

= O1

R2,

i.e. (2.2.11) is valid. 2

In the λ - plane we consider the contour γ = γ ∪ γ (with counterclockwise circuit),where γ is a bounded closed contour encircling the set Λ∪Λ∪0, and γ is the two-sidedcut along the arc λ : λ > 0, λ /∈ intγ (see fig. 2.1.1).

Theorem 2.2.2. The following relations hold

ϕ(x, λ) = ϕ(x, λ) +

1

2πi γ r(x,λ,µ)ϕ(x, µ) dµ, (2.2.12)

r(x,λ,µ) − r(x,λ,µ) +1

2πi

γ

r(x,λ,ξ )r(x, ξ, µ) dξ = 0. (2.2.13)

Equation (2.2.12) is called the main equation of the inverse problem.

Proof. It follows from (2.1.73), (2.1.87), (2.2.6) and (2.2.7) that for λ, µ ∈ γ, ±ReρReθ ≥0,

|r(x,λ,µ)

|,

|r(x,λ,µ)

| ≤C x

|µ|(|ρ ∓ θ| + 1),

|ϕ(x, λ)

| ≤C. (2.2.14)



124

In view of (2.2.10), it follows from (2.2.14) that the integrals in (2.2.12) and (2.2.13) convergeabsolutely and uniformly on γ for each fixed x ≥ 0.

Denote J γ = λ : λ /∈ γ ∪ intγ . Consider the contour γ R = γ ∩ λ : |λ| ≤ Rwith counterclockwise circuit, and also consider the contour γ 0R = γ R ∪ λ : |λ| = R withclockwise circuit (see fig 2.1.2). By Cauchy‘s integral formula [con1, p.84],

P 1k(x, λ) − δ 1k =1

2πi

γ 0R

P 1k(x, µ) − δ 1k

λ − µdµ, λ ∈ intγ 0R,


λ − µ=

1

2πi

γ 0R

P jk (x, ξ )

(λ − ξ )(ξ − µ)dξ, λ, µ ∈ intγ 0R.


limR→∞ |µ|=R

P 1k(x, µ)

−δ 1k

λ − µ dµ = 0, limR→∞ |ξ|=R

P jk (x, ξ )

(λ − ξ )(ξ − µ) dξ = 0,

and consequently

P 1k(x, λ) = δ 1k +1

2πi

γ

P 1k(x, µ)

λ − µdµ, λ ∈ J γ , (2.2.15)


λ − µ=

1

2πi

γ

P jk (x, ξ )

(λ − ξ )(ξ − µ)dξ, λ, µ ∈ J γ . (2.2.16)

Here (and everywhere below, where necessary) the integral is understood in the principal

value sense:

γ = lim

R→∞

γ R.

By virtue of (2.2.2) and (2.2.15),

ϕ(x, λ) = ϕ(x, λ) +1

2πi

γ

ϕ(x, λ)P 11(x, µ) + ϕ(x, λ)P 12(x, µ)

λ − µdµ, λ ∈ J γ .

Taking (2.2.1) into account we get

ϕ(x, λ) = ϕ(x, λ) + 12πi

γ (ϕ(x, λ)(ϕ(x, µ)Φ(x, µ) − Φ(x, µ)ϕ(x, µ))+

ϕ(x, λ)(Φ(x, µ)ϕ(x, µ) − ϕ(x, µ)Φ(x, µ))dµ

λ − µ.

In view of (2.1.70), this yields (2.2.12), since the terms with S (x, µ) vanish by Cauchy‘stheorem.

Using (2.2.16) and acting in the same way as in the proof of Lemma 1.6.3, we arrive at

D(x,λ,µ) − D(x,λ,µ) =

1

2πi

γ

ϕ(x, λ), Φ(x, ξ )ϕ(x, ξ ), ϕ(x, µ)(λ − ξ )(ξ − µ)

− ϕ(x, λ), ϕ(x, ξ )Φ(x, ξ ), ϕ(x, µ)(λ − ξ )(ξ − µ)

dξ.

In view of (2.1.70) and (2.2.6), this yields (2.2.13). 2

Analogously one can obtain the relation

Φ(x, λ) = Φ(x, λ) +1

2πi γ

Φ(x, λ), ϕ(x, µ)λ − µ

M (µ)ϕ(x, µ) dµ, λ∈

J γ . (2.2.17)



125

Let us consider the Banach space C (γ ) of continuous bounded functions z (λ), λ ∈ γ,with the norm z = sup

λ∈γ |z (λ)|.

Theorem 2.2.3. For each fixed x ≥ 0, the main equation (2.2.12) has a unique solution ϕ(x, λ)

∈C (γ ).

Proof. For a fixed x ≥ 0, we consider the following linear bounded operators in C (γ ) :

Az (λ) = z (λ) +1

2πi

γ

r(x,λ,µ)z (µ) dµ,

Az (λ) = z (λ) − 1

2πi

γ

r(x,λ,µ)z (µ) dµ.

Then

AAz (λ) = z (λ) +1

2πi

γ r(x,λ,µ)z (µ) dµ −

1

2πi

γ r(x,λ,µ)z (µ) dµ

− 1

2πi

γ

r(x,λ,ξ ) 1

2πi

γ

r(x,ξ,µ)z (µ) dµ

dξ

= z (λ) − 1

2πi

γ

r(x,λ,µ) − r(x,λ,µ) +

1

2πi

γ

r(x,λ,ξ )r(x,ξ,µ) dξ

z (µ) dµ.

By virtue of (2.2.13) this yields

AAz (λ) = z (λ), z (λ) ∈ C (γ ).

Interchanging places for L and L, we obtain analogously AAz (λ) = z (λ). Thus, AA =AA = E, where E is the identity operator. Hence the operator A has a bounded inverseoperator, and the main equation (2.2.12) is uniquely solvable for each fixed x ≥ 0. 2

Denote

ε0(x) =1

2πi

γ

ϕ(x, µ)ϕ(x, µ)M (µ) dµ, ε(x) = −2ε0(x). (2.2.18)


q (x) = q (x) + ε(x), (2.2.19)

h = h − ε0(0). (2.2.20)

Proof. Differentiating (2.2.12) twice with respect to x and using (2.2.6) and (2.2.18)we get

ϕ(x, λ) − ε0(x)ϕ(x, λ) = ϕ(x, λ) + 12πi

γ

r(x,λ,µ)ϕ(x, µ) dµ, (2.2.21)

ϕ(x, λ) = ϕ(x, λ) +1

2πi

γ

r(x,λ,µ)ϕ(x, µ) dµ

+1

2πi

γ

2ϕ(x, λ)ϕ(x, µ)M (µ)ϕ(x, µ) dµ +1

2πi

γ

(ϕ(x, λ)ϕ(x, µ))M (µ)ϕ(x, µ) dµ. (2.2.22)



126

In (2.2.22) we replace the second derivatives using equation (2.1.1), and then we replaceϕ(x, λ) using (2.2.12). This yields

q (x)ϕ(x, λ) = q (x)ϕ(x, λ) +1

2πi

γ ϕ(x, λ), ϕ(x, µ)M (µ)ϕ(x, µ) dµ

+1

2πi

γ

2ϕ(x, λ)ϕ(x, µ)M (µ)ϕ(x, µ) dµ +1

2πi

γ

(ϕ(x, λ)ϕ(x, µ))M (µ)ϕ(x, µ) dµ.

After canceling terms with ϕ(x, λ) we arrive at (2.2.19). Taking x = 0 in (2.2.21) we get(2.2.20). 2

Thus, we obtain the following algorithm for the solution of the inverse problem.

Algorithm 2.2.1. Let the function M (λ) be given. Then (1) Choose L

∈V N such that (2.2.4) holds.

(2) Find ϕ(x, λ) by solving equation (2.2.12).(3) Construct q (x) and h via (2.2.18)-(2.2.20).

Let us now formulate necessary and sufficient conditions for the solvability of the inverseproblem. Denote in the sequel by W the set of functions M (λ) such that(i) the functions M (λ) are analytic in Π with the exception of an at most countablebounded set Λ of poles, and are continuous in Π1 with the exception of bounded set Λ(in general, Λ and Λ are different for each function M (λ) );

(ii) for |λ| → ∞, (2.1.73) holds.Theorem 2.2.5. For a function M (λ) ∈ W to be the Weyl function for a certain

L ∈ V N , it is necessary and sufficient that the following conditions hold:1) (Asymptotics) There exists L ∈ V N such that (2.2.4) holds;2) (Condition S) For each fixed x ≥ 0, equation (2.2.12) has a unique solution ϕ(x, λ) ∈C (γ );3) ε(x) ∈ W N , where the function ε(x) is defined by (2.2.18).Under these conditions q (x) and h are constructed via (2.2.19)-(2.2.20).

As it is shown in Example 2.3.1, conditions 2) and 3) are essential and cannot be omitted.On the other hand, in Section 2.3 we provide classes of operators for which the uniquesolvability of the main equation can be proved.

The necessity part of Theorem 2.2.5 was proved above. We prove now the sufficiency .Let a function M (λ) ∈ W , satisfying the hypothesis of Theorem 2.2.5, be given, and letϕ(x, λ) be the solution of the main equation (2.2.12). Then (2.2.12) gives us the analyticcontinuation of ϕ(x, λ) to the whole λ - plane, and for each fixed x ≥ 0, the functionϕ(x, λ) is entire in λ of order 1/2. Using Lemma 1.5.1 one can show that the functionsϕ(ν )(x, λ), ν = 0, 1, are absolutely continuous with respect to x on compact sets, and

|ϕ(ν )(x, λ)| ≤ C |ρ|ν exp(|τ |x). (2.2.23)

We construct the function Φ(x, λ) via (2.2.17), and L = L(q (x), h) via (2.2.19)-(2.2.20).Obviously, L ∈ V N .


ϕ(x, λ) = λϕ(x, λ), Φ(x, λ) = λΦ(x, λ).



127

Proof. For simplicity, let ∞λ∗

ρ|V (λ)| dλ < ∞(the general case requires minor modifications). Then (2.2.23) is valid for ν = 0, 1, 2.Differentiating (2.2.12) twice with respect to x we obtain (2.2.21) and (2.2.22). It follows

from (2.2.22) and (2.2.12) that

−ϕ(x, λ) + q (x)ϕ(x, λ) = ϕ(x, λ) +1

2πi

γ


+1

2πi

γ ϕ(x, λ), ϕ(x, µ)M (µ)ϕ(x, µ) dµ − 2ϕ(x, λ)

1

2πi

γ

(ϕ(x, λ)ϕ(x, µ))M (µ)ϕ(x, µ) dµ.


ϕ(x, λ) = ϕ(x, λ) + 12πi

γ


+1

2πi

γ ϕ(x, λ), ϕ(x, µ)M (µ)ϕ(x, µ) dµ. (2.2.24)

Using (2.2.17) we calculate similarly

Φ(x, λ) − ε0(x)Φ(x, λ) = Φ(x, λ) +1

2πi γ

Φ(x, λ), ϕ(x, µ)λ

−µ

M (µ)ϕ(x, µ) dµ, (2.2.25)

Φ(x, λ) = Φ(x, λ) +1

2πi

γ

Φ(x, λ), ϕ(x, µ)λ − µ

M (µ)ϕ(x, µ) dµ

+1

2πi

γ Φ(x, λ), ϕ(x, µ)M (µ)ϕ(x, µ) dµ. (2.2.26)


λϕ(x, λ) = ϕ(x, λ) +1

2πi γ

r(x,λ,µ)ϕ(x, µ) dµ +1

2πi γ

(λ−

µ)r(x,λ,µ)ϕ(x, µ) dµ.

Taking (2.2.12) into account we deduce for a fixed x ≥ 0,

η(x, λ) +1

2πi

γ

r(x,λ,µ)η(x, µ) dµ = 0, λ ∈ γ, (2.2.27)

where η(x, λ) := ϕ(x, λ) − λϕ(x, λ). According to (2.2.23) we have

|η(x, λ)

| ≤C x

|ρ

|2, λ

∈γ. (2.2.28)

By virtue of (2.2.27), (2.2.6) and (2.2.7),

|η(x, λ)| ≤ C x

1 + ∞

λ∗

|V (µ)||ρ − θ| + 1

|η(x, µ)| dµ

, λ ∈ γ, θ > 0, Re ρ ≥ 0. (2.2.29)

Substituting (2.2.28) into the right-hand side of (2.2.29) we get

|η(x, λ)

| ≤C x1 +

∞

λ∗

θ2|V (µ)||ρ − θ| + 1

dµ, λ∈

γ, θ > 0, Re ρ≥

0.



128

Sinceθ

ρ(|ρ − θ| + 1)≤ 1 for θ, ρ ≥ 1,

this yields

|η(x, λ)

| ≤C x

|ρ

|, λ

∈γ. (2.2.30)

Using (2.2.30) instead of (2.2.28) and repeating the preceding arguments we infer

|η(x, λ)| ≤ C x, λ ∈ γ.

According to Condition S of Theorem 2.2.5, the homogeneous equation (2.2.27) has only thetrivial solution η(x, λ) ≡ 0. Consequently,

ϕ(x, λ) = λϕ(x, λ). (2.2.31)


λΦ(x, λ) = Φ(x, λ) +1

2πi

γ

Φ(x, λ), ϕ(x, µ)λ − µ

M (µ)µϕ(x, µ) dµ

+1

2πi

γ

(λ − µ)Φ(x, λ), ϕ(x, µ)

λ − µM (µ)ϕ(x, µ) dµ.

Together with (2.2.17) this yields Φ(x, λ) = λΦ(x, λ).2

Lemma 2.2.4 The following relations hold

ϕ(0, λ) = 1, ϕ(0, λ) = h. (2.2.32)

U (Φ) = 1, Φ(0, λ) = M (λ), (2.2.33)

Φ(x, λ) = O(exp(iρx)), x → ∞. (2.2.34)

Proof. Taking x = 0 in (2.2.12) and (2.2.21) and using (2.2.20) we get

ϕ(0, λ) = ϕ(0, λ) = 1,

ϕ(0, λ) = ϕ(0, λ) − ε0(0)ϕ(0, λ) = h + h − h = h,

i.e. (2.2.32) is valid. Using (2.2.17) and (2.2.25) we calculate

Φ(0, λ) = Φ(0, λ) +1

2πi

γ

M (µ)

λ

−µ

dµ, (2.2.35)

Φ(0, λ) = Φ(0, λ) − Φ(0, λ)ε0(0) +h

2πi

γ

M (µ)

λ − µdµ.

Consequently,

U (Φ) = Φ(0, λ) − hΦ(0, λ) = Φ(0, λ) − (ε0(0) + h)Φ(0, λ) =

Φ(0, λ) − hΦ(0, λ) = U (Φ) = 1.



129

Furthermore, since y, z = yz − yz, we rewrite (2.2.17) in the form

Φ(x, λ) = Φ(x, λ) +1

2πi

γ

Φ(x, λ)ϕ(x, µ) − Φ(x, λ)ϕ(x, µ)

λ − µM (µ)ϕ(x, µ) dµ, (2.2.36)

where λ ∈ J γ . The function ϕ(x, λ) is the solution of the Cauchy problem (2.2.31)-(2.2.32).Therefore, according to (2.1.87),

|ϕ(ν )(x, µ)| ≤ C |θ|ν , µ = θ2 ∈ γ, x ≥ 0, ν = 0, 1. (2.2.37)

Moreover, the estimates (2.1.87)-(2.1.88) are valid for ϕ(x, λ) and Φ(x, λ), i.e.

|ϕ(ν )(x, µ)| ≤ C |θ|ν , µ = θ2 ∈ γ, x ≥ 0, ν = 0, 1, (2.2.38)

|˜Φ

(ν )

(x, λ)| ≤ C |ρ|ν

−1

exp(−|Im ρ|x), x ≥ 0, ρ ∈ Ω. (2.2.39)By virtue of (2.1.73),

M (λ) = O1

λ

, |ρ| → ∞, ρ ∈ Ω. (2.2.40)

Fix λ ∈ J γ . Taking (2.2.37)-(2.2.40) into account we get from (2.2.36) that

|Φ(x, λ)exp(−iρx)| ≤ C

1 + ∞

ρ∗

dθ

θ|λ − µ|

≤ C 1,

i.e. (2.2.34) is valid.Furthermore, it follows from (2.2.35) that

Φ(0, λ) = M (λ) +1

2πi

γ

M (µ)

λ − µdµ.

By Cauchy‘s integral formula

M (λ) = 12πi

γ 0R

ˆM (µ)λ − µ dµ, λ ∈ intγ 0R.

Since

limR→∞

1

2πi

|µ|=R

M (µ)

λ − µdµ = 0,

we get

M (λ) =1

2πi γ

M (µ)

λ−

µdµ, λ ∈ J γ .

Consequently, Φ(0, λ) = M (λ) + M (λ) = M (λ), i.e. (2.2.33) is valid. 2

Thus, Φ(x, λ) is the Weyl solution, and M (λ) is the Weyl function for the constructedpair L(q (x), h), and Theorem 2.2.5 is proved. 2

The Gelfand-Levitan method. For Sturm-Liouville operators on a finite interval, theGelfand-Levitan method was presented in Section 1.5. For the case of the half-line there aresimilar results. Therefore here we confine ourselves to the derivation of the Gelfand-Levitan

equation. For further discussion see [mar1], [lev2] and [lev4].



130

Consider the differential equation and the linear form L = L(q (x), h) of the form (2.2.1)-(2.2.2). Let q (x) = 0, h = 0. Denote

F (x, t) =1

2πi

γ

cos ρx cos ρtM (λ) dλ, (2.2.41)

where γ is the contour defined in Section 2.1 (see fig. 2.1.1). We note that by virtue of (2.1.77) and (2.2.5),

1

2πi

γ

cos ρx cos ρtM (λ) dλ = ∞

λ∗cos ρx cos ρtV (λ) dλ < ∞.

Let G(x, t) and H (x, t) be the kernels of the transformation operators (1.3.11) and (1.5.12)respectively.

Theorem 2.2.6. For each fixed x, the function G(x, t) satisfies the following linearintegral equation

G(x, t) + F (x, t) + x

0G(x, s)F (s, t)ds = 0, 0 < t < x. (2.2.42)

Equation (2.2.42) is called the Gelfand-Levitan equation.

Proof. Using (1.3.11) and (1.5.12) we calculate

12πi

γ R

ϕ(x, λ)cos ρtM (λ) dλ = 12πi

γ R

cos ρx cos ρtM (λ) dλ

+1

2πi

γ R

x

0G(x, s)cos ρsds

cos ρtM (λ) dλ,

1

2πi

γ R

ϕ(x, λ)cos ρtM (λ) dλ =1

2πi

γ R

ϕ(x, λ)ϕ(t, λ)M (λ) dλ

+1

2πi γ R t

0

H (t, s)ϕ(s, λ) dsϕ(x, λ)M (λ) dλ,

where γ R = γ ∩ λ : |λ| ≤ R. This yields

ΦR(x, t) = I R1(x, t) + I R2(x, t) + I R3(x, t) + I R4(x, t),

where

ΦR(x, t) =1

2πi

γ R

ϕ(x, λ)ϕ(t, λ)M (λ) dλ − 1

2πi

γ R

cos ρx cos ρtM (λ) dλ,

I R1(x, t) =1

2πi

γ R

cos ρx cos ρtM (λ)dλ,

I R2(x, t) = x

0G(x, s)

1

2πi

γ R

cos ρt cos ρsM (λ)dλ

ds,

I R3(x, t) =1

2πi

γ R

cos ρt x

0G(x, s)cos ρsds

M (λ)dλ,

I R4(x, t) =−

1

2πi γ Rϕ(x, λ)

t

0H (t, s)ϕ(s, λ) dsM (λ) dλ.



131

Let ξ (t), t ≥ 0 be a twice continuously differentiable function with compact support. ByTheorem 2.1.8,

limR→∞

∞0

ξ (t)ΦR(x, t)dt = 0, limR→∞

∞0

ξ (t)I R1(x, t)dt = ∞

0ξ (t)F (x, t)dt,

limR→∞

∞0

ξ (t)I R2(x, t)dt = ∞

0ξ (t)

x

0G(x, s)F (s, t)ds

dt,

limR→∞

∞0

ξ (t)I R3(x, t)dt = x

0ξ (t)G(x, t)dt,

limR→∞

∞0

ξ (t)I R4(x, t)dt = − ∞

xξ (t)H (t, x)dt.

Put G(x, t) = H (x, t) = 0 for x < t. In view of the arbitrariness of ξ (t), we derive

G(x, t) + F (x, t) + x

0G(x, s)F (s, t)ds − H (t, x) = 0.

For t < x, this gives (2.2.42). 2

Thus, in order to solve the inverse problem of recovering L from the Weyl function M (λ) one can calculate F (x, t) by (2.2.41), find G(x, t) by solving the Gelfand-Levitan equation (2.2.42) and construct q (x) and h by (1.5.13).

Remark 2.2.1. We show the connection between the Gelfand-Levitan equation and

the main equation of inverse problem (2.2.12). For this purpose we use the cosine Fouriertransform. Let q (x) = h = 0. Then ϕ(x, λ) = cos

√ λx. Multiplying (2.2.42) by cos

√ λt

and integrating with respect to t, we obtain

x

0G(x, t)cos

√ λtdt +

x

0cos

√ λt

1

2πi

γ

cos√

µx cos√

µtM (µ) dµ

dt+

x

0cos

√ λt x

0G(x, s)

1

2πi

γ

cos√

µt cos√

µsM (µ) dµ

ds = 0.

Using (1.3.11) we arrive at (2.2.12).

Remark 2.2.2. If q (x) and h are real, and (1+ x)q (x) ∈ L(0, ∞) , then (2.2.41) takesthe form

F (x, t) = ∞−∞

cos ρx cos ρt dσ(λ),

where σ = σ − σ, and σ and σ are the spectral functions of L and L respectively.

2.3. RECOVERY OF THE DIFFERENTIAL EQUATION FROM THESPECTRAL DATA.

In Theorem 2.2.5, one of the conditions under which an arbitrary function M (λ) is theWeyl function for a certain pair L = L(q (x), h) of the form (2.1.1)-(2.1.2) is the require-ment that the main equation is uniquely solvable. This condition is difficult to check in thegeneral case. In this connection it is important to point out classes of operators for whichthe unique solvability of the main equation can be proved. The class of selfadjoint operators

is one of the most important classes having this property. For this class we introduce the



132

so-called spectral data which describe the set of singularities of the Weyl function M (λ) ,i.e. the discrete and continuous parts of the spectrum. In this section we investigate theinverse problem of recovering L from the given spectral data. We show that the specifica-tion of the spectral data uniquely determines the Weyl function. Thus, the inverse problemof recovering L from the Weyl function is equivalent to the inverse problem of recovering

L from the spectral data. The main equation, obtained in Section 2.2, can be constructeddirectly from the spectral data on the set λ : λ ≥ 0∪ Λ. We prove a unique solvability of the main equation in a suitable Banach space (see Theorem 2.3.9). At the end of the sectionwe consider the inverse problem from the spectral data also for the nonselfadjoint case.

2.3.1. The selfadjoint case. Consider the differential equation and the linear formL = L(q (x), h) of the form (2.1.1)-(2.1.2) and assume that q (x) and h are real. This meansthat the operator Lo, defined in Remark 2.1.3, is a selfadjoint operator. For the selfadjoint

case one can obtain additional properties of the spectrum to those which we established inSection 2.1.

Theorem 2.3.1. Let q (x) and h be real. Then Λ = ∅, i.e. there are no spectral singularities.

Proof. Since q (x) and h are real it follows from Theorem 2.1.1 and (2.1.63) that forreal ρ,

e(x, ρ) = e(x, −ρ), ∆(ρ) = ∆(−ρ). (2.3.1)

Suppose Λ = ∅, i.e. for a certain real ρ0

= 0, ∆(ρ0

) = 0. Then, according to (2.3.1),

∆(−ρ0) = ∆(ρ0) = 0.

Together with (2.1.6) this yields

−2iρ0 = e(x, ρ0), e(x, −ρ0)|x=0 = e(0, ρ0)∆(−ρ0) − e(0, −ρ0)∆(ρ0) = 0,

which is impossible. 2

Theorem 2.3.2. Let q (x) and h be real. Then the non-zero eigenvalues λk are real and negative, and

∆(ρk) = 0 ( where λk = ρ2k ∈ Λ). (2.3.2)

The eigenfunction e(x, ρk) and ϕ(x, λk) are real, and

e(x, ρk) = e(0, ρk)ϕ(x, λk), e(0, ρk) = 0. (2.3.3)

Moreover, eigenfunctions related to different eigenvalues are orthogonal in L2(0,

∞).

Proof. It follows from (2.1.66) and (2.1.71) that

ϕ(x, λ), e(x, ρ) = ∆(ρ). (2.3.4)

According to Theorem 2.1.7, (2.3.2) holds; therefore (2.3.4) implies

e(x, ρk) = ckϕ(x, λk), ck = 0.

Taking here x = 0 we get ck = e(0, ρk), i.e. (2.3.3) is valid.



133

Let λn and λk ( λn = λk ) be eigenvalues with eigenfunctions yn(x) = e(x, ρn) andyk(x) = e(x, ρk) respectively. Then integration by parts yields ∞

0yn(x) yk(x) dx =

∞0

yn(x)yk(x) dx,

and henceλn

∞0

yn(x)yk(x) dx = λk

∞0

yn(x)yk(x) dx,

consequently ∞0

yn(x)yk(x) dx = 0.

Assume now that λ0 = u + iv,v = 0 is a non-real eigenvalue with an eigenfunctiony0(x) ≡ 0. Since q (x) and h are real, we get that λ0 = u − iv is also the eigenvalue with

the eigenfunction y0

(x). Since λ0

= λ0

, we derive as before

y02L2

= ∞

0y0(x)y0(x) dx = 0,

which is impossible. Thus, all eigenvalues λk of L are real, and consequently the eigen-functions ϕ(x, λk) and e(x, ρk) are real too. Together with Theorem 2.1.6 this yieldsΛ ⊂ (−∞, 0). 2

Theorem 2.3.3. Let q (x) and h be real, and let Λ = λk, λk = ρ2k < 0. Denote

∆1(ρ) := ddλ

∆(ρ) (λ = ρ2). Then

∆1(ρk) = 0. (2.3.5)

Proof. Since e(x, ρ) satisfies (2.1.1) we have

d

dxe(x, ρ), e(x, ρk) = (ρ2 − ρ2

k)e(x, ρ)e(x, ρk).

Using (2.1.4), (2.1.63) and (2.3.2) we calculate

(ρ2 − ρ2k) ∞

0e(x, ρ)e(x, ρk) dx = e(x, ρ), e(x, ρk)

∞0

= e(0, ρk)∆(ρ), Im ρ > 0.

As ρ → ρk, this gives

∞0

e2(x, ρk)dx =1

2ρke(0, ρk)

d

dρ∆(ρ)

|ρ=ρk

= e(0, ρk)∆1(ρk). (2.3.6)

Since ∞0

e2(x, ρk)dx > 0,

we arrive at (2.3.5). 2

Let Λ = λk, λk = ρ2k < 0. It follows from (2.1.69), (2.3.2) and (2.3.5) that the Weyl

function M (λ) has simple poles at the points λ = λk, and

αk := Resλ=λk

M (λ) =e(0, ρk)

∆1(ρk). (2.3.7)



134

Taking (2.3.3) and (2.3.6) into account we infer

αk = ∞

0ϕ2(x, λk) dx

−1> 0.

Furthermore, the function V (λ), defined by (2.1.77), is continuous for λ = ρ2 > 0, and byvirtue of (2.1.69),

V (λ) :=1

2πi

M −(λ) − M +(λ)

=

1

2πi

e(0, −ρ)

∆(−ρ)− e(0, ρ)

∆(ρ)

.


V (λ) =ρ

π|∆(ρ)

|2

> 0, λ = ρ2 > 0. (2.3.8)

Denote by W N the set of functions q (x) ∈ W N such that ∞0

(1 + x)|q (x)| dx < ∞, (2.3.9)

We shall say that L ∈ V N if q (x) and h are real, and q (x) ∈ W N .

Theorem 2.3.4. Let L ∈ V N . Then Λ is a finite set.

Proof. For x ≥ 0, τ ≥ 0, the function e(x,iτ ) is real and by virtue of (2.1.24)

|e(x,iτ )exp(τ x) − 1| ≤ Q1(x) exp(Q1(x)), x ≥ 0, τ ≥ 0, (2.3.10)

whereQ1(x) =

∞x

(t − x)|q (t)| dt.

In particular, it follows from (2.3.10) that there exists a > 0 such that

e(x,iτ ) exp(τ x) ≥ 12

for x ≥ a, τ ≥ 0. (2.3.11)

Suppose that Λ = λk is an infinite set. Since Λ is bounded and λk = ρ2k < 0, it

follows that ρk = iτ k → 0, τ k > 0. Using (2.3.11) we calculate

∞a

e(x, ρk)e(x, ρn) dx ≥ 1

4

∞a

exp(−(τ k + τ n)x) dx

= exp(−(τ k + τ n)a)4(τ k + τ n) ≥ exp(−2aT )8T , (2.3.12)

where T = maxk

τ k. Since the eigenfunctions e(x, ρk) and e(x, ρn) are orthogonal in

L2(0, ∞) we get

0 = ∞

0e(x, ρk)e(x, ρn) dx =

∞a

e(x, ρk)e(x, ρn) dx

+ a

0 e

2

(x, ρk) dx + a

0 e(x, ρk)(e(x, ρn) − e(x, ρk)) dx. (2.3.13)



135

Taking (2.3.12) into account we get ∞a

e(x, ρk)e(x, ρn) dx ≥ C 0 > 0, a

0e2(x, ρk) dx ≥ 0. (2.3.14)

Let us show that a

0e(x, ρk)(e(x, ρn) − e(x, ρk)) dx → 0 as k, n → ∞. (2.3.15)

Indeed, according to (2.1.23),

|e(x, ρk)| ≤ exp(Q1(0)), x ≥ 0.

Then, using (2.1.33) we calculate

a

0e(x, ρk)(e(x, ρn) − e(x, ρk)) dx

≤ exp(Q1(0)) a

0| exp(−τ nx) − exp(−τ kx)| dx

+ a

0

∞x

|A(x, t)(exp(−τ nx) − exp(−τ kx))| dt

dx

. (2.3.16)


|A(x, t)| ≤ 1

2Q0

x + t

2 exp(Q1(x)) ≤ 1

2Q0

t

2exp(Q1(0)), 0 ≤ x ≤ t,

and consequently, (2.3.16) yields

a

0e(x, ρk)(e(x, ρn) − e(x, ρk)) dx

≤ C a

0| exp(−τ nx) − exp(−τ kx)| dx

+ ∞

0Q0

t

2

| exp(−τ nt) − exp(−τ kt)| dt

. (2.3.17)

Clearly,

a

0 | exp(−τ nx) − exp(−τ kx)| dx → 0 as k, n → ∞. (2.3.18)

Take a fixed ε > 0. There exists aε > 0 such that ∞aε

Q0

t

2

dt <

ε

2.

On the other hand, for sufficiently large k and n,

aε

0 Q0 t

2| exp(−τ nt) − exp(−τ kt)| dt ≤ Q0(0) aε

0 | exp(−τ nt) − exp(−τ kt)| dt <

ε

2 .

Thus, for sufficiently large k and n, ∞0

Q0

t

2

| exp(−τ nt) − exp(−τ kt)| dt < ε.

By virtue of arbitrariness of ε we obtain

∞0 Q0 t

2| exp(−τ nt) − exp(−τ kt)| dt → 0 as k, n → ∞.



136

Together with (2.3.17) and (2.3.18) this implies (2.3.15). The relations (2.3.13)-(2.3.15) giveus a contradiction. This implies that Λ is a finite set. 2

Theorem 2.3.5. Let L ∈ V N . Then

ρ

∆(ρ) = O(1), ρ → 0, I m ρ ≥ 0. (2.3.19)

Proof. Denote

g(ρ) =2iρ

∆(ρ).

By virtue of (2.1.6) and (2.1.63),

∆(ρ)e(0,

−ρ)

−e(0, ρ)∆(

−ρ) = 2iρ.

Hence, for real ρ = 0,g(ρ) = e(0, −ρ) − ξ (ρ)e(0, ρ), (2.3.20)

where

ξ (ρ) :=∆(−ρ)

∆(ρ).


|ξ (ρ)

|= 1 for real ρ

= 0. (2.3.21)

Let Λ = λkk=1,m, λk = ρ2k, ρk = iτ k, 0 < τ 1 < . . . < τ m. Denote

D = ρ : Im ρ > 0, |ρ| < τ ∗, (2.3.22)

where τ ∗ = τ 1/2. The function g(ρ) is analytic in D and continuous in D \ 0. Byvirtue of (2.3.20), (2.3.21) and (2.1.23),

|g(ρ)| ≤ C for real ρ = 0. (2.3.23)

Suppose that ∆(ρ) is analytic in the origin. Then, using (2.3.23), we get that the functiong(ρ) has a removable singularity in the origin, and consequently (after extending g(ρ)continuously to the origin) g(ρ) is continuous in D, i.e. (2.3.19) is proved.

In the general case we cannot use these arguments. Therefore, we introduce the potentialsq r(x) by (2.1.49) and the corresponding Jost solutions by (2.1.50). Denote

∆r(ρ) = er(0, ρ) − her(0, ρ), r ≥ 0.

By virtue of (2.1.53), the functions ∆r(ρ) are entire in ρ, and according to Lemma 2.1.3,

limr→∞∆r(ρ) = ∆(ρ) (2.3.24)

uniformly for Im ρ ≥ 0. Let δ r be the infimum of the distances between the zeros of ∆r(ρ)in the upper half-plane Im ρ ≥ 0. Let us show that

δ ∗ := inf r>0

δ r > 0. (2.3.25)



137

Indeed, suppose on the contrary that there exists a sequence rk → ∞ such that δ rk → 0.Let ρk1 = iτ k1, ρk2 = iτ k2 (τ k1, τ k2 ≥ 0) be zeros of ∆rk(ρ) such that ρk1 − ρk2 → 0 ask → ∞. It follows from (2.1.51) that there exists a > 0 such that

er(x,iτ )exp(τ x)

≥1

2

for x

≥a, τ

≥0, r

≥0. (2.3.26)

Similarly to (2.3.13) we can write

0 = ∞

0erk(x, ρk1)erk(x, ρk2) dx =

∞a

erk(x, ρk1)erk(x, ρk2) dx

+ a

0e2

rk(x, ρk1) dx +

a

0erk(x, ρk1)(erk(x, ρk2) − erk(x, ρk1)) dx.

Taking (2.3.26) into account we get as before

∞a

erk(x, ρk1)erk(x, ρk2) dx ≥ exp(−(τ k1 + τ k2)a)

4(τ k1 + τ k2).

By virtue of (2.3.24), |τ k1|, |τ k2| ≤ C, and consequently, ∞a

erk(x, ρk1)erk(x, ρk2) dx ≥ C 0 > 0.

Moreover, a

0e2

rk(x, ρk1) dx ≥ 0.

On the other hand, using (2.1.50) and (2.1.52) one can easily verify (similar to the proof of (2.3.15)) that a

0erk(x, ρk1)(erk(x, ρk2) − erk(x, ρk1)) dx → 0 as k → ∞,

and we get a contradiction. This means that (2.3.25) is valid.

Define D via (2.3.22) with τ ∗ = min( τ 12 , δ∗

2 ). Then, the function ∆r(ρ) has in D atmost one zero ρ = iτ 0r , 0 ≤ τ 0r ≤ τ ∗. Consider the function

γ r(ρ) := gr(ρ)g0r (ρ),

where

gr(ρ) =2iρ

∆r(ρ), g0

r (ρ) =ρ − iτ 0rρ + iτ 0r

(if ∆r(ρ) has no zeros in D we put g0

r

(ρ) := 1 ). Clearly

|g0r (ρ)| ≤ 1 for ρ ∈ D. (2.3.27)

Analogously to (2.3.20)-(2.3.21), for real ρ = 0,

gr(ρ) = er(0, −ρ) − ξ r(ρ)er(0, ρ), |ξ r(ρ)| = 1. (2.3.28)


|gr(ρ)| ≤ C for ρ ∈ ∂D, r ≥ 0,



138

where ∂D = D \ D is the boundary of D, and C does not depend on r. Together with(2.3.27) this yields

|γ r(ρ)| ≤ C for ρ ∈ ∂D, r ≥ 0.

Since the functions γ r(ρ) are analytic in D, we have by the maximum principle [con1,

p.128], |γ r(ρ)| ≤ C for ρ ∈ D, r ≥ 0, (2.3.29)

where C does not depend on r.Fix δ ∈ (0, τ ∗) and denote Dδ := ρ : Im ρ > 0, δ < |ρ| < τ ∗. By virtue of (2.3.24),

limr→∞ gr(ρ) = g(ρ) uniformly in Dδ.

Moreover, (2.3.24) implies limr→∞ τ 0r = 0, and consequently,

limr→∞ γ r(ρ) = g(ρ) uniformly in Dδ.

Then, it follows from (2.3.29) that

|g(ρ)| ≤ C for ρ ∈ Dδ.

By virtue of arbitrariness of δ we get

|g(ρ)

| ≤C for ρ

∈D

\ 0

,

i.e. (2.3.19) is proved. 2

Theorem 2.3.6. Let L ∈ V N . Then λ = 0 is not an eigenvalue of L.

Proof. The function e(x) := e(x, 0) is a solution of (2.1.1) for λ = 0, and accordingto Theorem 2.1.2,

limx→∞ e(x) = 1. (2.3.30)

Take a > 0 such that

e(x) ≥ 12

for x ≥ a, (2.3.31)

and consider the function

z (x) := e(x) x

a

dt

e2(t). (2.3.32)

It is easy to check thatz (x) = q (x)z (x)

and

e(x)z (x) − e(x)z (x) ≡ 1.

It follows from (2.3.30)-(2.3.32) that

limx→∞ z (x) = +∞. (2.3.33)

Suppose that λ = 0 is an eigenvalue, and let y0(x) be a corresponding eigenfunction. Sincethe functions e(x), z (x) form a fundamental system of solutions of (2.1.1) for λ = 0, wehave

y0(x) = C 01 e(x) + C

02 z (x).



139

By virtue of (2.3.30) and (2.3.33), this is possible only if C 01 = C 02 = 0. 2

Remark 2.3.1. Let

q (x) =2a2

(1 + ax)2, h = −a,

where a is a complex number such that a /∈ (−∞, 0]. Then q (x) ∈ L(0, ∞), but xq (x) /∈L(0, ∞). In this case λ = 0 is an eigenvalue, and by differentiation one verifies that

y(x) =1

1 + ax

is the corresponding eigenfunction.

According to Theorem 2.1.2, e(0, ρ) = O(1) as ρ → 0, I m ρ ≥ 0. Therefore, Theorem

2.3.5 together with (2.1.69) yields

M (λ) = O(ρ−1), |ρ| → 0. (2.3.34)

It is shown below in Example 2.3.1 that if the condition (2.3.9) is not fulfilled, then (2.3.34)is not valid in general.

Combining the results obtained above we arrive at the following theorem.

Theorem 2.3.7. Let L ∈ V N , and let M (λ) be the Weyl function for L. Then

M (λ) is analytic in Π with the exception of an at most finite number of simple poles Λ = λkk=1,m, λk = ρ2k < 0, and continuous in Π1 \ Λ. Moreover,

αk := Resλ=λk

M (λ) > 0, k = 1, m,

and (2.3.34) is valid. The function ρV (λ) is continuous and bounded for λ = ρ2 > 0, and

V (λ) > 0 for λ = ρ2 > 0.

Definition 2.3.1. The data S := (V (λ)λ>0, λk, αkk=1,m) are called the spectral data of L.

Lemma 2.3.1. The Weyl function is uniquely determined by the specification of the spectral data S via the formula

M (λ) = ∞

0

V (µ)

λ − µdµ +

mk=1

αk

λ − λk, λ ∈ Π \ Λ. (2.3.35)

Proof. Consider the function

I R(λ) :=1

2πi

|µ|=R

M (µ)

λ − µdµ.

It follows from (2.1.73) thatlim

R→∞I R(λ) = 0 (2.3.36)



140

uniformly on compact subsets of Π \ Λ. On the other hand, moving the contour |µ| = Rto the real axis and using the residue theorem, we get

I R(λ) = −M (λ) + R

0

V (µ)

λ

−µ

dµ +m

k=1

αk

λ

−λk

.

Together with (2.3.36) this yields (2.3.35). 2

It follows from Theorem 2.2.1 and Lemma 2.3.1 that the specification of the spectral data uniquely determines the potential q (x), x ≥ 0 and the coefficient h.

Let L ∈ V N , and let L ∈ V N be chosen such that (2.2.4) holds. Denote


p = m + m, θ(x) = [θk(x)]k=1,p,

θk(x) = ϕk0(x), k = 1, m, θk+m(x) = ϕk1(x), k = 1, m.

Analogously we define θ(x). It follows from (2.2.12), (2.2.17) and (2.2.18) that

ϕ(x, λ) = ϕ(x, λ) + ∞

0D(x,λ,µ)V (µ)ϕ(x, µ) dµ+

mk=1

D(x,λ,λk0)αk0ϕk0(x) −m

k=1

D(x,λ,λk1)αk1ϕk1(x), (2.3.37)

ϕni(x) = ϕni(x) + ∞

0D(x, λni, µ)V (µ)ϕ(x, µ) dµ+

mk=1

D(x, λni, λk0)αk0ϕk0(x) −m

k=1

D(x, λni, λk1)αk1ϕk1(x), (2.3.38)

Φ(x, λ) = Φ(x, λ) + ∞0

Φ(x, λ), ϕ(x, µ)

λ − µ V (µ)ϕ(x, µ) dµ+

mk=1

Φ(x, λ), ϕk0(x)λ − λk0

αk0ϕk0(x) −m

k=1


αk1ϕk1(x), (2.3.39)

ε0(x) = ∞

0ϕ(x, µ)ϕ(x, µ)V (µ) dµ +

mk=1

ϕk0(x)ϕk0(x)αk0 −m

k=1

ϕk1(x)ϕk1(x)αk1,

ε(x) = −2ε0(x). (2.3.40)

For each fixed x ≥ 0, the relations (2.3.37)-(2.3.38) can be considered as a system of linearequations with respect to ϕ(x, λ), λ > 0 and θk(x), k = 1, p. We rewrite (2.3.37)-(2.3.38)as a linear equation in a corresponding Banach space.

Let C = C (0, ∞) be the Banach space of continuous bounded functions f : [0, ∞) →C, λ → f (λ) on the half-line λ ≥ 0 with the norm f C = supλ≥0 |f λ|. Obviously, foreach fixed x ≥ 0, ϕ(x, ·), ϕ(x, ·) ∈ C. Consider the Banach space B of vectors

F = f

f 0 , f

∈C, f 0 = [f 0k ]k=1,p

∈R

p



141

with the norm F B = max(f C , f 0Rp

). Denote

ψ(x) =

ϕ(x, ·)

θ(x)

, ψ(x) =

ϕ(x, ·)

θ(x)

.

Then ψ(x), ψ(x) ∈ B for each fixed x ≥ 0. Let

H λ,µ(x) = D(x,λ,µ)V (µ),

H λ,k(x) = D(x,λ,λk0)αk0, k = 1, m, H λ,k+m(x) = −D(x,λ,λk1)αk1, k = 1, m,

H n,µ(x) = D(x, λn0, µ)V (µ), n = 1, m, H n+m,µ(x) = D(x, λn1, µ)V (µ), n = 1, m,

H n,k(x) = D(x, λn0, λk0)αk0, n, k = 1, m,

H n,k+m(x) = −D(x, λn0, λk1)αk1, n = 1,m, k = 1, m,H n+m,k(x) = D(x, λn1, λk0)αk0, n = 1, m, k = 1, m,

H n+m,k+m(x) = −D(x, λn1, λk1)αk1, n,k = 1, m.

Denote by H : B → B the operator defined by

F = HF, F =

f

f 0

∈ B, F =

f

f 0

∈ B,

f (λ) = ∞

0H λ,µf (µ) dµ +

pk=1

H λ,kf 0k ,

f 0n = ∞

0H n,µf (µ) dµ +

pk=1

H n,kf 0k .

Then for each fixed x ≥ 0, the operator E + H (x) (here E is the identity operator),acting from B to B, is a linear bounded operator. Taking into account our notations wecan rewrite (2.3.37)-(2.3.38) to the form

ψ(x) = (E + H (x))ψ(x). (2.3.41)

Thus, we proved the following assertion.

Theorem 2.3.8. For each fixed x ≥ 0, the vector ψ(x) ∈ B is a solution of equation (2.3.41).

Now let us go on to necessary and sufficient conditions for the solvability of the inverseproblem under consideration. Denote by W the set of vectors S = (

V (λ)

λ>0,

λk, αk

k=1,

(in general, m is different for each S ) such that(1) αk > 0, λk < 0, for all k, and λk = λs for k = s;(2) the function ρV (λ) is continuous and bounded for λ > 0, V (λ) > 0, and M (λ) =O(ρ−1), ρ → 0, where M (λ) is defined by (2.3.35);(3) there exists L such that (2.2.4) holds.

Clearly, if S is the vector of spectral data for a certain L ∈ V N , then S ∈ W.

Theorem 2.3.9. Let S ∈ W. Then for each fixed x ≥ 0, equation (2.3.41) has a

unique solution in B, i.e. the operator E + H (x) is invertible.



142

Proof. As in the proof of Lemma 1.6.6 it is sufficient to prove that for each fixed x ≥ 0the homogeneous equation

(E + H (x))β (x) = 0, β (x) ∈ B, (2.3.42)

has only the zero solution. Let

β (x) =

β (x, ·)β 0(x)

∈ B, β 0(x) = [β 0k(x)]k=1,p,

be a solution of (2.3.42), i.e.

β (x, λ) + ∞

0D(x,λ,µ)V (µ)β (x, µ) dµ

+m

k=1

D(x,λ,λk0)αk0β k0(x) −m

k=1

D(x,λ,λk1)αk1β k1(x) = 0, (2.3.43)

β ni(x) + ∞

0D(x, λni, µ)V (µ)β (x, µ) dµ

+m

k=1

D(x, λni, λk0)αk0β k0(x) −m

k=1

D(x, λni, λk1)αk1β k1(x) = 0, (2.3.44)

where β k0(x) = β 0

k (x), k = 1, m, β k1(x) = β 0

k+m(x), k = 1, m. Then (2.3.43) gives us theanalytic continuation of β (x, λ) to the whole λ - plane, and for each fixed x ≥ 0, thefunction β (x, λ) is entire in λ. Moreover, according to (2.3.43), (2.3.44),

β (x, λni) = β ni(x). (2.3.45)

Let us show that for each fixed x ≥ 0,

|β (x, λ)

| ≤C x

|ρ|exp(

|τ

|x), λ = ρ2, τ := Imρ. (2.3.46)

Indeed, using (2.2.6) and (2.1.87), we get

|D(x,λ,λki)| ≤ C x|ρ| exp(|τ |x). (2.3.47)

For definiteness, let σ := Re ρ ≥ 0. It follows from (2.3.43), (2.2.7) and (2.3.47) that

|β (x, λ)| ≤ C x exp(|τ |x) ∞1

|ˆV (µ)|θ|ρ − θ| + 1 dθ + 1|ρ|

, µ = θ2. (2.3.48)

In view of (2.2.4),

∞1

|V (µ)|θ|ρ − θ| + 1

dθ2 ≤

∞1

|V (µ)|2θ4 dθ ∞

1

dθ

θ2(|ρ − θ| + 1)2

≤C

∞

1

dθ

θ2(|ρ − θ| + 1)2

. (2.3.49)



143

Since|ρ − θ| = σ2 + τ 2 + θ2 − 2σθ, ||ρ| − θ| = σ2 + τ 2 + θ2 − 2|ρ|θ,

we have|ρ − θ| ≥ ||ρ| − θ|. (2.3.50)

By virtue of (2.3.49), (2.3.50) and (2.2.11), ∞1

|V (µ)|θ|ρ − θ| + 1

dθ ≤ C

|ρ| . (2.3.51)

Using (2.3.48) and (2.3.51) we arrive at (2.3.46).Furthermore, we construct the function Γ(x, λ) by the formula

Γ(x, λ) =−

∞

0

Φ(x, λ), ϕ(x, µ)λ − µ

V (µ)β (x, µ) dµ

−m

k=1


αk0β k0(x) +m

k=1


αk1β k1(x). (2.3.52)


Γ(x, λ) = M (λ)β (x, λ) − ∞

0

S (x, λ), ϕ(x, µ)λ − µ

V (µ)β (x, µ) dµ

−m

k=1

S (x, λ), ϕk0(x)λ − λk0

αk0β k0(x) +m

k=1

S (x, λ), ϕk1(x)λ − λk1

αk1β k1(x). (2.3.53)

Since S (x, λ), ϕ(x, µ)|x=0 = −1, we infer from (1.6.1) that

S (x, λ), ϕ(x, µ)λ − µ

= − 1

λ − µ+ x

0S (t, λ)ϕ(t, µ) dt.

Hence, (2.3.53) takes the form

Γ(x, λ) = M (λ)β (x, λ) + ∞

0

V (µ)β (x, µ)

λ − µdµ

+m

k=1

αk0β k0(x)

λ − λk0

−m

k=1

αk1β k1(x)

λ − λk1+ Γ1(x, λ), (2.3.54)

whereΓ1(x, λ) =

∞

0 x

0

S (t, λ)ϕ(t, µ) dtV (µ)β (x, µ) dµ

−m

k=1

x

0S (t, λ)ϕk0(t) dt

αk0β k0(x) +

mk=1

x

0S (t, λ)ϕk1(t) dt

αk1β k1(x).

The function Γ1(x, λ) is entire in λ for each fixed x ≥ 0. Using (2.3.35) we derive from(2.3.54):

Γ(x, λ) = M (λ)β (x, λ) + Γ0(x, λ), (2.3.55)

where

Γ0(x, λ) = Γ1(x, λ) + Γ2(x, λ) + Γ3(x, λ),



144

Γ2(x, λ) = − ∞

0

V (µ)(β (x, λ) − β (x, µ))

λ − µdµ,

Γ3(x, λ) = −m

k=1

αk0

λ − λk0(β (x, λ) − β k0(x)) +

mk=1

αk1

λ − λk1(β (x, λ) − β k1(x)).

In view of (2.3.45), the function Γ0(x, λ) is entire in λ for each fixed x ≥ 0. Using (2.3.55),(2.3.52), (2.3.46) and (2.1.88) we obtain the following properties of the function Γ(x, λ).

(1) For each fixed x ≥ 0, the function Γ(x, λ) is analytic in Π \ Λ with respect to λ(with the simple poles λk0, k = 1, m ), and continuous in Π1 \ Λ. Moreover,

Resλ=λk0

Γ(x, λ) = αk0β k0(x), k = 1, m. (2.3.56)

(2) DenoteΓ±(λ) = lim

z→0, R e z >0Γ(λ ± iz ), λ > 0.

Then1

2πi

Γ−(λ) − Γ+(λ)

= V (λ)β (x, λ), λ > 0. (2.3.57)

(3) For |λ| → ∞,

|Γ(x, λ)| ≤ C x|ρ|2

exp(−|τ |x). (2.3.58)

Now we construct the function B(x, λ) via the formula

B(x, λ) = β (x, λ)Γ(x, λ). (2.3.59)

By Cauchy‘s theorem1

2πi

γ 0R

B(x, λ) dλ = 0,

where the contour γ 0R is defined in Section 2.1 (see fig. 2.1.2). By virtue of (2.3.46), (2.3.58)

and (2.3.59) we get

limR→∞

1

2πi

|λ|=R

B(x, λ) dλ = 0,

and consequently1

2πi

γ

B(x, λ) dλ = 0, (2.3.60)

where the contour γ is defined in Section 2.1 (see fig. 2.1.1). Moving the contour γ in(2.3.60) to the real axis and using the residue theorem and (2.3.56) we get for sufficiently

small ε > 0,

mn=1

αn0|β n0(x)|2 +1

2πi

|λ|=ε

B(x, λ) dλ +1

2πi

γ ε

B(x, λ) dλ = 0, (2.3.61)

where γ ε is the two-sided cut along the arc λ : λ ≥ ε. Since M (λ) = O(ρ−1) as|λ| → 0, we get in view of (2.3.55) and (2.3.59) that for each fixed x ≥ 0,

B(x, λ) = O(ρ−1) as|λ| →

0,



145

and consequently

limε→∞

1

2πi

|λ|=ε

B(x, λ) dλ = 0.

Together with (2.3.61), (2.3.57) and (2.3.59) this yields

mk=1

αk0|β k0(x)|2 + ∞

0|β (x, λ)|2V (λ) dλ = 0.

Since αk0 > 0, V (λ) > 0, we get

β (x, λ) = 0, β n0(x) = 0, β n1(x) = β (x, λn1) = 0.

Thus, β (x) = 0, and Theorem 2.3.9 is proved. 2

Theorem 2.3.10. For a vector S = (V (λ)λ>0, λk, αkk=1,m) ∈ W to be the spectral data for a certain L ∈ V N , it is necessary and sufficient that ε(x) ∈ W N , where ε(x) is defined via (2.3.40), and ψ(x) is the solution of (2.3.41). The function q (x) and the number h can be constructed via (2.2.19)-(2.2.20).

Denote by W the set of functions M (λ) such that:

(1) M (λ) is analytic in Π with the exception of an at most finite number of simple polesΛ = λkk=1,m, λk = ρ2

k < 0, and αk := Resλ=λk

M (λ) > 0;

(2) M (λ) is continuous in Π1 \ Λ, M (λ) = O(ρ−1) for ρ → 0, and

V (λ) :=1

2πi

M −(λ) − M +(λ)

> 0 for λ > 0;

(3) there exists L such that (2.2.4) holds.

Theorem 2.3.11. For a function M (λ) ∈ W to be the Weyl function for a certain L ∈ V N , it is necessary and sufficient that ε(x) ∈ W N , where ε(x) is defined via (2.3.40).

We omit the proofs of Theorems 2.3.10-2.3.11, since they are similar to correspondingfacts for Sturm-Liouville operators on a finite interval presented in Section 1.6 (see also theproof of Theorem 2.2.5).

2.3.2. The nonselfadjoint case. One can consider the inverse problem from thespectral data also for the nonselfadjoint case when q (x) ∈ L(0, ∞) is a complex-valuedfunction, and h is a complex number. For simplicity we confine ourselves here to thecase of a simple spectrum (see Definition 2.3.2). For nonselfadjoint differential operators

with a simple spectrum we introduce the spectral data and study the inverse problem of recovering L from the given spectral data. We also consider a particular case when onlythe discrete spectrum is perturbed. Then the main equation of the inverse problem becomesa linear algebraic system, and the solvability condition is equivalent to the condition thatthe determinant of this system differs from zero.

Definition 2.3.2. We shall say that L has simple spectrum if ∆(ρ) has a finitenumber of simple zeros in Ω, and M (λ) = O(ρ−1), ρ → 0. We shall write L ∈ V N if Lhas simple spectrum and q ∈ W N .



146

Clearly, V N ⊂ V N , since the selfadjoint operators considered in Subsection 2.3.1 havesimple spectrum.

Let L ∈ V N . Then Λ = λkk=1,m, λk = ρ2k is a finite set, and the Weyl function M (λ)

is analytic in Π \ Λ and continuous in Π1 \ Λ; Λ = λkk=1,r are the eigenvalues, andΛ =

λk

k=r+1,m are the spectral singularities of L. The next assertion is proved by the

same arguments as in Lemma 2.3.1.


M (λ) = ∞

0

V (µ)

λ − µdµ +

mk=1

αk

λ − λk,

where

V (λ) :=1

2πi M −(λ) − M +(λ)

, M ±(λ) := lim

z

→0, R e z >0

M (λ ± iz ),

αk :=

e(0, ρk)(∆1(ρk))−1, k = 1, r,1

2e(0, ρk)(∆1(ρk))−1, k = r + 1, m,

∆1(ρ) :=d

dλ∆(ρ).

Definition 2.3.3. The data S = (V (λ)λ>0, λk, αkk=1,m) are called the spectraldata of L.

Using Theorem 2.2.1 and Lemma 2.3.2 we obtain the following uniqueness theorem.

Theorem 2.3.12. Let S and S be the spectral data for L and L respectively. If S = S, then L = L. Thus, the specification of the spectral data uniquely determines L.

For L ∈ V N the main equation (2.3.41) of the inverse problem remains valid, andTheorem 2.3.8 also holds true. Moreover, the potential q and the coefficient h can beconstructed by (2.3.40), (2.2.19)-(2.2.20).

Perturbation of the discrete spectrum. Let L ∈ V N , and let M (λ) be the Weylfunction for L. Consider the function

M (λ) = M (λ) + λ0∈J

aλ0

λ − λ0 , (2.3.62)

where J is a finite set of the λ - plane, and aλ0 , λ0 ∈ J are complex numbers. In thiscase V (λ) = 0. Then the main equation (2.2.12) becomes the linear algebraic system

ϕ(x, z 0) = ϕ(x, z 0) +

λ0∈J

D(x, z 0, λ0)aλ0ϕ(x, λ0), z 0 ∈ J, (2.3.63)

with the determinant det(E + G(x)), where

G(x) = [D(x, z 0, λ0)aλ0 ]z0,λ0∈J .

The solvability condition (Condition S in Theorem 2.2.5) takes here the form

det(E + G(x)) = 0 for all x ≥ 0. (2.3.64)

The potential q and the coefficient h can be constructed by the formulae

q (x) = q (x) + ε(x), h = h

− λ0∈J

aλ0 , (2.3.65)



147

ε(x) = −2

λ0∈J

aλ0

d

dx

ϕ(x, λ0)ϕ(x, λ0)

. (2.3.66)

From Theorem 2.2.5 we have

Theorem 2.3.13. Let L ∈ V N . For a function M (λ) of the form (2.3.62) to be

the Weyl function for a certain L ∈ V N it is necessary and sufficient that (2.3.64) holds,ε(x) ∈ W N , where ε(x) is defined via (2.3.66), and that ϕ(x, λ0λ0∈J is a solution of (2.3.63). Under these conditions the potential q and the coefficient h are constructed by (2.3.65).

Example 2.3.1. Let q (x) = 0 and h = 0. Then M (λ) =1

iρ. Consider the function

M (λ) = M (λ) +a

λ−

λ0,

where a and λ0 are complex numbers. Then the main equation (2.3.63) becomes

ϕ(x, λ0) = F (x)ϕ(x, λ0),

whereϕ(x, λ0) = cos ρ0x, F (x) = 1 + a

x

0cos2 ρ0t dt, λ0 = ρ2

0.

The solvability condition (2.3.64) takes the form

F (x) = 0 for all x ≥ 0, (2.3.67)

and the function ε(x) can be found by the formula

ε(x) =2aρ0 sin2ρ0x

F (x)+

2a2 cos4 ρ0x

F 2(x).

Case 1. Let λ0 = 0. ThenF (x) = 1 + ax,

and (2.3.67) is equivalent to the condition

a /∈ (−∞, 0). (2.3.68)

If (2.3.68) is fulfilled then M (λ) is the Weyl function for L of the form (2.2.1)-(2.2.2) with

q (x) =2a2

(1 + ax)2, h = −a,

ϕ(x, λ) = cos ρx −a

1 + ax ·sin ρx

ρ , e(x, ρ) = exp(iρx)1 −a

iρ(1 + ax),

∆(ρ) = iρ, V (λ) =1

πρ, ϕ(x, 0) =

1

1 + ax.

If a < 0, then the solvability condition is not fulfilled, and the function M (λ) is not aWeyl function.

Case 2. Let λ0 = 0 be a real number, and let a > 0. Then F (x) ≥ 1, and (2.3.67) isfulfilled. But in this case ε(x) /∈ L(0, ∞), i.e. ε(x) /∈ W N for any N ≥ 0.



148

2.4. AN INVERSE PROBLEM FOR A WAVE EQUATION

In this section we consider an inverse problem for a wave equation with a focused sourceof disturbance. In applied problems the data are often functions of compact support localizedwithin a relative small area of space. It is convenient to model such situations mathematically

as problems with a focused source of disturbance (see [rom1]).

Consider the following boundary value problem B(q (x), h) :

utt = uxx − q (x)u, 0 ≤ x ≤ t, (2.4.1)

u(x, x) = 1, (ux − hu)|x=0, (2.4.2)

where q (x) is a complex-valued locally integrable function (i.e. it is integrable on every

finite interval), and h is a complex number. Denote r(t) := u(0, t). The function r iscalled the trace of the solution. In this section we study the following inverse problem.

Inverse Problem 2.4.1. Given the trace r(t), t ≥ 0, of the solution of B(q (x), h),construct q (x), x ≥ 0, and h.

We prove an uniqueness theorem for Inverse Problem 2.4.1 (Theorem 2.4.3), provide analgorithm for the solution of this inverse problem (Algorithm 2.4.1) and give necessary andsufficient conditions for its solvability (Theorem 2.4.4). Furthermore we show connectionsbetween Inverse Problem 2.4.1 and the inverse spectral problems considered in Sections

2.2-2.3.

Remark 2.4.1. Let us note here that the boundary value problem B(q (x), h) is equiva-lent to a Cauchy problem with a focused source of disturbance. For simplicity, we assume herethat h = 0. We define u(x, t) = 0 for 0 < t < x, and u(x, t) = u(−x, t), q (x) = q (−x) forx < 0. Then, using symmetry, it follows that u(x, t) is a solution of the Goursat problem

utt = uxx − q (x)u, 0 ≤ |x| ≤ t,

u(x, |x|) = 1.

Moreover, it can be shown that this Goursat problem is equivalent to the Cauchy problem

utt = uxx − q (x)u, −∞ < x < ∞, t > 0,

u|t=0 = 0, ut|t=0 = 2δ (x),

where δ (x) is the Dirac delta-function. Similarly, for h = 0, the boundary value problem(2.4.1)-(2.4.2) also corresponds to a problem with a focused source of disturbance.

Let us return to the boundary value problem (2.4.1)-(2.4.2). Denote

Q(x) = x

0|q (t)| dt, Q∗(x) =

x

0Q(t) dt, d = max(0, −h).

Theorem 2.4.1. The boundary value problem (2.4.1)-(2.4.2) has a unique solution u(x, t), and

|u(x, t)| ≤ exp(d(t − x))exp2Q∗t + x

2 , 0 ≤ x ≤ t. (2.4.3)



149

Proof. We transform (2.4.1)-(2.4.2) by means of the replacement

ξ = t + x, η = t − x, v(ξ, η) = uξ − η

2,

ξ + η

2

to the boundary value problem

vξη (ξ, η) = −1

4q ξ − η

2

v(ξ, η), 0 ≤ η ≤ ξ, (2.4.4)

v(ξ, 0) = 1, (vξ(ξ, η) − vη(ξ, η) − hv(ξ, η))|ξ=η = 0. (2.4.5)

Since vξ(ξ, 0) = 0, integration of (2.4.4) with respect to η gives

vξ(ξ, η) =

−

1

4 η

0q

ξ − α

2 v(ξ, α) dα. (2.4.6)

In particular, we have

vξ(ξ, η)|ξ=η = −1

4

η

0q η − α

2

v(η, α) dα. (2.4.7)


v(ξ, η) = v(η, η)

−

1

4 ξ

η η

0

q β − α

2 v(β, α) dα dβ. (2.4.8)

Let us calculate v(η, η). Since

d

dη(v(η, η) exp(hη)) = (vξ(ξ, η) + vη(ξ, η) + hv(ξ, η))|ξ=η exp(hη),

we get by virtue of (2.4.5) and (2.4.7),

d

dη (v(η, η)exp(hη)) = 2vξ(ξ, η)|ξ=η exp(hη) = −1

2 exp(hη) η

0 q η

−α

2 v(η, α) dα.

This yields (with v(0, 0) = 1 )

v(η, η)exp(hη) − 1 = −1

2

η

0exp(hβ )

β

0q β − α

2

v(β, α) dα

dβ,

and consequently

v(η, η) = exp(−hη) −1

2 η

0 exp(−h(η − β )) β

0 q β −

α

2v(β, α) dα dβ. (2.4.9)

Substituting (2.4.9) into (2.4.8) we deduce that the function v(ξ, η) satisfies the integralequation

v(ξ, η) = exp(−hη) − 1

2

η

0exp(−h(η − β ))

β

0q β − α

2

v(β, α) dα

dβ

−1

4 ξ

η η

0q β − α

2 v(β, α) dα dβ. (2.4.10)



150

Conversely, if v(ξ, η) is a solution of (2.4.10) then one can verify that v(ξ, η) satisfies(2.4.4)-(2.4.5).

We solve the integral equation (2.4.10) by the method of successive approximations. Thecalculations are slightly different for h ≥ 0 and h < 0.

Case 1. Let h ≥ 0. Denote

v0(ξ, η) = exp(−hη), vk+1(ξ, η) = −1

2

η

0exp(−h(η − β ))

β

0q β − α

2

vk(β, α) dα

dβ

−1

4

ξ

η

η

0q β − α

2

vk(β, α) dα

dβ. (2.4.11)

Let us show by induction that

|vk(ξ, η)| ≤1

k!2Q∗ξ

2k

, k ≥ 0, 0 ≤ η ≤ ξ. (2.4.12)

Indeed, for k = 0, (2.4.12) is obvious. Suppose that (2.4.12) is valid for a certain k ≥ 0.It follows from (2.4.11) that

|vk+1(ξ, η)| ≤ 1

2

ξ

0

η

0

q β − α

2

vk(β, α)

dα

dβ. (2.4.13)

Substituting (2.4.12) into the right-hand side of (2.4.13) we obtain

|vk+1(ξ, η)| ≤ 1

2k!

ξ

0

2Q∗

β

2

k η

0

q β − α

2

dα

dβ

≤ 1

k!

ξ

0

2Q∗

β

2

k β/2

0|q (s)| ds

dβ =

1

k!

ξ

0

2Q∗

β

2

kQβ

2

dβ

=1

k!

ξ/2

0(2Q∗(s))k(2Q∗(s)) ds =

1

(k + 1)!

2Q∗

ξ

2

k+1;

hence (2.4.12) is valid.It follows from (2.4.12) that the series

v(ξ, η) =∞

k=0

vk(ξ, η)

converges absolutely and uniformly on compact sets 0 ≤ η ≤ ξ ≤ T, and

|v(ξ, η)| ≤ exp 2Q∗ξ

2.

The function v(ξ, η) is the unique solution of the integral equation (2.4.10). Consequently,the function u(x, t) = v(t + x, t − x) is the unique solution of the boundary value problem(2.4.1)-(2.4.2), and (2.4.3) holds.

Case 2. Let h < 0. we transform (2.4.10) by means of the replacement

w(ξ, η) = v(ξ, η)exp(hη)



151

to the integral equation

w(ξ, η) = 1 − 1

2

η

0

β

0q β − α

2

exp(h(β − α))w(β, α) dα

dβ

−1

4 ξ

η η

0 q β

−α

2 exp(h(η − α))w(β, α) dα dβ. (2.4.14)By the method of successive approximations we get similarly to Case 1 that the integralequation (2.4.14) has a unique solution, and that

|w(ξ, η)| ≤ exp

2Q∗ξ

2

,

i.e. Theorem 2.4.1 is proved also for h < 0. 2

Remark 2.4.2. It follows from the proof of Theorem 2.4.1 that the solution u(x, t) of (2.4.1)-(2.4.2) in the domain ΘT := (x, t) : 0 ≤ x ≤ t, 0 ≤ x + t ≤ 2T

6

-x

t

T

T

2T ΘT

fig. 2.4.1.

is uniquely determined by the specification of h and q (x) f or 0 ≤ x ≤ T, i.e. if q (x) =q (x), x

∈[0, T ] and h = h, then u(x, t) = u(x, t) f o r (x, t)

∈ΘT . Therefore, one can

also consider the boundary value problem (2.4.1)-(2.4.2) in the domains ΘT and study theinverse problem of recovering q (x), 0 ≤ x ≤ T and h from the given trace r(t), t ∈ [0, 2T ].

Denote by DN (N ≥ 0) the set of functions f (x), x ≥ 0 such that for each fixed T > 0,the functions f ( j)(x), j = 0, N − 1 are absolutely continuous on [0, T ], i.e. f ( j)(x) ∈L(0, T ), j = 0, N. It follows from the proof of Theorem 2.4.1 that r(t) ∈ D2, r(0) =1, r(0) = −h. Moreover, the function r has the same smoothness properties as thepotential q. For example, if q ∈ DN then r ∈ DN +2.

In order to solve Inverse Problem 2.4.1 we will use the Riemann formula for the solutionof the Cauchy problem

utt − p(t)u = uxx − q (x)u + p1(x, t), −∞ < x < ∞, t > 0,

u|t=0 = r(x), ut|t=0 = s(x).

(2.4.15)

It is known (see, for example, [hel1, p.150]) that the solution of (2.4.15) has the form

u(x, t) =1

2r(x + t) + r(x−

t) +1

2 x+t

x−t s(ξ )R(ξ, 0, x , t)−

r(ξ )R2(ξ, 0, x , t) dξ +



152

1

2

t

0dτ

x+t−τ

x+τ −tR(ξ, τ, x, t) p1(ξ, τ ) dξ,

where R(ξ, τ, x, t) is the Riemann function, and R2 =∂R

∂τ . Note that if q (x) ≡ const,

then R(ξ, τ, x, t) = R(ξ

−x, τ, t). In particular, the solution of the Cauchy problem

utt = uxx − q (x)u, −∞ < t < ∞, x > 0,

u|x=0 = r(t), ux|x=0 = hr(t),

(2.4.16)

has the form

u(x, t) =1

2

r(t + x) + r(t − x)

− 1

2

t+x

t−xr(ξ )(R2(ξ − t, 0, x) − hR(ξ − t, 0, x)) dξ.

The change of variables τ = t − ξ leads to

u(x, t) =1

2

r(t + x) + r(t − x)

+

1

2

x

−xr(t − τ )G(x, τ ) dτ, (2.4.17)

where G(x, τ ) = −R2(−τ, 0, x) + hR(−τ, 0, x).

Remark 2.4.3. In (2.4.16) take r(t) = cos ρt. Obviously, the function u(x, t) =ϕ(x, λ)cos ρt, where ϕ(x, λ) was defined in Section 1.1, is a solution of problem (2.4.16).Therefore, (2.4.17) yields for t = 0,

ϕ(x, λ) = cos ρx +1

2

x

−xG(x, τ )cos ρτ dτ.

Since G(x, −τ ) = G(x, τ ), we have


0G(x, τ )cos ρτ dτ,

i.e. we obtained another derivation of the representation (1.3.11). Thus, the function G(x, t)is the kernel of the transformation operator. In particular, it was shown in Section 1.3 that

G(x, x) = h +1

2

x

0q (t) dt. (2.4.18)

Let us go on to the solution of Inverse Problem 2.4.1. Let u(x, t) be the solution of the boundary value problem (2.4.1)-(2.4.2). We define u(x, t) = 0 f o r 0 ≤ t < x, andu(x, t) =

−u(x,

−t), r(t) =

−r(

−t) for t < 0. Then u(x, t) is the solution of the Cauchy

problem (2.4.16), and consequently (2.4.17) holds. But u(x, t) = 0 for x > |t| (this is aconnection between q (x) and r(t) ), and hence

1

2

r(t + x) + r(t − x)

+

1

2

x

−xr(t − τ )G(x, τ ) dτ = 0, |t| < x. (2.4.19)

Denote a(t) = r(t). Differentiating (2.4.19) with respect to t and using the relations

r(0+) = 1, r(0

−) =

−1, (2.4.20)



153

we obtainG(x, t) + F (x, t) +

x

0G(x, τ )F (t, τ ) dτ = 0, 0 < t < x, (2.4.21)

where

F (x, t) =1

2a(t + x) + a(t

−x) with a(t) = r(t). (2.4.22)

Equation (2.4.21) is called the Gelfand-Levitan equation.

Theorem 2.4.2. For each fixed x > 0, equation (2.4.21) has a unique solution.

Proof. Fix x0 > 0. It is sufficient to prove that the homogeneous equation

g(t) + x0

0g(τ )F (t, τ ) dτ = 0, 0 ≤ t ≤ x0 (2.4.23)

has only the trivial solution g(t) = 0.Let g(t), 0 ≤ t ≤ x0 be a solution of (2.4.23). Since a(t) = r(t) ∈ D1, it follows from

(2.4.22) and (2.4.23) that g(t) is an absolutely continuous function on [0, x0]. We defineg(−t) = g(t) for t ∈ [0, x0], and furthermore g(t) = 0 for |t| > x0.

Let us show that x0

−x0r(t − τ )g(τ ) dτ = 0, t ∈ [−x0, x0]. (2.4.24)

Indeed, by virtue of (2.4.20) and (2.4.23), we have

ddt x0

−x0r(t − τ )g(τ ) dτ = ddt

t

−x0r(t − τ )g(τ ) dτ + x0

tr(t − τ )g(τ ) dτ

= r(+0)g(t) − r(−0)g(t) + x0

−x0a(t − τ )g(τ ) dτ = 2

g(t) +

x0

0g(τ )F (t, τ ) dτ

= 0.

Consequently, x0

−x0r(t − τ )g(τ ) dτ ≡ C 0.

Taking here t = 0 and using that r(

−τ )g(τ ) is an odd function, we calculate C 0 = 0, i.e.

(2.4.24) is valid.

Denote ∆0 = (x, t) : x − x0 ≤ t ≤ x0 − x, 0 ≤ x ≤ x0

6

-

∆0

x0 x

x0

−x0

t

fig 2.4.2.



154

and consider the function

w(x, t) := ∞−∞

u(x, t − τ )g(τ ) dτ, (x, t) ∈ ∆0, (2.4.25)

where u(x, t) is the solution of the boundary value problem (2.4.1)-(2.4.2). Let us show

thatw(x, t) = 0, (x, t) ∈ ∆0. (2.4.26)

Since u(x, t) = 0 for x > |t|, (2.4.25) takes the form

w(x, t) = t−x

−∞u(x, t − τ )g(τ ) dτ +

∞t+x

u(x, t − τ )g(τ ) dτ. (2.4.27)

Differentiating (2.4.27) and using the relations

u(x, x) = 1, u(x, −x) = −1, (2.4.28)

we calculate

wx(x, t) = g(t + x) − g(t − x) + t−x

−∞ux(x, t − τ )g(τ ) dτ +

∞t+x

ux(x, t − τ )g(τ ) dτ, (2.4.29)

wt(x, t) = g(t + x) + g(t − x) + t−x

−∞ut(x, t − τ )g(τ ) dτ +

∞t+x

ut(x, t − τ )g(τ ) dτ. (2.4.30)

Since, in view of (2.4.28),ux(x, t) ± ut(x, t)

|t=±x

=d

dxu(x, ±x) = 0,

it follows from (2.4.29)-(2.4.30) that

wxx(x, t) − wtt(x, t) − q (x)w(x, t) = ∞−∞

[uxx − utt − q (x)u](x, t − τ )g(τ ) dτ,

and consequentlywtt(x, t) = wxx(x, t) − q (x)w(x, t), (x, t) ∈ ∆0. (2.4.31)

Furthermore, (2.4.25) and (2.4.29) yield

w(0, t) = x0

−x0r(t − τ )g(τ ) dτ, wx(0, t) = h

x0

−x0r(t − τ )g(τ ) dτ = 0, t ∈ [−x0, x0].

Therefore, according to (2.4.24), we have

w(0, t) = wx(0, t) = 0, t ∈ [−x0, x0]. (2.4.32)

Since the Cauchy problem (2.4.31)-(2.4.32) has only the trivial solution, we arrive at (2.4.26).

Denote u1(x, t) := ut(x, t). It follows from (2.4.30) that

wt(x, 0) = 2g(x) + −x

−∞u1(x, τ )g(τ ) dτ +

∞x

u1(x, τ )g(τ ) dτ

= 2g(x) + ∞x u1(x, τ )g(τ ) dτ = 2g(x) + x0

x u1(x, τ )g(τ ) dτ .



155


g(x) + x0

xu1(x, τ )g(τ ) dτ = 0, 0 ≤ x ≤ x0.

This integral equation has only the trivial solution g(x) = 0, and consequently Theorem2.4.2 is proved. 2

Let r and r be the traces for the boundary value problems B(q (x), h) and B(q (x), h)respectively.

Theorem 2.4.3. If r(t) = r(t), t ≥ 0, then q (x) = q (x), x ≥ 0 and h = h. Thus,the specification of the trace r uniquely determines the potential q and the coefficient h.

Proof. Since r(t) = r(t), t ≥ 0, we have, by virtue of (2.4.22),

F (x, t) = F (x, t).

Therefore, Theorem 2.4.2 gives us

G(x, t) = G(x, t), 0 ≤ t ≤ x. (2.4.33)


q (x) = 2d

dxG(x, x), h = G(0, 0) = −r(0). (2.4.34)

Together with (2.4.33) this yields q (x) = q (x), x ≥ 0 and h = h. 2

The Gelfand-Levitan equation (2.4.21) and Theorem 2.4.2 yield finally the followingalgorithm for the solution of Inverse Problem 2.4.1.

Algorithm 2.4.1. Let r(t), t ≥ 0 be given. Then(1) Construct the function F (x, t) using (2.4.22).

(2) Find the function G(x, t) by solving equation (2.4.21).(3) Calculate q (x) and h by the formulae (2.4.34).

Remark 2.4.4. We now explain briefly the connection of Inverse Problem 2.4.1 withthe inverse spectral problem considered in Sections 2.2-2.3. Let q (x) ∈ L(0, ∞), and letu(x, t) be the solution of (2.4.1)-(2.4.2). Then, according to (2.4.3), |u(x, t)| ≤ C 1 exp(C 2t).Denote

Φ(x, λ) = − ∞

xu(x, t) exp(iρt)dt, x ≥ 0, Im ρ ≥ 0,

M (λ) = − ∞0

r(t)exp(iρt)dt, Im ρ ≥ 0.

One can verify that Φ(x, λ) and M (λ) are the Weyl solution and the Weyl function forthe pair L = L(q (x), h) of the form (2.1.1)-(2.1.2). The inversion formula for the Laplacetransform gives

r(t) =1

2π

γ 1

M (λ) exp(−iρt) dρ =1

2πi

γ 1

sin ρt

ρ(2ρ)M (λ) dρ =

1

2πi

γ

sin ρt

ρM (λ) dλ,



156

where γ is the contour defined in Section 2.1 (see fig. 2.1.1), and γ 1 is the image of γ under the map ρ → λ = ρ2. Let q (x) = h = 0. Since then

1

2πi

γ

sin ρt

ρM (λ) dλ = r(t) = 1,

we infer

r(t) = 1 +1

2πi

γ

sin ρt

ρM (λ) dλ.

If (2.2.5) is valid we calculate

a(t) =1

2πi

γ

cos ρt M (λ)dλ,

and consequently

F (x, t) = 12πi

γ

cos ρx cos ρt M (λ)dλ,

i.e. equation (2.4.21) coincides with equation (2.2.42).

Let us now formulate necessary and sufficient conditions for the solvability of InverseProblem 2.4.1.

Theorem 2.4.4. For a function r(t), t ≥ 0 to be the trace for a certain boundary value problem B(q (x), h) of the form (2.4.1)-(2.4.2) with q

∈DN , it is necessary and sufficient

that r(t) ∈ DN +2, r(0) = 1, and that for each fixed x > 0 the integral equation (2.4.21) is uniquely solvable.

Proof. The necessity part of Theorem 2.4.4 was proved above, here we prove thesufficiency. For simplicity let N ≥ 1 (the case N = 0 requires small modifications).Let a function r(t), t ≥ 0, satisfying the hypothesis of Theorem 2.4.4, be given, and letG(x, t), 0 ≤ t ≤ x, be the solution of (2.4.21). We define G(x, t) = G(x, −t), r(t) = −r(−t)for t < 0, and consider the function

u(x, t) := 12

r(t + x) + r(t − x)

+ 12 x

−xr(t− τ )G(x, τ ) dτ, −∞ < t < ∞, x ≥ 0. (2.4.35)

Furthermore, we construct q and h via (2.4.34) and consider the boundary value problem(2.4.1)-(2.4.2) with these q and h . Let u(x, t) be the solution of (2.4.1)-(2.4.2), and letr(t) := u(0, t). Our goal is to prove that u = u, r = r.

Differentiating (2.4.35) and taking (2.4.20) into account, we get

ut(x, t) =1

2a(t + x) + a(t

−x) + G(x, t) +

1

2 x

−x

a(t

−τ )G(x, τ ) dτ, (2.4.36)

ux(x, t) =1

2

a(t + x) − a(t − x)

+

1

2

r(t − x)G(x, x) + r(t + x)G(x, −x)

+1

2

x

−xr(t − τ )Gx(x, τ ) dτ. (2.4.37)

Since a(0+) = a(0−), it follows from (2.4.36) that

utt

(x, t) =1

2a(t + x) + a(t−

x) + Gt(x, t) +

1

2 x

−xa(t

−τ )G(x, τ ) dτ. (2.4.38)



157

Integration by parts yields

utt(x, t) =1

2

a(t + x) + a(t − x)

+ Gt(x, t) − 1

2

a(t − τ )G(x, τ )

t−x+ a(t − τ )G(x, τ )

xt

+

1

2 x

−x a(t − τ )Gt(x, τ ) dτ =

1

2a(t + x) + a(t − x)

+Gt(x, t) +1

2

a(t + x)G(x, −x) − a(t − x)G(x, x)

+

1

2

x

−xr(t − τ )Gt(x, τ ) dτ.

Integrating by parts again and using (2.4.20) we calculate

utt(x, t) =1

2

a(t + x) + a(t − x)

+ Gt(x, t) +

1

2

a(t + x)G(x, −x) − a(t − x)G(x, x)

−12r(t − τ )Gt(x, τ )t−x

+ r(t − τ )Gt(x, τ )xt

+ 12 x

−xr(t − τ )Gtt(x, τ ) dτ

=1

2

a(t + x) + a(t − x)

+

1

2

a(t + x)G(x, −x) − a(t − x)G(x, x)

+1

2

r(t + x)Gt(x, −x) − r(t − x)Gt(x, x)

+

1

2

x

−xr(t − τ )Gtt(x, τ ) dτ. (2.4.39)

Differentiating (2.4.37) we obtain

uxx(x, t) = 12a(t + x) + a(t − x) + 1

2a(t + x)G(x, −x) − a(t − x)G(x, x)

+1

2

r(t + x)

d

dxG(x, −x) + r(t − x)

d

dxG(x, x)

+

1

2

r(t + x)Gx(x, −x) + r(t − x)Gx(x, x)

+1

2

x

−xr(t − τ )Gxx(x, τ ) dτ.

Together with (2.4.35), (2.4.39) and (2.4.34) this yields

uxx(x, t) − q (x)u(x, t) − utt(x, t) = 12

x

−xr(t − τ )g(x, τ ) dτ, −∞ < t < ∞, x ≥ 0, (2.4.40)

whereg(x, t) = Gxx(x, t) − Gtt(x, t) − q (x)G(x, t).

Let us show thatu(x, t) = 0, x > |t|. (2.4.41)

Indeed, it follows from (2.4.36) and (2.4.21) that ut(x, t) = 0 for x >|t|, and consequently

u(x, t) ≡ C 0(x) for x > |t|. Taking t = 0 in (2.4.35) we infer like above

C 0(x) =1

2

r(x) + r(−x)

+

1

2

x

−xr(−τ )G(x, τ ) dτ = 0,

i.e. (2.4.41) holds.It follows from (2.4.40) and (2.4.41) that

1

2 x

−xr(t

−τ )g(x, τ ) dτ = 0, x >

|t|. (2.4.42)



158

Differentiating (2.4.42) with respect to t and taking (2.4.20) into account we deduce

1

2

r(0+)g(x, t) − r(0−)g(x, t)

+

1

2

x

−xa(t − τ )g(x, τ ) dτ = 0,

or

g(x, t) + x

0F (t, τ )g(x, τ ) dτ = 0.

According to Theorem 2.4.2 this homogeneous equation has only the trivial solution g(x, t) =0, i.e.

Gtt = Gxx − q (x)G, 0 < |t| < x. (2.4.43)

Furthermore, it follows from (2.4.38) for t = 0 and (2.4.41) that

0 =1

2a(x) + a(−

x) + Gt(x, 0) +1

2 x

−x

a(−

τ )G(x, τ ) dτ.

Since a(x) = −a(−x), G(x, t) = G(x, −t), we infer

∂G(x, t)

∂tt=0

= 0. (2.4.44)

According to (2.4.34) the function G(x, t) satisfies also (2.4.18).


utt(x, t) = uxx(x, t) − q (x)u(x, t), −∞ < t < ∞, x ≥ 0.

Moreover, (2.4.35) and (2.4.37) imply (with h=G(0,0))

u|x=0 = r(t), ux|x=0 = hr(t).

Let us show thatu(x, x) = 1, x

≥0. (2.4.45)

Since the function G(x, t) satisfies (2.4.43), (2.4.44) and (2.4.18), we get according to(2.4.17),

u(x, t) =1

2

r(t + x) + r(t − x)

+

1

2

x

−xr(t − τ )G(x, τ ) dτ. (2.4.46)

Comparing (2.4.35) with (2.4.46) we get

u(x, t) =1

2r(t + x) + r(t − x) +1

2 x

−x

r(t − τ )G(x, τ ) dτ,

where u = u − u, r = r − r. Since the function r(t) is continuous for −∞ < t < ∞, itfollows that the function u(x, t) is also continuous for −∞ < t < ∞, x > 0. On the otherhand, according to (2.4.41), u(x, t) = 0 for x > |t|, and consequently u(x, x) = 0. By(2.4.2), u(x, x) = 1, and we arrive at (2.4.45).

Thus, the function u(x, t) is a solution of the boundary value problem (2.4.1)-(2.4.2).By virtue of Theorem 2.4.1 we obtain u(x, t) = u(x, t), and consequently r(t) = r(t). The-orem 2.4.4 is proved. 2



159

2.5. THE GENERALIZED WEYL FUNCTION

Let us consider the differential equation and the linear form L = L(q (x), h) :

y := −y + q (x)y = λy, x > 0,

U (y) := y(0) − hy(0).

In this section we study the inverse spectral problem for L in the case when q (x) is alocally integrable complex-valued function, and h is a complex number. In this case weintroduce the so-called generalized Weyl function as a main spectral characteristics.

For this purpose we define a space of generalized functions (distributions). Let D bethe set of all integrable and bounded on the real line entire functions of exponential typewith ordinary operations of addition and multiplication by complex numbers and with thefollowing convergence: z

k(ρ) is said to converge to z (ρ) if the types σ

kof the functions

z k(ρ) are bounded ( sup σk < ∞ ), and z k(ρ) − z (ρ)L(−∞,∞) → 0 as k → ∞. The linearmanifold D with this convergence is our space of test functions.

Definition 2.5.1. All linear and continuous functionals

R : D → C, z (ρ) → R(z (ρ)) = (z (ρ), R),

are called generalized functions (GF). The set of these GF is denoted by D. A sequence of GF Rk ∈ D converges to R ∈ D, if lim(z (ρ), Rk) = (z (ρ), R), k → ∞ for any z (ρ) ∈ D.

A GF R ∈ D is called regular if it is determined by R(ρ) ∈ L∞ via

(z (ρ), R) = ∞−∞

z (ρ)R(ρ) dρ.

Definition 2.5.2. Let a function f (t) be locally integrable for t > 0 (i.e. it isintegrable on every finite segment [0, T ] ). The GF Lf (ρ) ∈ D defined by the equality

(z (ρ), Lf (ρ)) :=

∞

0f (t)

∞

−∞

z (ρ)exp(iρt) dρ

dt, z (ρ) ∈ D, (2.5.1)

is called the generalized Fourier-Laplace transform for the function f (t).

Since z (ρ) ∈ D, we have ∞−∞

|z (ρ)|2 dρ ≤ sup−∞<ρ<∞

|z (ρ)| · ∞−∞

|z (ρ)| dρ,

i.e. z (ρ) ∈ L2(−∞, ∞). Therefore, by virtue of the Paley-Wiener theorem [zyg1], thefunction

B(t) :=1

2π ∞−∞ z (ρ)exp(iρt) dρ

is continuous and has compact support, i.e. there exists a d > 0 such that B(t) = 0 for|t| > d, and

z (ρ) = d

−dB(t)exp(−iρt) dt. (2.5.2)

Consequently, the integral in (2.5.1) exists. We note that f (t) ∈ L(0, ∞) implies

(z (ρ), Lf

(ρ)) := ∞

−∞z (ρ)

∞

0f (t)exp(iρt) dt dρ,



160

i.e. Lf (ρ) is a regular GF (defined by ∞

0 f (t)exp(iρt) dt ) and coincides with the ordinaryFourier-Laplace transform for the function f (t). Since

1

π

∞−∞

1 − cos ρx

ρ2exp(iρt)dρ =

x − t, t < x,

0, t > x,

the following inversion formula is valid:

x

0(x − t)f (t)dt =

1

π· 1 − cos ρx

ρ2, Lf (ρ)

. (2.5.3)

Let now u(x, t) be the solution of (2.4.1)-(2.4.2) with a locally integrable complex-valued function q (x). Define u(x, t) = 0 f o r 0 < t < x , and denote (with λ = ρ2 )Φ(x, λ) := −Lu(ρ), i.e.

(z (ρ), Φ(x, λ)) = − ∞

xu(x, t)

∞−∞

z (ρ) exp(iρt)dρ

dt. (2.5.4)

For z (ρ) ∈ D, ρ2z (ρ) ∈ L(−∞, ∞), ν = 1, 2, we put

(z (ρ), (iρ)ν Φ(x, λ)) := ((iρ)ν z (ρ), Φ(x, λ)),

(z (ρ), Φ(ν )(x, λ)) :=dν

dxν (z (ρ), Φ(x, λ)).


Φ(x, λ) = λΦ(x, λ), U (Φ) = 1.

Proof. We calculate

(z (ρ), Φ(x, λ)) = (z (ρ), −Φ(x, λ) + q (x)Φ(x, λ))

= − ∞−∞(iρ)z (ρ)exp(iρx)dρ − ux(x, x) ∞−∞ z (ρ) exp(iρx)dρ

+ ∞

x

uxx(x, t) − q (x)u(x, t)

∞−∞

z (ρ)exp(iρt) dρ

dt,

(z (ρ), λΦ(x, λ)) = − ∞−∞

(iρ)z (ρ) exp(iρx)dρ − ut(x, x) ∞−∞

z (ρ) exp(iρx)dρ

+ ∞

xutt(x, t)

∞−∞

z (ρ)exp(iρt) dρ

dt.

Using ut(x, x) + ux(x, x) = ddx

u(x, x) ≡ 0, we infer (z (ρ), Φ(x, λ) − λΦ(x, λ)) = 0. Fur-

thermore, since

(z (ρ), Φ(x, λ)) = ∞−∞

z (ρ)exp(iρx) dρ − ∞

xux(x, t)

∞−∞

z (ρ) exp(iρt) dρ

dt,

we get

(z (ρ), U (Φ)) = (z (ρ), Φ(0, λ) − hΦ(0, λ)) = ∞−∞

z (ρ) dρ.

2



161

Definition 2.5.3. The GF Φ(x, λ) is called the generalized Weyl solution , and theGF M (λ) := Φ(0, λ) is called the generalized Weyl function (GWF) for L(q (x), h).

Note that if q (x) ∈ L(0, ∞), then |u(x, t)| ≤ C 1 exp(C 2t), and Φ(x, λ) and M (λ)coincide with the ordinary Weyl solution and Weyl function (see Remark 2.4.4).

The inverse problem considered here is formulated as follows:Inverse Problem 2.5.1. Given the GWF M (λ), construct the potential q (x) and

the coefficient h.

Denote r(t) := u(0, t). It follows from (2.5.4) that

(z (ρ), M (λ)) = − ∞

0r(t)

∞−∞

z (ρ)exp(iρt) dρ

dt,

i.e. M (λ) = −Lr(ρ). In view of (2.5.3), we get by differentiation

r(t) = − d2

dt2

1

π· 1 − cos ρt

ρ2, M (λ)

, (2.5.5)

and Inverse Problem 2.5.1 has been reduced to Inverse Problem 2.4.1 from the trace rconsidered in Section 2.4. Thus, the following theorems hold.

Theorem 2.5.2. Let M (λ) and M (λ) be the GWF‘s for L = L(q (x), h) and L =

L(q (x), h) respectively. If M (λ) = M (λ), then L = L. Thus, the specification of the GWF uniquely determines the potential q and the coefficient h.

Theorem 2.5.3. Let ϕ(x, λ) be the solution of the differential equation ϕ = λϕ under the initial conditions ϕ(0, λ) = 1, ϕ(0, λ) = h. Then the following representation holds


0G(x, t)cos ρtdt,

and the function G(x, t) satisfies the integral equation

G(x, t) + F (x, t) + x

0G(x, τ )F (t, τ ) dτ = 0, 0 < t < x, (2.5.6)

where

F (x, t) =1

2

r(t + x) + r(t − x)

.

The function r is defined via (2.5.5), and r ∈ D2. If q ∈ DN then r ∈ DN +2. Moreover, for each fixed x > 0, the integral equation (2.5.6) is uniquely solvable.

Theorem 2.5.4. For a generalized function M ∈ D to be the GWF for a certain L(q (x), h) with q ∈ DN , it is necessary and sufficient that 1) r ∈ DN +2, r(0) = 1, where r is defined via (2.5.5);2) for each x > 0, the integral equation (2.5.6) is uniquely solvable.

The potential q and the coefficient h can be constructed by the following algorithm.

Algorithm 2.5.1. Let the GWF M (λ) be given. Then(1) Construct the function r(t) by (2.5.5).



162

(2) Find the function G(x, t) by solving the integral equation (2.5.6).(3) Calculate q (x) and h by

q (x) = 2dG(x, x)

dx, h = G(0, 0).

Let us now prove an expansion theorem for the case of locally integrable complex-valuedpotentials q.

Theorem 2.5.4. Let f (x) ∈ W 2. Then, uniformly on compact sets,

f (x) =1

π

ϕ(x, λ)F (λ)(iρ), M (λ)

, (2.5.7)

where F (λ) = ∞

0f (t)ϕ(t, λ) dt. (2.5.8)

Proof. First we assume that q (x) ∈ L(0, ∞). Let f (x) ∈ Q, where Q = f ∈ W 2 :U (f ) = 0, f ∈ L2(0, ∞) (the general case when f ∈ W 2 requires small modifications).

Let D+ = z (ρ) ∈ D : ρz (ρ) ∈ L2(−∞, ∞). Clearly, z (ρ) ∈ D+ if and only if B(t) ∈ W 12 [−d, d] in (2.5.2). For z (ρ) ∈ D1, integration by parts in (2.5.2) yields

z (ρ) = d

−d B(t)exp(−iρt) dt =

1

iρ d

−d B(t)exp(−iρt) dt.


F (λ) =1

λ

∞0

f (t)

− ϕ(t, λ) + q (t)ϕ(t, λ)

dt =1

λ

∞0

ϕ(t, λ)f (t) dt,

and consequently F (λ)(iρ) ∈ D+. According to Theorem 2.1.8 we have

f (x) =1

2πi

γ ϕ(x, λ)F (λ)M (λ) dλ = −1

π

γ 1ϕ(x, λ)F (λ)(iρ)M (λ) dρ, (2.5.9)

where the contour γ 1 in the ρ - plane is the image of γ under the mapping ρ → λ = ρ2.In view of Remark 2.4.4,

M (λ) = − ∞

0r(t)exp(iρt) dt, |r(t)| ≤ C 1 exp(C 2t). (2.5.10)

Take b > C 2. Then, by virtue of Cauchy‘s theorem, (2.5.9)-(2.5.10) imply

f (x) =1

π

∞+ib

−∞+ibϕ(x, λ)F (λ)(iρ)

− ∞

0r(t) exp(iρt) dt

dρ

= − 1

π

∞0

r(t) ∞+ib

−∞+ibϕ(x, λ)F (λ)(iρ)exp(iρt) dρ

dt.

Using Cauchy‘s theorem again we get

f (x) =−

1

π ∞

0r(t)

∞

−∞ϕ(x, λ)F (λ)(iρ) exp(iρt) dρ dt =

1

πϕ(x, λ)F (λ)(iρ), M (λ),



163

i.e. (2.5.7) is valid.

Let now q (x) be a locally integrable complex-valued function. Denote

q R(x) =

q (x), 0 ≤ x ≤ R,0, x > R.

Let rR(t) be the trace for the potential q R. According to Remark 2.4.2,

rR(t) = r(t) for t ≤ 2R. (2.5.11)

Since q R(x) ∈ L(0, ∞) we have by virtue of (2.5.7),

f (x) = − 1

π

∞0

rR(t) ∞

−∞ϕ(x, λ)F (λ)(iρ) exp(iρt) dρ

dt.

Let x ∈ [0, T ] for a certain T > 0. Then there exists a d > 0 such that

f (x) = − 1

π

d

0rR(t)

∞−∞

ϕ(x, λ)F (λ)(iρ) exp(iρt) dρ

dt, 0 ≤ x ≤ T.

For sufficiently large R (R > d/2) we have in view of (2.5.11),

f (x) = − 1

π

d

0r(t)

∞−∞

ϕ(x, λ)F (λ)(iρ) exp(iρt) dρ

dt

= − 1

π

∞0

r(t) ∞

−∞ϕ(x, λ)F (λ)(iρ)exp(iρt) dρ

dt

=1

π

ϕ(x, λ)F (λ)(iρ), M (λ)

,

i.e. (2.5.7) is valid, and Theorem 2.5.4 is proved. 2

2.6. THE WEYL SEQUENCE

We consider in this section the differential equation and the linear form

y := −y + q (x)y = λy, x > 0, U 0(y) := y(0).

Let λ = ρ2, Im ρ > 0. Then the Weyl solution Φ0(x, λ) and the Weyl function M 0(λ) aredefined by the conditions

Φ0(0, λ) = 1, Φ0(x, λ) = O(exp(iρx)), x → ∞, M 0(λ) = Φ0(0, λ),

and (2.1.81) is valid. Let the function q be analytic at the point x = 0 (in this case weshall write in the sequel q ∈ A ). It follows from (2.1.21) and (2.1.81) that for q ∈ A thefollowing relations are valid

Φ0(x, λ) = exp(iρx)

1 +∞

k=1

bk(x)

(2iρ)k

, |ρ| → ∞, (2.6.1)

M 0(λ) = iρ +∞

k=1

M k

(2iρ)k

,

|ρ

| → ∞, (2.6.2)



164

in the sense of asymptotic formulae. The sequence m := M kk≥1 is called the Weyl sequence for the potential q.

Substituting (2.6.1)-(2.6.2) into the relations

−Φ0(x, λ) + q (x)Φ0(x, λ) = λΦ0(x, λ), Φ0(0, λ) = 1, Φ

0(0, λ) = M 0(λ),

we obtain the following recurrent formulae for calculating bk(x) and M k :

bk+1(x) = −bk(x) + q (x)bk(x), k ≥ 0, b0(x) := 1, (2.6.3)

bk(0) = 0, M k = bk(0), k ≥ 1. (2.6.4)

Hencebk+1(x) = bk(0) − bk(x) +

x

0q (t)bk(t) dt, k ≥ 0.

In particular, this gives

b1(x) = x

0q (t) dt, b2(x) = q (0) − q (x) +

1

2

x

0q (t) dt

2,

b3(x) = q (x) − q (0) + (q (x) + q (0)) x

0q (t) dt −

x

0q 2(t) dt +

1

3!

x

0q (t) dt

3, · · · .

Moreover, (2.6.3)-(2.6.4) imply

M 1 = q (0), M 2 = −q (0), M 3 = q (0) − q

2

(0), M 4 = −q (0) + 4q (0)q (0),M 5 = q (4)(0) − 6q (0)q (0) − 5(q (0))2 + 2q 3(0), · · · .

We note that the Weyl sequence m = M kk≥1 depends only on the Taylor coefficientsQ = q ( j)(0) j≥0 of the potential q at the point x = 0, and this dependence is nonlinear.

In this section we study the following inverse problem

Inverse Problem 2.6.1. Given the Weyl sequence m := M kk≥1, construct q.

We prove the uniqueness theorem for Inverse problem 2.6.1, provide an algorithm forthe solution of this inverse problem and give necessary and sufficient conditions for its solv-ability. We note that the specification of the Weyl sequence allows us to construct q in aneighbourhood of the original. However, if q (x) is analytic for x ≥ 0, then one can obtainthe global solution.

The following theorem gives us necessary and sufficient conditions on the Weyl sequence.

Theorem 2.6.1. 1) If q ∈ A, then there exists a δ > 0 such that

M k = Okδ k for k ≥ 0. (2.6.5)

2) Let an arbitrary sequence M kk≥1, satisfying (2.6.5) for a certain δ > 0, be given.Then there exists a unique function q ∈ A for which M kk≥1 is the corresponding Weyl sequence.

First we prove auxiliary assertions. Denote

C ν

k

= k

ν =k!

ν !(k − ν )!, 0

≤ν

≤k.



165


C ν k ≤ kk

ν ν (k − ν )k−ν , 0 ≤ ν ≤ k, (2.6.6)

where we put 00

= 1.Proof. By virtue of Stirling‘s formula [Hen1, p.42],

√ 2πk

k

e

k ≤ k! ≤ 1.1√

2πkk

e

k, k ≥ 1. (2.6.7)

For ν = 0 and ν = k, (2.6.6) is obvious. Let 1 ≤ ν ≤ k − 1. Then, using (2.6.7) wecalculate

C

ν

k =

k!

ν !(k − ν )! ≤1.1

√ 2π ·kk

ν ν (k − ν )k−ν · k

ν (k − ν ) .

It is easy to check that

(1.1)2

2π· k

ν (k − ν )< 1, 1 ≤ ν ≤ k − 1,

and consequently, (2.6.6) is valid. 2


(k − s)k−s

kk

k−s j=1

( j + s − 2) j+s−2

j j≤ 1, k − 1 ≥ s ≥ 1, (2.6.8)

k−1s=1

(k − s)k−s

kk

k−s j=1

( j + s − 2) j+s−2

j j≤ 3, k ≥ 2. (2.6.9)

Proof. For s = 1 and s = 2, (2.6.8) is obvious since

( j + s − 2) j+s−2

j j≤ 1.

Let now s ≥ 3. Since the function

f (x) =(x + a)x+a

xx

is monotonically increasing for x > 0, we get

(k − s)k−s

kk

k−s j=1

( j + s − 2) j+s−2

j j≤ (k − s)k−s

kk

(k − 2)k−2

(k − s)k−s(k − s) ≤ (k − 2)k−1

kk< 1,

i.e. (2.6.8) is valid. Furthermore,

k−1

s=3

(k − s)k−s

kk

k−s

j=1

( j + s − 2) j+s−2

j j

≤

(k − 2)k−2

kk

k−1

s=3

(k

−s)



166

=(k − 2)k−2

kk· (k − 3)(k − 2)

2<

1

2,

and consequently (2.6.9) is valid. 2

Proof of Theorem 2.6.1. 1) Let q ∈ A. We denote q k = q (k)(0), bkν = b(ν )k (0).

Differentiating k − ν times the equality bν +1(x) = −bν (x) + q (x)bν (x) and inserting thenx = 0 we get together with (2.6.4),

bν +1,k−ν +1 = −bν,k−ν +2 +k−ν j=0

C jk−ν q k−ν − jbνj , 0 ≤ ν ≤ k,

bk+1,0 = 0, bk+1,1 = M k+1, b0,k = δ 0k, k ≥ 0,

(2.6.10)

where δ jk is the Kronecker delta. Since q ∈ A, there exist δ 0 > 0 and C > 0 such that

|q k| ≤ C k

δ 0

k.

We denote a = 1 + Cδ 20 . Let us show by induction with respect to ν that

|bν +1,k−ν +1| ≤ Caν k

δ 0

k, 0 ≤ ν ≤ k. (2.6.11)

For ν = 0 (2.6.11) is obvious by virtue of b1,k+1 = q k. We assume that (2.6.11) holds

for ν = 1, . . . , s − 1. Then, using (2.6.10) and Lemma 2.6.1 we obtain

|bs+1,k−s+1| ≤ Cas−1 k

δ 0

k+

k−s j=1

(k − s)k−s

j j(k − s − j)k−s− j·C k − s − j

δ 0

k−s− j ·Cas−1s + j − 2

δ 0

s+ j−2

= Cas−1 k

δ 0

k1 + Cδ 20

(k − s)k−s

kk

k−s j=1

(s + j − 2)s+ j−2

j j

.

By virtue of (2.6.8) this yields

|bs+1,k−s+1| ≤ Cas−1 k

δ 0

k(1 + Cδ 20) = Cas

k

δ 0

k,

i.e. (2.6.11) holds for ν = s.

It follows from (2.6.11) for ν = k that

|M k+1| ≤ Ca

k k

δ 0k

, k ≥ 0,

and consequently (2.6.5) holds for δ = δ 0/a.

2) Let the sequence M kk≥1, satisfying (2.6.5) for a certain δ > 0, be given. Solv-ing (2.6.10) successively for k = 0, 1, 2, . . . with respect to q k and bν +1,k−ν +1ν =0,k, wecalculate

bk

− j+1,j+1 = (

−1) j M k+1 +

j−1

ν =1

ν

ξ=1

(

−1) j−ν −1C ξν q ν

−ξbk

−ν,ξ, 0

≤j

≤k, (2.6.12)



167

q k = (−1)kM k+1 +k−1ν =1

ν ξ=1

(−1)k−ν −1C ξν q ν −ξbk−ν,ξ. (2.6.13)

By virtue of (2.6.5) there exist C > 0 and δ 1 > 0 such that

|M k+1| ≤ C k

δ 1k

. (2.6.14)

Let us show by induction with respect to k ≥ 0 that

|q k| ≤ Cak0

k

δ 1

k, (2.6.15)

|bk− j+1,j+1| ≤ Cak0

k

δ 1

k, 0 ≤ j ≤ k, (2.6.16)

where a0 = max(2, √ 6Cδ 1).Since q 0 = b11 = M 1, (2.6.15)-(2.6.16) are obvious for k = 0. Suppose that (2.6.15)-

(2.6.16) hold for k = 0, . . . , n−1. Then, using Lemma 2.6.1 and (2.6.13)-(2.6.14) for k = n,we obtain

|q n| ≤ C n

δ 1

n+

n−1ν =1

ν ξ=1

ν ν

ξ ξ(ν − ξ )ν −ξ· Caν −ξ

0

ν − ξ

δ 1

ν −ξ · Can−ν +ξ−20

n − ν + ξ − 2

δ 1

n−ν +ξ−2

= C nδ 1n

an0 1an

0+ Cδ

2

1a2

0

n

−1

ν =1ν

ν

nn

ν ξ=1

(n − ν + ξ − 2)

n

−ν +ξ

−2

ξ ξ.


n−1ν =1

ν ν

nn

ν ξ=1

(n − ν + ξ − 2)n−ν +ξ−2

ξ ξ=

n−1s=1

(n − s)n−s

nn

n−sξ=1

(s + ξ − 2)s+ξ−2

ξ ξ≤ 3,

and consequently

|q n| ≤ C n

δ 1n

an0 1

an0

+3Cδ 2

1a2

0

≤ C n

δ 1n

an0 ,

i.e. (2.6.15) holds for k = n. Similarly one can obtain that (2.6.16) also holds for k = n.

In particular, it follows from (2.6.15) that for k ≥ 0,

q k = O k

δ 2

k, δ 2 =

δ 1a0

.

We construct the function q

∈A by

q (x) =∞

k=0

q kxk

k!.

It is easy to verify that M kk≥1 is the Weyl sequence for this function. Notice that theuniqueness of the solution of the inverse problem is obvious. Theorem 2.6.1 is proved. 2

Remark 2.6.1. The relations (2.6.12)-(2.6.13) give us an algorithm for the solution of Inverse Problem 2.6.1. Moreover, it follows from (2.6.12)-(2.6.13) that the specification of

the Weyl sequence m := M kk≥1 uniquely determines q in a neighbourhood of the origin.



168

III. INVERSE SCATTERING ON THE LINE

In this chapter the inverse scattering problem for the Sturm-Liouville operator on the lineis considered. In Section 3.1 we introduce the scattering data and study their properties. In

Section 3.2, using the transformation operator method, we give a derivation of the so-calledmain equation and prove its unique solvability. In Section 3.3, using the main equation, weprovide an algorithm for the solution of the inverse scattering problem along with necessaryand sufficient conditions for its solvability. In Section 3.4 a class of reflectionless potentials,which is important for applications, is studied, and an explicit formula for constructing suchpotentials is given. We note that the inverse scattering problem for the Sturm-Liouvilleoperator on the line was considered by many authors (see, for example, [mar1], [lev2], [fad1],[dei1]).

3.1. SCATTERING DATA

3.1.1. Let us consider the differential equation

y := −y + q (x)y = λy, −∞ < x < ∞. (3.1.1)

Everywhere below in this chapter we will assume that the function q (x) is real, and that ∞−∞

(1 + |x|)|q (x)| dx < ∞. (3.1.2)

Let λ = ρ2, ρ = σ + iτ, and let for definiteness τ := Im ρ ≥ 0. Denote Ω+ = ρ : Im ρ >0,

Q+0 (x) =

∞x

|q (t)| dt, Q+1 (x) =

∞x

Q+0 (t) dt =

∞x

(t − x)|q (t)| dt,

Q−0 (x) = x

−∞ |q (t)| dt, Q−1 (x) = x

−∞ Q−0 (t) dt = x

−∞(t − x)|q (t)| dt.

Clearly,lim

x→±∞Q± j (x) = 0.

The following theorem introduces the Jost solutions e(x, ρ) and g(x, ρ) with prescribedbehavior in ±∞.

Theorem 3.1.1. Equation (3.1.1) has unique solutions y = e(x, ρ) and y = g(x, ρ),satisfying the integral equations


x

sin ρ(t − x)

ρq (t)e(t, ρ) dt,

g(x, ρ) = exp(−iρx) + x

−∞sin ρ(x − t)

ρq (t)g(t, ρ) dt.

The functions e(x, ρ) and g(x, ρ) have the following properties:1) For each fixed x, the functions e(ν )(x, ρ) and g(ν )(x, ρ) (ν = 0, 1) are analytic in

Ω+ and continuous in Ω+.



169

2) For ν = 0, 1,

e(ν )(x, ρ) = (iρ)ν exp(iρx)(1 + o(1)), x → +∞,

g(ν )(x, ρ) = (−iρ)ν exp(−iρx)(1 + o(1)), x → −∞,

(3.1.3)

uniformly in Ω+. Moreover, for ρ ∈ Ω+,

|e(x, ρ) exp(−iρx)| ≤ exp(Q+1 (x)),

|e(x, ρ)exp(−iρx) − 1| ≤ Q+1 (x)exp(Q+

1 (x)),

|e(x, ρ)exp(−iρx) − iρ| ≤ Q+0 (x)exp(Q+

1 (x)),

(3.1.4)

|g(x, ρ)exp(iρx)

| ≤exp(Q−

1 (x)),

|g(x, ρ)exp(iρx) − 1| ≤ Q−1 (x) exp(Q−

1 (x)),

|g(x, ρ)exp(iρx) + iρ| ≤ Q−0 (x) exp(Q−

1 (x)).

(3.1.5)

3) For each fixed ρ ∈ Ω+ and each real α, e(x, ρ) ∈ L2(α, ∞), g(x, ρ) ∈ L2(−∞, α).Moreover, e(x, ρ) and g(x, ρ) are the unique solutions of (3.1.1) (up to a multiplicative constant) having this property.

4) For |ρ| → ∞, ρ ∈ Ω+, ν = 0, 1,

e(ν )(x, ρ) = (iρ)ν exp(iρx)

1 + ω+(x)iρ

+ o1

ρ

, ω+(x) = −1

2

∞x

q (t) dt,

g(ν )(x, ρ) = (−iρ)ν exp(−iρx)

1 +ω−(x)

iρ+ o

1

ρ

, ω−(x) = −1

2

x

−∞q (t) dt,

(3.1.6)

uniformly for x ≥ α and x ≤ α respectivrly.5) For real ρ = 0, the functions e(x, ρ), e(x, −ρ) and g(x, ρ), g(x, −ρ) form fun-

damental systems of solutions for (3.1.1), and

e(x, ρ), e(x, −ρ) = −g(x, ρ), g(x, −ρ) ≡ −2iρ, (3.1.7)

where y, z := yz − yz.6) The functions e(x, ρ) and g(x, ρ) have the representations


xA+(x, t)exp(iρt) dt,

g(x, ρ) = exp(−iρx) + x

−∞ A−(x, t)exp(−iρt) dt,

(3.1.8)

where A±(x, t) are real continuous functions, and

A+(x, x) =1

2

∞x

q (t)dt, A−(x, x) =1

2

x

−∞q (t)dt, (3.1.9)

|A±(x, t)| ≤ 1

2Q±

0

x + t

2

exp

Q±

1 (x) − Q±1

x + t

2

. (3.1.10)



170

The functions A±(x, t) have first derivatives A±1 := ∂A±

∂x, A±

2 := ∂A±

∂t; the functions

A±i (x, t) ± 1

4q x + t

2

are absolutely continuous with respect to x and t , and

A±i (x, t) ± 1

4q x + t

2

≤ 1

2Q±

0 (x)Q±0

x + t

2

exp(Q±

1 (x)), i = 1, 2. (3.1.11)

For the function e(x, ρ), Theorem 3.1.1 was proved in Section 2.1 (see Theorems 2.1.1-2.1.3). For g(x, ρ) the arguments are the same. Moreover, all assertions of Theorem 3.1.1for g(x, ρ) can be obtained from the corresponding assertions for e(x, ρ) by the replacementx → −x.

In the next lemma we describe properties of the Jost solutions e j(x, ρ) and g j(x, ρ)related to the potentials q j, which approximate q.

Lemma 3.1.1. If (1 + |x|)|q (x)| ∈ L(a, ∞), a > −∞, and

lim j→∞

∞a

(1 + |x|)|q j(x) − q (x)| dx = 0, (3.1.12)

then

lim j→∞ supρ∈Ω+ supx≥a |(e

(ν )

j (x, ρ) − e

(ν )

(x, ρ)) exp(−iρx)| = 0, ν = 0, 1. (3.1.13)

If (1 + |x|)|q (x)| ∈ L(−∞, a), a < ∞, and

lim j→∞

a

−∞(1 + |x|)|q j(x) − q (x)| dx = 0,

then lim

j→∞ supρ

∈Ω+

supx

≤a|(g

(ν ) j (x, ρ) − g(ν )(x, ρ)) exp(iρx)| = 0, ν = 0, 1. (3.1.14)

Here e j(x, ρ) and g j(x, ρ) are the Jost solutions for the potentials q j.

Proof. Denote

z j(x, ρ) = e j(x, ρ) exp(−iρx), z (x, ρ) = e(x, ρ) exp(−iρx), u j(x, ρ) = |z j (x, ρ) − z (x, ρ)|.

Then, it follows from (2.1.8) that

z j(x, ρ) − z (x, ρ) =

1

2iρ ∞x (1 − exp(2iρ(t − x)))(q (t)z (t, ρ) − q j(t)z j(t, ρ)) dt.

From this, taking (2.1.29) into account, we infer

u j(x, ρ) ≤ ∞

x(t − x)|(q (t) − q j(t))z (t, ρ)| dt +

∞x

(t − x)|q j(t)|u j(t, ρ) dt.


|z (x, ρ)| ≤ exp(Q+

1 (x)) ≤ exp(Q+

1 (a)), x ≥ a, (3.1.15)



171

and consequently

u j(x, ρ) ≤ exp(Q+1 (a))

∞a

(t − a)|q (t) − q j(t)| dt + ∞

x(t − x)|q j(t)|u j(t, ρ) dt.

By virtue of Lemma 2.1.2 this yields

u j(x, ρ) ≤ exp(Q+1 (a))

∞a

(t − a)|q (t) − q j(t)| dt exp ∞

x(t − x)|q j(t)| dt

≤ exp(Q+1 (a))

∞a

(t − a)|q (t) − q j(t)| dt exp ∞

a(t − a)|q j(t)| dt

.

Henceu j(x, ρ) ≤ C a

∞a

(t − a)|q (t) − q j(t)| dt. (3.1.16)

In particular, (3.1.16) and (3.1.12) implylim

j→∞ supρ∈Ω+

supx≥a

u j(x, ρ) = 0,

and we arrive at (3.1.13) for ν = 0.Denote

v j(x, ρ) = |(e j(x, ρ) − e(x, ρ)) exp(−iρx)|.It follows from (2.1.20) that

v j(x, ρ) ≤ ∞

x|q (t)z (t, ρ) − q j (t)z j(t, ρ)| dt,

and consequently,

v j(x, ρ) ≤ ∞

a|(q (t) − q j (t))z (t, ρ)| dt +

∞a

|q j(t)|u j (t, ρ) dt. (3.1.17)

By virtue of (3.1.15)-(3.1.17) we obtain

v j(x, ρ) ≤ C a ∞

a|q (t) − q j(t)| dt +

∞a

(t − a)|q (t) − q j(t)| dt · ∞

a|q j(t)| dt

.


lim j→∞ sup

ρ∈Ω+

supx≥a

v j(x, ρ) = 0,

and we arrive at (3.1.13) for ν = 1. The relations (3.1.14) is proved analogously. 2

3.1.2. For real ρ = 0, the functions e(x, ρ), e(x, −ρ) and g(x, ρ), g(x, −ρ) formfundamental systems of solutions for (3.1.1). Therefore, we have for real ρ = 0 :

e(x, ρ) = a(ρ)g(x, −ρ) + b(ρ)g(x, ρ), g(x, ρ) = c(ρ)e(x, ρ) + d(ρ)e(x, −ρ). (3.1.18)

Let us study the properties of the coefficients a(ρ), b(ρ), c(ρ) and d(ρ).

Lemma 3.1.2. For real ρ = 0, the following relations hold

c(ρ) = −b(−ρ), d(ρ) = a(ρ), (3.1.19)



172

a(ρ) = a(−ρ), b(ρ) = b(−ρ), (3.1.20)

|a(ρ)|2 = 1 + |b(ρ)|2, (3.1.21)

a(ρ) = − 1

2iρe(x, ρ), g(x, ρ), b(ρ) =

1

2iρe(x, ρ), g(x, −ρ). (3.1.22)

Proof. Since e(x, ρ) = e(x, −ρ), g(x, ρ) = g(x, −ρ), then (3.1.20) follows from (3.1.18).Using (3.1.18) we also calculate

e(x, ρ), g(x, ρ) = a(ρ)g(x, −ρ) + b(ρ)g(x, ρ), g(x, ρ) = −2iρa(ρ),

e(x, ρ), g(x, −ρ) = a(ρ)g(x, −ρ) + b(ρ)g(x, ρ), g(x, −ρ) = 2iρb(ρ),

e(x, ρ), g(x, ρ) = e(x, ρ), c(ρ)e(x, ρ) + d(ρ)e(x, −ρ) = 2iρd(ρ),

e(x, −ρ), g(x, ρ) = e(x, −ρ), c(ρ)e(x, ρ) + d(ρ)e(x, −ρ) = 2iρc(ρ),

i.e. (3.1.19) and (3.1.22) are valid. Furthermore,

−2iρ = e(x, ρ), e(x, −ρ) = a(ρ)g(x, −ρ) + b(ρ)g(x, ρ), a(−ρ)g(x, ρ) + b(−ρ)g(x, −ρ) =

a(ρ)a(−ρ)g(x, −ρ), g(x, ρ) + b(ρ)b(−ρ)g(x, ρ), g(x, −ρ) = −2iρ|a(ρ)|2 − |b(ρ)|2

,

and we arrive at (3.1.21). 2

We note that (3.1.22) gives the analytic continuation for a(ρ) to Ω+. Hence, the functiona(ρ) is analytic in Ω+, and ρa(ρ) is continuous in Ω+. The function ρb(ρ) is continuousfor real ρ. Moreover, it follows from (3.1.22) and (3.1.6) that

a(ρ) = 1 − 1

2iρ

∞−∞

q (t) dt + o

1

ρ

, b(ρ) = o

1

ρ

, |ρ| → ∞ (3.1.23)

(in the domains of definition), and consequently the function ρ(a(ρ)−1) is bounded in Ω+.

Using (3.1.22) and (3.1.8) one can calculate more precisely

a(ρ) = 1 − 12iρ

∞−∞ q (t) dt + 1

2iρ

∞0 A(t)exp(iρt) dt,

b(ρ) = 12iρ

∞−∞ B(t) exp(iρt) dt,

(3.1.24)

where A(t) ∈ L(0, ∞) and B(t) ∈ L(−∞, ∞) are real functions.Indeed,

2iρa(ρ) = g(0, ρ)e(0, ρ)

−e(0, ρ)g(0, ρ)

=

1 + 0

−∞A−(0, t)exp(−iρt) dt

iρ − A+(0, 0) +

∞0

A+1 (0, t) exp(iρt) dt

+

1 + ∞

0A+(0, t)exp(iρt) dt

iρ − A−(0, 0) −

0

−∞A−

1 (0, t) exp(−iρt) dt

.

Integration by parts yields

iρ 0

−∞

A−(0, t)exp(

−iρt) dt =

−A−(0, 0) +

0

−∞

A−2 (0, t)exp(

−iρt) dt,



173

iρ ∞

0A+(0, t) exp(iρt) dt = −A+(0, 0) −

∞0

A+2 (0, t) exp(iρt) dt.

Furthermore, 0

−∞A−(0, t)exp(−iρt) dt

∞0

A+1 (0, s)exp(iρs) ds

= 0

−∞A−(0, t)

∞−t

A+1 (0, ξ + t)exp(iρξ ) dξ

dt

= ∞

0

0

−ξA−(0, t)A+

1 (0, ξ + t) dt

exp(iρξ ) dξ.

Analogously, 0

−∞A−

1 (0, t)exp(−iρt) dt ∞

0A+(0, s)exp(iρs) ds

= ∞0

0

−ξ A−1 (0, t)A+(0, ξ + t) dt exp(iρξ ) dξ.

Since2(A+(0, 0) + A−(0, 0)) =

∞−∞

q (t) dt,

we arrive at (3.1.24) for a(ρ), where

A(t) = A+1 (0, t) − A−

1 (0, −t) + A−2 (0, −t) − A+

2 (0, t) − A+(0, 0)A−(0, −t) − A−(0, 0)A+(0, t)

+ 0

−tA−(0, ξ )A+

1 (0, ξ + t) dξ − 0

−tA−

1 (0, ξ )A+(0, ξ + t) dξ.

It follows from (3.1.10)-(3.1.11) that A(t) ∈ L(0, ∞). For the function b(ρ) the argumentsare similar.

Denote

e0(x, ρ) =e(x, ρ)

a(ρ), g0(x, ρ) =

g(x, ρ)

a(ρ), (3.1.25)

s+(ρ) = −b(

−ρ)

a(ρ) , s−(ρ) =b(ρ)

a(ρ) . (3.1.26)

The functions s+(ρ) and s−(ρ) are called the reflection coefficients (right and left, respec-tively). It follows from (3.1.18), (3.1.25) and (3.1.26) that

e0(x, ρ) = g(x, −ρ) + s−(ρ)g(x, ρ), g0(x, ρ) = e(x, −ρ) + s+(ρ)e(x, ρ). (3.1.27)

Using (3.1.25), (3.1.27) and (3.1.3) we get

e0(x, ρ) ∼ exp(iρx) + s−(ρ)exp(−iρx) (x → −∞), e0(x, ρ) ∼ t(ρ)exp(iρx) (x → ∞),

g0(x, ρ) ∼ t(ρ)exp(iρx) (x → −∞), g0(x, ρ) ∼ exp(−iρx) + s+(ρ)exp(iρx) (x → ∞),

where t(ρ) = (a(ρ))−1 is called the transmission coefficient .

We point out the main properties of the functions s±(ρ). By virtue of (3.1.20)-(3.1.22)and (3.1.26), the functions s±(ρ) are continuous for real ρ = 0, and

s±(ρ) = s±(−ρ).



174

Moreover, (3.1.21) implies

|s±(ρ)|2 = 1 − 1

|a(ρ)|2,

and consequently,

|s±(ρ)

|< 1 for real ρ

= 0.

Furthermore, according to (3.1.23) and (3.1.26),

s±(ρ) = o

1

ρ

as |ρ| → ∞.

Denote by R±(x) the Fourier transform for s±(ρ) :

R±(x) :=1

2π ∞

−∞

s±(ρ)exp(±iρx) dρ. (3.1.28)

Then R±(x) ∈ L2(−∞, ∞) are real, and

s±(ρ) = ∞−∞

R±(x)exp(∓iρx) dx. (3.1.29)


ρe(x, ρ) = ρa(ρ)

(s−(ρ) + 1)g(x, ρ) + g(x, −ρ) − g(x, ρ)

,

ρg(x, ρ) = ρa(ρ)(s+

(ρ) + 1)e(x, ρ) + e(x, −ρ) − e(x, ρ),and consequently,

limρ→0

ρa(ρ)(s±(ρ) + 1) = 0.

3.1.3. Let us now study the properties of the discrete spectrum.

Definition 3.1.1. The values of the parameter λ, for which equation (3.1.1) hasnonzero solutions y(x) ∈ L2(−∞, ∞), are called eigenvalues of (3.1.1), and the correspond-

ing solutions are called eigenfunctions .The properties of the eigenvalues are similar to the properties of the discrete spectrum

of the Sturm-Liouville operator on the half-line (see Chapter 2).

Theorem 3.1.2. There are no eigenvalues for λ ≥ 0.

Proof. Repeat the arguments in the proof of Theorems 2.1.6 and 2.3.6. 2

Let Λ+ := λ, λ = ρ2, ρ ∈ Ω+ : a(ρ) = 0 be the set of zeros of a(ρ) in the upperhalf-plane Ω+. Since the function a(ρ) is analytic in Ω+ and, by virtue of (3.1.23),

a(ρ) = 1 + O1

ρ

, |ρ| → ∞, I m ρ ≥ 0,

we get that Λ+ is an at most countable bounded set.

Theorem 3.1.3. The set of eigenvalues coincides with Λ+. The eigenvalues λk are real and negative (i.e. Λ+ ⊂ (−∞, 0) ). For each eigenvalue λk = ρ2

k, there exists only one (up to a multiplicative constant) eigenfunction, namely

g(x, ρk) = dke(x, ρk), dk = 0. (3.1.30)



175

The eigenfunctions e(x, ρk) and g(x, ρk) are real. Eigenfunctions related to different eigen-values are orthogonal in L2(−∞, ∞).

Proof. Let λk = ρ2k ∈ Λ+. By virtue of (3.1.22),

e(x, ρk), g(x, ρk)

= 0, (3.1.31)

i.e. (3.1.30) is valid. According to Theorem 3.1.1, e(x, ρk) ∈ L2(α, ∞), g(x, ρk) ∈ L2(−∞, α)for each real α. Therefore, (3.1.30) implies

e(x, ρk), g(x, ρk) ∈ L2(−∞, ∞).

Thus, e(x, ρk) and g(x, ρk) are eigenfunctions, and λk = ρ2k is an eigenvalue.

Conversly, let λk = ρ2k, ρk ∈ Ω+ be an eigenvalue, and let yk(x) be a corresponding

eigenfunction. Since yk(x)

∈L2(

−∞,

∞), we have

yk(x) = ck1e(x, ρk), yk(x) = ck2g(x, ρk), ck1, ck2 = 0,

and consequently, (3.1.31) holds. Using (3.1.22) we obtain a(ρk) = 0, i.e. λk ∈ Λ+.Let λn and λk ( λn = λk ) be eigenvalues with eigenfunctions yn(x) = e(x, ρn) and

yk(x) = e(x, ρk) respectively. Then integration by parts yields ∞−∞

yn(x) yk(x) dx = ∞−∞

yn(x)yk(x) dx,

and henceλn

∞−∞

yn(x)yk(x) dx = λk

∞−∞

yn(x)yk(x) dx

or ∞−∞

yn(x)yk(x) dx = 0.

Furthermore, let λ0 = u + iv,v = 0 be a non-real eigenvalue with an eigenfunctiony0(x) = 0. Since q (x) is real, we get that λ0 = u − iv is also the eigenvalue with theeigenfunction y0(x). Since λ0

= λ0, we derive as before

y02L2

= ∞−∞

y0(x)y0(x) dx = 0,

which is impossible. Thus, all eigenvalues λk are real, and consequently the eigenfunctionse(x, ρk) and g(x, ρk) are real too. Together with Theorem 3.1.2 this yields Λ+ ⊂ (−∞, 0).Theorem 3.1.3 is proved. 2

For λk = ρ2k ∈ Λ+ we denote

α+k =

∞−∞

e2(x, ρk) dx−1

, α−k = ∞

−∞g2(x, ρk) dx

−1.

Theorem 3.1.4. Λ+ is a finite set, i.e. in Ω+ the function a(ρ) has at most a finite number of zeros. All zeros of a(ρ) in Ω+ are simple, i.e. a1(ρk) = 0, where a1(ρ) := d

dρa(ρ). Moreover,

α+

k

=dk

ia1(ρk), α−

k

=1

idka1(ρk), (3.1.32)



176

where the numbers dk are defined by (3.1.30).

Proof. 1) Let us show that

−2ρ

x

−A

e(t, ρ)g(t, ρ) dt = e(t, ρ), g(t, ρ)

x

−A,

2ρ A

xe(t, ρ)g(t, ρ) dt = e(t, ρ), g(t, ρ)

Ax

,

(3.1.33)

where in this subsection

e(t, ρ) :=d

dρe(t, ρ), g(t, ρ) :=

d

dρg(t, ρ).

Indeed,

ddx

e(x, ρ), g(x, ρ) = e(x, ρ)g(x, ρ) − e(x, ρ)g(x, ρ).

Since

−e(x, ρ) + q (x)e(x, ρ) = ρ2e(x, ρ), −g(x, ρ) + q (x)g(x, ρ) = ρ2g(x, ρ) + 2ρg(x, ρ),

we getd

dxe(x, ρ), g(x, ρ) = −2ρe(x, ρ)g(x, ρ).

Similarly,d

dxe(x, ρ), g(x, ρ) = 2ρe(x, ρ)g(x, ρ),

and we arrive at (3.1.33).It follows from (3.1.33) that

2ρ A

−Ae(t, ρ)g(t, ρ) dt = −e(x, ρ), g(x, ρ) − e(x, ρ), g(x, ρ)

+e(x, ρ), g(x, ρ)x=A+ e(x, ρ), g(x, ρ)x=−A

.

On the other hand, differentiating (3.1.22) with respect to ρ , we obtain

2iρa1(ρ) + 2ia(ρ) = −e(x, ρ), g(x, ρ) − e(x, ρ), g(x, ρ).

For ρ = ρk this yields with the preceding formula

ia1(ρk) = A

−Ae(t, ρk)g(t, ρk) dt + δ k(A), (3.1.34)

where

δ k(A) = − 1

2ρk

e(x, ρk), g(x, ρk)x=A

+ e(x, ρk), g(x, ρk)x=−A

.

Since ρk = iτ k, τ k > 0, we have by virtue of (3.1.4),

e(x, ρk), e(x, ρk) = O(exp(

−τ kx)), x

→+

∞.



177


e(x, ρk) = ix exp(−τ kx) + ∞

xitA+(x, t)exp(−τ kt) dt,

e(x, ρk) = i exp(−

τ kx)−

ixτ k exp(−

τ kx)−

ixA+(x, x)exp(−

τ kx)

+ ∞

xitA+

1 (x, t) exp(−τ kt) dt.

Hencee(x, ρk), e(x, ρk) = O(1), x → +∞.

From this, using (3.1.30), we calculate

e(x, ρk), g(x, ρk) = dke(x, ρk), e(x, ρk) = o(1) as x → +∞,

e(x, ρk), g(x, ρk) =1

dk

g(x, ρk), g(x, ρk) = o(1) as x → −∞.

Consequently,lim

A→ +∞δ k(A) = 0.

Then (3.1.34) implies

ia1(ρk) = ∞−∞

e(t, ρk)g(t, ρk) dt.

Using (3.1.30) again we obtain

ia1(ρk) = dk

∞−∞

e2(t, ρk) dt =1

dk

∞−∞

g2(t, ρk) dt.

Hence a1(ρk) = 0, and (3.1.32) is valid.

2) Suppose that Λ+ = λk is an infinite set. Since Λ+ is bounded and λk = ρ2k < 0,

it follows that ρk = iτ k → 0, τ k > 0. By virtue of (3.1.4)-(3.1.5), there exists A > 0 such

that e(x,iτ ) ≥ 12

exp(−τ x) for x ≥ A, τ ≥ 0,

g(x,iτ ) ≥ 12

exp(τ x) for x ≤ −A, τ ≥ 0,

(3.1.35)

and consequently

∞A

e(x, ρk)e(x, ρn) dx ≥ exp(−(τ k + τ n)A)

4(τ k + τ n)≥ exp(−2AT )

8T ,

−A

−∞g(x, ρk)g(x, ρn) dx

≥exp(−(τ k + τ n)A)

4(τ k + τ n) ≥exp(−2AT )

8T ,

(3.1.36)

where T = maxk

τ k. Since the eigenfunctions e(x, ρk) and e(x, ρn) are orthogonal in

L2(−∞, ∞) we get

0 = ∞−∞

e(x, ρk)e(x, ρn) dx = ∞

Ae(x, ρk)e(x, ρn) dx +

1

dkdn

−A

−∞g(x, ρk)g(x, ρn) dx

+ A

−Ae2(x, ρ

k) dx +

A

−Ae(x, ρ

k)(e(x, ρ

n)−

e(x, ρk

)) dx. (3.1.37)



178

Take x0 ≤ −A such that e(x0, 0) = 0. According to (3.1.30),

1

dkdn=

e(x0, ρk)e(x0, ρn)

g(x0, ρk)g(x0, ρn).

Since the functions e(x, ρ) and g(x, ρ) are continuous for Im ρ≥

0, we calculate with thehelp of (3.1.35),

limk,n→∞

g(x0, ρk)g(x0, ρn) = g2(x0, 0) > 0,

limk,n→∞

e(x0, ρk)e(x0, ρn) = e2(x0, 0) > 0.

Therefore,

limk,n→∞

1

dkdn> 0.

Together with (3.1.36) this yields ∞A

e(x, ρk)e(x, ρn) dx +1

dkdn

−A

−∞g(x, ρk)g(x, ρn) dx +

A

−Ae2(x, ρk) dx ≥ C > 0 (3.1.38)

for sufficiently large k and n. On the other hand, acting in the same way as in the proof of Theorem 2.3.4 one can easily verify that

A

−A

e(x, ρk)(e(x, ρn) − e(x, ρk)) dx → 0 as k, n → ∞. (3.1.39)

The relations (3.1.37)-(3.1.39) give us a contrudiction.This means that Λ+ is a finite set. 2

Thus, the set of eigenvalues has the form

Λ+ = λkk=1,N , λk = ρ2k, ρk = iτ k, 0 < τ 1 < .. . < τ m.

Definition 3.1.2. The data J + =

s+(ρ), λk, α+

k ; ρ

∈R, k = 1, N

are called the

right scattering data, and the data J − = s−(ρ), λk, α−k ; ρ ∈ R, k = 1, N are called theleft scattering data.

Example 3.1.1. Let q (x) ≡ 0. Then

e(x, ρ) = exp(iρx), g(x, ρ) = exp(−iρx), a(ρ) = 1, b(ρ) = 0, s±(ρ) = 0, N = 0,

i.e. there are no eigenvalues at all.

3.1.4. In this subsection we study connections between the scattering data J + andJ −. Consider the function

γ (ρ) =1

a(ρ)

N k=1

ρ − iτ kρ + iτ k

. (3.1.40)

Lemma 3.1.3. (i1) The function γ (ρ) is analytic in Ω+ and continuous in Ω+ \0.(i2) γ (ρ) has no zeros in Ω+ \ 0.(i3) For |ρ| → ∞, ρ ∈ Ω+,

γ (ρ) = 1 + O1

ρ. (3.1.41)



179

(i4) |γ (ρ)| ≤ 1 for ρ ∈ Ω+.

Proof. The assertions (i1) − (i3) are obvious consequences of the preceding discussion,and only (i4) needs to be proved. By virtue of (3.1.21), |a(ρ)| ≥ 1 for real ρ = 0, andconsequently,

|γ (ρ)| ≤ 1 for real ρ = 0. (3.1.42)Suppose that the function ρa(ρ) is analytic in the origin. Then, using (3.1.40) and

(3.1.42) we deduce that the function γ (ρ) has a removable singularity in the origin, and γ (ρ)(after extending continuously to the origin) is continuous in Ω+. Using (3.1.41), (3.1.42)and the maximum principle we arrive at (i4).

In the general case we cannot use these arguments for γ (ρ). Therefore, we introduce thepotentials

q r(x) = q (x), |x| ≤ r,

0, |x| > r,

r

≥0,

and consider the corresponding Jost solutions er(x, ρ) and gr(x, ρ). Clearly, er(x, ρ) ≡exp(iρx) for x ≥ r and gr(x, ρ) ≡ exp(−iρx) for x ≤ −r. For each fixed x, the functionse(ν )

r (x, ρ) and g(ν )r (x, ρ) (ν = 0, 1) are entire in ρ. Take

ar(ρ) = − 1

2iρer(x, ρ), gr(x, ρ), γ r(ρ) =

1

ar(ρ)

N rk=1

ρ − iτ kr

ρ + iτ kr,

where ρkr = iτ kr, k = 1, N r are zeros of ar(ρ) in the upper half-plane Ω+. The functionρar(ρ) is entire in ρ, and (see Lemma 3.1.2) |ar(ρ)| ≥ 1 for real ρ. The function γ r(ρ)is analytic in Ω+, and

|γ r(ρ)| ≤ 1 for ρ ∈ Ω+. (3.1.43)

By virtue of Lemma 3.1.1,

limr→∞ sup

ρ∈Ω+

supx≥a

|(e(ν )r (x, ρ) − e(ν )(x, ρ)) exp(−iρx)| = 0,

limr→∞ sup

ρ∈Ω+

supx≤a

|(g(ν )r (x, ρ) − g(ν )(x, ρ)) exp(iρx)| = 0,

for ν = 0, 1 and each real a. Therefore

limr→∞ sup

ρ∈Ω+

|ρ(ar(ρ) − a(ρ))| = 0,

i.e.lim

r→∞ρar(ρ) = ρa(ρ) uniformly in Ω+. (3.1.44)

In particular, (3.1.44) yields that 0 < τ kr ≤ C for all k and r.Let δ r be the infimum of distances between the zeros ρkr of ar(ρ) in the upper

half-plane Im ρ > 0. Let us show that

δ ∗ := inf r>0

δ r > 0. (3.1.45)



180

Indeed, suppose on the contrary that there exists a sequence rk → ∞ such that δ rk → 0.

Let ρ(1)k = iτ

(1)k , ρ

(2)k = iτ

(2)k (τ

(1)k , τ

(2)k ≥ 0) be zeros of ark(ρ) such that ρ

(1)k − ρ

(2)k → 0 as

k → ∞. It follows from (3.1.4)-(3.1.5) that there exists A > 0 such that

er(x,iτ )

≥12 exp(

−τ x) for x

≥A, τ

≥0, r

≥0,

gr(x,iτ ) ≥ 12

exp(τ x) for x ≤ −A, τ ≥ 0, r ≥ 0. (3.1.46)

Since the functions erk(x, ρ(1)k ) and erk(x, ρ

(2)k ) are orthogonal in L2(−∞, ∞), we get

0 = ∞−∞

erk(x, ρ(1)k )erk(x, ρ

(2)k ) dx

= ∞

A


(2)k ) dx +

1

d(1)k d

(2)k

−A

−∞grk(x, ρ

(1)k )grk(x, ρ

(2)k ) dx

+ A

−Ae2

rk(x, ρ

(1)k ) dx +

A

−Aerk(x, ρ

(1)k )(erk(x, ρ

(2)k ) − erk(x, ρ

(1)k )) dx, (3.1.47)

where the numbers d( j)k are defined by

grk(x, ρ( j)k ) = d

( j)k erk(x, ρ

( j)k ), d

( j)k = 0.

Take x0

≤ −A. Then, by virtue of (3.1.46),

grk(x0, ρ(1)k )grk(x0, ρ

(2)k ) ≥ C > 0,

and1

d(1)k d

(2)k

=erk(x0, ρ

(1)k )erk(x0, ρ

(2)k )

grk(x0, ρ(1)k )grk(x0, ρ

(2)k )

.

Using Lemma 3.1.1 we get

limk→∞ erk(x0, ρ(1)k )erk(x0, ρ(2)k ) ≥ 0;

hence

limk→∞

1

d(1)k d

(2)k

≥ 0.

Then ∞A


(2)k ) dx +

1

d(1)k d

(2)k

−A

−∞grk(x, ρ

(1)k )grk(x, ρ

(2)k ) dx

+ A

−Ae2

rk(x, ρ

(1)k ) dx ≥ C > 0.

On the other hand,

A

−Aerk(x, ρ

(1)k )(erk(x, ρ

(2)k ) − erk(x, ρ

(1)k )) dx → 0 as k → ∞.

From this and (3.1.47) we arrive at a contradiction, i.e. (3.1.45) is proved.



181

Let Dδ,R := ρ ∈ Ω+ : δ < |ρ| < R, where 0 < δ < min(δ ∗, τ 1), R > τ N . Using(3.1.44) one can show that

limr→∞γ r(ρ) = γ (ρ) uniformly in Dδ,R. (3.1.48)


|γ (ρ)

| ≤1 for ρ

∈Dδ,R. By virtue of arbitrariness

of δ and R we obtain |γ (ρ)| ≤ 1 for ρ ∈ Ω+, i.e. (i4) is proved. 2

It follows from Lemma 3.1.3 that

(a(ρ))−1 = O(1) as |ρ| → 0, ρ ∈ Ω+. (3.1.49)

We also note that since the function σa(σ) is continuous at the origin, it follows that forsufficiently small real σ,

1 ≤ |a(σ)| =1

|γ (σ)

|

≤ C

|σ

|

.

The properties of the function γ (ρ) obtained in Lemma 3.1.3 allow one to recover γ (ρ)in Ω+ from its modulus |γ (σ)| given for real σ.


γ (ρ) = exp 1

πi

∞−∞

ln |γ (ξ )|ξ − ρ

dξ

, ρ ∈ Ω+. (3.1.50)

Proof. 1) The function ln γ (ρ) is analytic in Ω+ and ln γ (ρ) = O(ρ−1) for |ρ| →

∞, ρ

∈Ω+. Consider the closed contour C R (with counterclockwise circuit) which is the

boundary of the domain DR = ρ ∈ Ω+ : |ρ| < R (see fig. 3.1.1). By Cauchy’s integralformula [con1, p.84],

ln γ (ρ) =1

2πi

C R

ln γ (ξ )

ξ − ρdξ, ρ ∈ DR.

Since

limR→∞

1

2πi

|ξ|=R

ξ∈Ω+

ln γ (ξ )

ξ − ρdξ = 0,

we obtain

ln γ (ρ) = 12πi

∞−∞

ln γ (ξ )ξ − ρ

dξ, ρ ∈ Ω+. (3.1.51)

2) Take a real σ and the closed contour C σR,δ (with counterclockwise circuit) consistingof the semicircles C 0R = ξ : ξ = R exp(iϕ), ϕ ∈ [0, π], Γσ

δ = ξ : ξ − σ = δ exp(iϕ), ϕ ∈[0, π], δ > 0 and the intervals [−R, R] \ [σ − δ, σ + δ ] (see fig.3.1.1).

6

-

6

E

G

©

−R R 6

-

6

E

G

©

−R Rσ

fig. 3.1.1.



182

By Cauchy’s theorem,1

2πi

C σR,δ

ln γ (ξ )

ξ − σdξ = 0.

Since

limR→∞1

2πi C 0R

ln γ (ξ )

ξ − σ dξ = 0, limδ→0

1

2πi Γσδ

ln γ (ξ )

ξ − σ dξ = −1

2 ln γ (σ),

we get for real σ,

ln γ (σ) =1

πi

∞−∞

ln γ (ξ )

ξ − σdξ. (3.1.52)

In (3.1.52) (and everywhere below where necessary) the integral is understood in the principalvalue sense.

3) Let γ (σ) = |γ (σ)| exp(−iβ (σ)). Separating in (3.1.52) real and imaginary parts we

obtainβ (σ) =

1

π

∞−∞

ln |γ (ξ )|ξ − σ

dξ, ln |γ (σ))| = − 1

π

∞−∞

β (ξ )

ξ − σdξ.

Then, using (3.1.51) we calculate for ρ ∈ Ω+ :

ln γ (ρ) =1

2πi

∞−∞

ln |γ (ξ )|ξ − ρ

dξ − 1

2π

∞−∞

β (ξ )

ξ − ρdξ

=1

2πi ∞

−∞

ln |γ (ξ )|ξ − ρ

dξ −

1

2π2 ∞

−∞ ∞

−∞

dξ

(ξ − ρ)(s − ξ ) ln|γ (s)

|ds.

Since1

(ξ − ρ)(s − ξ )=

1

s − ρ

1

ξ − ρ− 1

ξ − s

,

it follows that for ρ ∈ Ω+ and real s,

∞−∞

dξ

(ξ − ρ)(s − ξ )=

πi

s − ρ.

Consequently,

ln γ (ρ) =1

πi

∞−∞

ln |γ (ξ )|ξ − ρ

dξ, ρ ∈ Ω+,

and we arrive at (3.1.50). 2

It follows from (3.1.21) and (3.1.26) that for real ρ = 0,

1

|a(ρ)|2

= 1

− |s±(ρ)

|2.

By virtue of (3.1.40) this yields for real ρ = 0,

|γ (ρ)| =

1 − |s±(ρ)|2.

Using (3.1.40) and (3.1.50) we obtain

a(ρ) =N

k=1

ρ − iτ k

ρ + iτ kexp −

1

2πi ∞

−∞

ln(1 − |s±(ξ )|2)

ξ − ρ

dξ , ρ

∈Ω+. (3.1.53)



183

We note that since the function ρa(ρ) is continuous in Ω+, it follows that

ρ2

1 − |s±(ρ)|2= O(1) as |ρ| → 0.

Relation (3.1.53) allows one to establish connections between the scattering data J +and J −. More precisely, from the given data J + one can uniquely reconstruct J − (andvice versa) by the following algorithm.

Algorithm 3.1.1. Let J + be given. Then1) Construct the function a(ρ) by (3.1.53);2) Calculate dk and α−k , k = 1, N by (3.1.32);3) Find b(ρ) and s−(ρ) by (3.1.26).

3.2. THE MAIN EQUATION

3.2.1. The inverse scattering problem is formulated as follows:

Given the scattering data J + (or J − ), construct the potential q.

The central role for constructing the solution of the inverse scattering problem is playedby the so-called main equation which is a linear integral equation of Fredholm type. In thissection we give a derivation of the main equation and study its properties. In Section 3.3we provide the solution of the inverse scattering problem along with necessary and sufficientconditions of its solvability.

Theorem 3.2.1. For each fixed x, the functions A±(x, t), defined in (3.1.8), satisfy the integral equations

F +(x + y) + A+(x, y) + ∞

xA+(x, t)F +(t + y) dt = 0, y > x, (3.2.1)

F −(x + y) + A−(x, y) + x

−∞A−(x, t)F −(t + y) dt = 0, y < x, (3.2.2)

where

F ±(x) = R±(x) +N

k=1

α±k exp(∓τ kx), (3.2.3)

and the functions R±(x) are defined by (3.1.28).Equations (3.2.1) and (3.2.2) are called the main equations or Gelfand-Levitan-Marchenko

equations.

Proof. By virtue of (3.1.18) and (3.1.19),1

a(ρ)− 1

g(x, ρ) = s+(ρ)e(x, ρ) + e(x, −ρ) − g(x, ρ). (3.2.4)

Put A+(x, t) = 0 for t < x, and A−(x, t) = 0 for t > x. Then, using (3.1.8) and (3.1.29),we get

s+

(ρ)e(x, ρ) + e(x, −ρ) − g(x, ρ)



184

= ∞

−∞R+(y)exp(iρy)dy

exp(iρx) +

∞−∞

A+(x, t) exp(iρt) dt

+ ∞−∞

(A+(x, t) − A−(x, t)) exp(−iρt) dt = ∞−∞

H (x, y) exp(−iρy) dy,

where

H (x, y) = A+(x, y) − A−(x, y) + R+(x + y) + ∞

xA+(x, t)R+(t + y) dt. (3.2.5)

Thus, for each fixed x, the right-hand side in (3.2.4) is the Fourier transform of the functionH (x, y). Hence

H (x, y) =1

2π

∞−∞

1

a(ρ)− 1

g(x, ρ)exp(iρy) dρ. (3.2.6)

Fix x and y (y > x) and consider the function

f (ρ) :=

1

a(ρ)− 1

g(x, ρ)exp(iρy). (3.2.7)


f (ρ) =c

ρexp(iρ(y − x))(1 + o(1)), |ρ| → ∞, ρ ∈ Ω+. (3.2.8)

Let C δ,R be a closed contour (with counterclockwise circuit) which is the boundary of thedomain Dδ,R = ρ ∈ Ω+ : δ < |ρ| < R, where δ < τ 1 < . . . < τ N < R. Thus, all zerosρk = iτ k, k = 1, N of a(ρ) are contained in Dδ,R. By the residue theorem,

1

2πi

C δ,R

f (ρ) dρ =N

k=1

Resρ=ρk

f (ρ).

On the other hand, it follows from (3.2.7)-(3.2.8), (3.1.5) and (3.1.49) that

limR→∞1

2πi |ρ|=R

ρ∈Ω+

f (ρ) dρ = 0, limδ→0

1

2πi |ρ|=δ

ρ∈Ω+

f (ρ) dρ = 0.

Hence1

2πi

∞−∞

f (ρ) dρ =N

k=1

Resρ=ρk

f (ρ).

From this and (3.2.6)-(3.2.7) it follows that

H (x, y) = i

N k=1

g(x,iτ k)exp(

−τ ky)

a1(iτ k) .

Using the fact that all eigenvalues are simple, (3.1.30), (3.1.8) and (3.1.32) we obtain

H (x, y) = iN

k=1

dke(x,iτ k)exp(−τ ky)

a1(iτ k)

=

−

N

k=1

α+k exp(

−τ k(x + y)) +

∞

x

A+(x, t) exp(

−τ k(t + y)) dt . (3.2.9)



185

Since A−(x, y) = 0 for y > x, (3.2.5) and (3.2.9) yield (3.2.1).Relation (3.2.2) is proved analogously. 2

Lemma 3.2.1. Let nonnegative functions v(x), u(x) (a ≤ x ≤ T ≤ ∞) be given such that v(x) ∈ L(a, T ), u(x)v(x) ∈ L(a, T ), and let c1 ≥ 0. If

u(x) ≤ c1 + T

xv(t)u(t) dt, (3.2.10)

then

u(x) ≤ c1 exp T

xv(t) dt

. (3.2.11)

Proof. Denote

ξ (x) := c1 + T

x

v(t)u(t) dt.

Then ξ (T ) = c1, −ξ (x) = v(x)u(x), and (3.2.10) yields

0 ≤ −ξ (x) ≤ v(x)ξ (x).

Let c1 > 0. Then ξ (x) > 0, and

0 ≤ −ξ (x)

ξ (x)≤ v(x).

Integrating this inequality we obtain

lnξ (x)

ξ (T )≤ T

xv(t) dt,

and consequently,

ξ (x) ≤ c1 exp T

xv(t) dt

.

According to (3.2.10) u(x) ≤ ξ (x), and we arrive at (3.2.11).If c1 = 0, then ξ (x) = 0. Indeed, suppose on the contrary that ξ (x) = 0. Then, there

exists T 0 ≤ T such that ξ (x) > 0 for x < T 0, and ξ (x) ≡ 0 for x ∈ [T 0, T ]. Repeatingthe arguments we get for x < T 0 and sufficiently small ε > 0,

lnξ (x)

ξ (T 0 − ε)≤ T 0−ε

xv(t) dt ≤

T 0

xv(t) dt,

which is impossible. Thus, ξ (x)

≡0, and (3.2.11) becomes obvious. 2

Lemma 3.2.2. The functions F ±(x) are absolutely continuous and for each fixed a >−∞, ∞

a|F ±(±x)| dx < ∞,

∞a

(1 + |x|)|F ±(±x)| dx < ∞. (3.2.12)

Proof. 1) According to (3.2.3) and (3.1.28), F +(x) ∈ L2(a, ∞) for each fixed a > −∞.By continuity, (3.2.1) is also valid for y = x :

F

+

(2x) + A

+

(x, x) + ∞x A

+

(x, t)F

+

(t + x) dt = 0. (3.2.13)



186

Rewrite (3.2.13) to the form

F +(2x) + A+(x, x) + 2 ∞

xA+(x, 2ξ − x)F +(2ξ ) dξ = 0. (3.2.14)

It follows from (3.2.14) and (3.1.10) that the function F +(x) is continuous, and for x≥

a,

|F +(2x)| ≤ 1

2Q+

0 (x) + exp(Q+1 (a))

∞x

Q+0 (ξ )|F +(2ξ )| dξ. (3.2.15)

Fix r ≥ a. Then for x ≥ r, (3.2.15) yields

|F +(2x)| ≤ 1

2Q+

0 (r) + exp(Q+1 (a))

∞x

Q+0 (ξ )|F +(2ξ )| dξ.

Applying Lemma 3.2.1 we obtain

|F +(2x)| ≤ 1

2Q+

0 (r)exp(Q+1 (a)exp(Q+

1 (a))), x ≥ r ≥ a,

and consequently|F +(2x)| ≤ C aQ+

0 (x), x ≥ a. (3.2.16)

It follows from (3.2.16) that for each fixed a > −∞,

∞a

|F +(x)| dx < ∞.

2) By virtue of (3.2.14), the function F +(x) is absolutely continuous, and

2F +(2x) +

d

dxA+(x, x) − 2A+(x, x)F +(2x)

+2 ∞

x A+

1 (x, 2ξ − x) + A+2 (x, 2ξ − x)

F +(2ξ ) dξ = 0,

where

A+1 (x, t) =

∂A+(x, t)

∂x, A+

2 (x, t) =∂A+(x, t)

∂t.


F +(2x) =

1

4q (x) + P (x), (3.2.17)

where

P (x) = − ∞

x

A+

1 (x, 2ξ − x) + A+2 (x, 2ξ − x)

F +(2ξ ) dξ − 1

2F +(2x)

∞x

q (t) dt.


|P (x)| ≤ C a(Q+0 (x))2, x ≥ a. (3.2.18)

Since

xQ

+

0 (x) ≤ ∞x t|q (t)| dt,



187

it follows from (3.2.17) and (3.2.18) that for each fixed a > −∞, ∞a

(1 + |x|)|F +(x)| dx < ∞,

and (3.2.12) is proved for the function F +(x). For F −(x) the arguments are similar. 2

3.2.2. Now we are going to study the solvability of the main equations (3.2.1) and (3.2.2).Let sets J ± = s±(ρ), λk, α±k ; ρ ∈ R, k = 1, N be given satisfying the following condition.

Condition A. For real ρ = 0, the functions s±(ρ) are continuous, |s±(ρ)| < 1, s±(ρ) =

s±(−ρ) and s±(ρ) = o

1ρ

as |ρ| → ∞. The real functions R±(x), defined by (3.1.28),

are absolutely continuous, R±(x) ∈ L2(−∞, ∞), and for each fixed a > −∞,

∞

a |R±(

±x)

|dx <

∞,

∞

a

(1 +

|x

|)

|R±(

±x)

|dx <

∞. (3.2.19)

Moreover, λk = −τ 2k < 0, α±k > 0, k = 1, N.

Theorem 3.2.2. Let sets J + ( J − ) be given satisfying Condition A. Then for each fixed x, the integral equation (3.2.1) ((3.2.2) respectively) has a unique solution A+(x, y) ∈L(x, ∞) ( A−(x, y) ∈ L(−∞, x) respectively).

Proof. For definiteness we consider equation (3.2.1). For (3.2.2) the arguments are thesame. It is easy to check that for each fixed x, the operator

(J xf )(y) = ∞

xF +(t + y)f (t) dt, y > x

is compact in L(x, ∞). Therefore, it is sufficient to prove that the homogeneous equation

f (y) + ∞

xF +(t + y)f (t) dt = 0 (3.2.20)

has only the zero solution. Let f (y) ∈ L(x, ∞) be a real function satisfying (3.2.20). It

follows from (3.2.20) and Condition A that the functions F +

(y) and f (y) are boundedon the half-line y > x, and consequently f (y) ∈ L2(x, ∞). Using (3.2.3) and (3.1.28) wecalculate

0 = ∞

xf 2(y) dy +

∞x

∞x

F +(t + y)f (t)f (y) dtdy

= ∞

xf 2(y) dy +

N k=1

α+k

∞x

f (y) exp(−τ ky) dy2

+1

2π

∞−∞

s+(ρ)Φ2(ρ) dρ,

where

Φ(ρ) = ∞x f (y)exp(iρy) dy.

According to Parseval’s equality ∞x

f 2(y) dy =1

2π

∞−∞

|Φ(ρ)|2 dρ,

and hence

N

k=1

α+k

∞

x

f (y) exp(

−τ ky) dy

2+

1

2π ∞

−∞ |Φ(ρ)

|21

− |s+(ρ)

|exp(i(2θ(ρ) + η(ρ))) dρ = 0,



188

where θ(ρ) = arg Φ(ρ), η(ρ) = arg(−s+(ρ)). In this equality we take the real part:

N k=1

α+k

∞x

f (y) exp(−τ ky) dy2

+1

2π

∞−∞

|Φ(ρ)|2

1 − |s+(ρ)| cos((2θ(ρ) + η(ρ)))

dρ = 0.

Since |s+

(ρ)| < 1, this is possible only if Φ(ρ) ≡ 0. Then f (y) = 0, and Theorem 3.2.2 isproved. 2

Remark 3.2.1. The main equations (3.2.1)-(3.2.2) can be rewritten in the form

F +(2x + y) + B+(x, y) + ∞

0B+(x, t)F +(2x + y + t) dt = 0, y > 0,

F −(2x + y) + B−(x, y) + 0

−∞B−(x, t)F −(2x + y + t) dt = 0, y < 0,

(3.2.21)

where B±(x, y) = A±(x, x + y).

3.3. THE INVERSE SCATTERING PROBLEM

3.3.1. In this section, using the main equations (3.2.1)-(3.2.2), we provide the solutionof the inverse scattering problem of recovering the potential q from the given scatteringdata J + (or J − ). First we prove the uniqueness theorem.

Theorem 3.3.1. The specification of the scattering data J + (or J − ) uniquely deter-mines the potential q.

Proof. Let J + and J + be the right scattering data for the potentials q and q respectively, and let J + = J +. Then, it follows from (3.2.3) and (3.1.28) that F +(x) =F +(x). By virtue of Theorems 3.2.1 and 3.2.2, A+(x, y) = A+(x, y). Therefore, taking(3.1.9) into account, we get q = q. For J − the arguments are the same. 2

The solution of the inverse scattering problem can be constructed by the following algo-rithm.

Algorithm 3.3.1. Let the scattering data J + (or J − ) be given. Then1) Calculate the function F +(x) (or F −(x) ) by (3.2.3) and (3.1.28).2) Find A+(x, y) (or A−(x, y) ) by solving the main equation (3.2.1) (or (3.2.2) respec-tively).3) Construct q (x) = −2 d

dxA+(x, x) (or q (x) = 2 d

dxA−(x, x) ).

Now we going to study necessary and sufficient conditions for the solvability of the inverse

scatering problem. In other words we will describe conditions under which a set J + (orJ − ) represents the scattering data for a certain potential q. First we prove the followingauxiliary assertion.

Lemma 3.3.1. Let sets J ± = s±(ρ), λk, α±k ; ρ ∈ R, k = 1, N be given satisfying Condition A, and let A±(x, y) be the solutions of the integral equations (3.2.1)-(3.2.2).Construct the functions e(x, ρ) and g(x, ρ) by (3.1.8) and the functions q ±(x) by the

formulae

q

+

(x) = −2

d

dxA

+

(x, x), q −(x) = 2

d

dxA−(x, x). (3.3.1)



189

Then, for each fixed a > −∞, ∞a

(1 + |x|)|q ±(±x)| dx < ∞, (3.3.2)

and

−e(x, ρ) + q +(x)e(x, ρ) = ρ2e(x, ρ), −g(x, ρ) + q −(x)g(x, ρ) = ρ2g(x, ρ). (3.3.3)

Proof. 1) It follows from (3.2.19) and (3.2.3) that ∞a

|F ±(±x)| dx < ∞, ∞

a(1 + |x|)|F ±(±x)| dx < ∞. (3.3.4)

We rewrite (3.2.1) in the form

F +(y + 2x) + A+(x, x + y) + ∞

0A+(x, x + t)F +(t + y + 2x) dt = 0, y > 0, (3.3.5)

and for each fixed x consider the operator

(F xf )(y) = ∞

0F +(t + y + 2x)f (t) dt, y ≥ 0

in L(0,

∞). It follows from Theorem 3.2.2 that there exists (E +

F x)−1, where E is the

identity operator, and (E + F x)−1 < ∞. Using Lemma 1.5.1 (or an analog of Lemma1.5.2 for an infinite interval) it is easy to verify that A+(x, y), y ≥ x and (E + F x)−1are continuous functions. Since

F x = supy

∞0

|F +(t + y + 2x)| dt = supy

∞y+2x

|F +(ξ )| dξ ≤ ∞

2x|F +(ξ )| dξ,

we getlim

x

→∞ F x

= 0.

Therefore,C 0a := sup

x≥a(E + F x)−1 < ∞.

Denote

τ 0(x) = ∞

x|F +

(t)| dt, τ 1(x) =

∞x

τ 0(t) dt = ∞

x(t − x)|F +

(t)| dt.

Then

|F +(x)| ≤ τ 0(x). (3.3.6)


A+(x, x + y) = −(E + F x)−1F +(y + 2x),

consequently the preceding formulae yield,

∞

0|A+(x, x + y)| dy ≤ C 0a

∞

0|F +(y + 2x)| dy ≤ C 0a τ 1(2x), x ≥ a. (3.3.7)



190

Using (3.3.5), (3.3.6) and (3.3.7) we calculate

|A+(x, x + y)| ≤ τ 0(y + 2x) + ∞

0|A+(x, x + t)|τ 0(t + y + 2x) dt

≤ 1 + C 0a τ 1(2x)τ 0(y + 2x). (3.3.8)

Applying Lemma 1.5.1 one can show that the function A+(x, y) has the first derivatives

A+1 (x, y) :=

∂A+(x, y)

∂x, A+

2 (x, y) :=∂A+(x, y)

∂y,

and therefore, differentiating (3.2.1), we get

F +(x + y) + A+

1 (x, y)

−A+(x, x)F +(x + y) +

∞

xA+

1 (x, t)F +(t + y) dt = 0,

F +(x + y) + A+

2 (x, y) + ∞

xA+(x, t)F +

(t + y) dt = 0.

(3.3.9)

Denote A+0 (x, x + y) := d

dxA+(x, x + y). Differentiating (3.3.5) with respect to x we obtain

2F +(y + 2x) + A+

0 (x, x + y)

+

∞

0A+

0 (x, x + t)F +(t + y + 2x) dt + 2

∞

0A+(x, x + t)F +

(t + y + 2x) dt = 0, (3.3.10)

and consequently ∞0

|A+0 (x, x + y)| dy ≤ 2C 0a

∞0

|F +(y + 2x)| dy

+ ∞

0|A+(x, x + t)|

∞0

|F +(t + y + 2x)| dy

dt

.

By virtue of (3.3.8), this yields

∞0

|A+0 (x, x + y)| dy ≤ 2C 0aτ 0(2x) + (1 + C 0a τ 1(2x)) ∞0

τ 20 (t + 2x) dt≤ 2C 0a τ 0(2x)

1 + (1 + C 0a τ 1(2x))τ 1(2x)

. (3.3.11)


|A+0 (x, x + y) + 2F +

(y + 2x)| ≤

∞

0 |A+

0 (x, x + t)F +(t + y + 2x)

|dt + 2

∞

0 |A+(x, x + t)F +

(t + y + 2x)

|dt.

Using (3.3.6), (3.3.11) and (3.3.8) we get

|A+0 (x, x + y) + 2F +

(y + 2x)| ≤ 2C 0a τ 0(y + 2x)τ 0(2x)τ 1(2x)(1 + C 0a τ 1(2x))

+2τ 0(2x)τ 0(y + 2x)(1 + C 0a τ 1(2x)).

For y = 0 this implies

|2d

dxA

+

(x, x) + 4F

+

(2x)| ≤ 2C aτ

2

0 (2x), x ≥ a,



191

where C a = 4(1 + C 0a τ 1(2a))2. Taking (3.3.1) into account we derive ∞a

(1 + |x|)|q +(x)| dx ≤ 4 ∞

a(1 + |x|)|F +

(2x)| dx + C a

∞a

(1 + |x|)τ 20 (2x) dx.

Since

xτ 0(2x) ≤ ∞2x

t|F +(t)| dt ≤ ∞0

t|F +(t)| dt,

we arrive at (3.3.2) for q +. For q − the arguments are the same.

2) Let us now prove (3.3.3). For definiteness we will prove (3.3.3) for the functione(x, ρ). First we additionally assume that the function F +

(x) is absolutely continuous and

F +

(x) ∈ L(a, ∞) for each fixed a > −∞. Differentiating the equality

J (x, y) := F +(x + y) + A+(x, y) +

∞

xA+(x, t)F +(t + y) dt = 0, y > x, (3.3.12)

we calculate

J yy (x, y) = F +

(x + y) + A+yy (x, y) +

∞x

A+(x, t)F +

(t + y) dt = 0, (3.3.13)

J xx(x, y) = F +

(x + y) + A+xx(x, y) − d

dxA+(x, x)F +(x + y)

−A+(x, x)F +(x + y) − A+

1 (x, t)|t=xF +(x + y) + ∞

xA+

xx(x, t)F +(t + y) dt = 0. (3.3.14)

Integration by parts in (3.3.13) yields

J yy (x, y) = F +

(x + y) + A+yy (x, y) +

A+(x, t)F +

(t + y) − A+

2 (x, t)F +(t + y)∞

x

+ ∞

xA+

tt(x, t)F +(t + y) dt = 0.

It follows from (3.3.8) and (3.3.9) that here the substitution at infinity is equal to zero, andconsequently,

J yy (x, y) = F +(x + y) + A+yy (x, y) − A+(x, x)F +(x + y) + A+

2 (x, x)F +(x + y)

+ ∞

xA+

tt(x, t)F +(t + y) dt = 0. (3.3.15)

Using (3.3.14), (3.3.15), (3.3.12), (3.3.1) and the equality

J xx(x, y) − J yy (x, y) − q +(x)J (x, y) = 0,

which follows from J (x, y) = 0, we obtain

f (x, y) + ∞

xf (x, t)F +(t + y) dt = 0, y ≥ x, (3.3.16)

wheref (x, y) := A+

xx(x, y) − A+yy (x, y) − q +(x)A+(x, y).

It is easy to verify that for each fixed x ≥ a, f (x, y) ∈ L(x, ∞). According to Theorem3.2.2, the homogeneous equation (3.3.16) has only the zero solution, i.e.

A+xx(x, y) − A

+yy (x, y) − q

+

(x)A+

(x, y) = 0, y ≥ x. (3.3.17)



192

Differentiating (3.1.8) twice we get

e(x, ρ) = (iρ)2 exp(iρx) − (iρ)A+(x, x)exp(iρx)

−d

dx

A+(x, x) + A+1 (x, t)

|t=x exp(iρx) +

∞

x

A+xx(x, t)exp(iρt) dt. (3.3.18)

On the other hand, integrating by parts twice we obtain

ρ2e(x, ρ) = −(iρ)2 exp(iρx) − (iρ)2 ∞

xA+(x, t)exp(iρt) dt

= −(iρ)2 exp(iρx) + (iρ)A+(x, x)exp(iρx) − A+2 (x, t)|t=x exp(iρx) −

∞x

A+tt(x, t)exp(iρt) dt.

Together with (3.1.8) and (3.3.18) this gives

e(x, ρ) + ρ2e(x, ρ) − q +(x)e(x, ρ) =

− 2d

dxA+(x, x) − q +(x)

exp(iρx)

+ ∞

x(A+

xx(x, t) − A+tt(x, t) − q +(x)A+(x, t)) exp(iρt) dt.

Taking (3.3.1) and (3.3.17) into account we arrive at (3.3.3) for e(x, ρ).Let us now consider the general case when (3.3.4) holds. Denote by e(x, ρ) the Jost

solution for the potential q +. Our goal is to prove that e(x, ρ)

≡e(x, ρ). For this purpose

we choose the functions F + j (x) such that F + j (x), F + j (x) are absolutely continuous, andfor each fixed a > −∞, F + j

(x) ∈ L(a, ∞) and

lim j→∞

∞a

|F + j (x) − F +(x)| dx = 0, lim j→∞

∞a

(1 + |x|)|F + j(x) − F +

(x)| dx = 0. (3.3.19)

Denote

τ 0 j(x) = ∞

x

|F + j

(t)

−F +

(t)

|dt, τ 1 j(x) =

∞

xτ 0 j (t) dt =

∞

x(t

−x)

|F + j

(t)

−F +

(t)

|dt.

Using (3.3.19) and Lemma 1.5.1 one can show that for sufficiently large j, the integralequation

F + j (x + y) + A+( j)(x, y) +

∞x

A+( j)(x, t)F + j (t + y) dt = 0, y > x,

has a unique solution A+( j)(x, y), and

∞

x |A+

( j)(x, y)

−A+(x, y)

|dy

≤C aτ 1 j(2x), x

≥a. (3.3.20)

Consequently,

lim j→∞max

x≥a

∞x

|A+( j)(x, y) − A+(x, y)| dy = 0. (3.3.21)

Denotee j(x, ρ) = exp(iρx) +

∞x

A+( j)(x, t)exp(iρt) dt, (3.3.22)

q + j (x) =

−2

d

dx

A+( j)(x, x).



193

It was proved above that

−e j (x, ρ) + q + j (x)e j(x, ρ) = ρ2e j(x, ρ),

i.e. e j(x, ρ) is the Jost solution for the potential q + j . Using (3.3.19)-(3.3.20) and acting in

the same way as in the first part of the proof of Lemma 3.3.1 we get

lim j→∞

∞a

(1 + |x|)|q + j (x) − q +(x)| dx = 0.

By virtue of Lemma 3.1.1, this yields

lim j→∞

maxρ∈Ω+

maxx≥a

|(e j(x, ρ) − e(x, ρ)) exp(−iρx)| = 0.

On the other hand, using (3.1.8), (3.3.21) and (3.3.22) we infer

lim j→∞ max

ρ∈Ω+

maxx≥a

|(e j(x, ρ) − e(x, ρ)) exp(−iρx)| = 0.

Consequently, e(x, ρ) ≡ e(x, ρ), and (3.3.3) is proved for the function e(x, ρ). For g(x, ρ)the arguments are similar. Therefore, Lemma 3.3.1 is proved. 2

3.3.2. Let us now formulate necessary and sufficient conditions for the solvability of the inverse scattering problem.

Theorem 3.3.1. For data J + = s+(ρ), λk, α+k ; ρ ∈ R, k = 1, N to be the right

scattering data for a certain real potential q satisfying (3.1.2), it is necessary and sufficient that the following conditions hold:

1) λk = −τ 2k , 0 < τ 1 < .. . < τ N ; α+k > 0, k = 1, N .

2) For real ρ = 0, the function s+(ρ) is continuous, s+(ρ) = s+(−ρ), |s+(ρ)| < 1, and

s+(ρ) = o1

ρ

as |ρ| → ∞, (3.3.23)

ρ2

1 − |s+(ρ)|2= O(1) as |ρ| → 0. (3.3.24)

3) The function ρ(a(ρ) − 1), where a(ρ) is defined by

a(ρ) :=N

k=1

ρ − iτ kρ + iτ k

exp(B(ρ)), B(ρ) := − 1

2πi

∞−∞

ln(1 − |s+(ξ )|2)

ξ − ρdξ, ρ ∈ Ω+, (3.3.25)

is continuous and bounded in Ω+, and

1

a(ρ)= O(1) as |ρ| → 0, ρ ∈ Ω+, (3.3.26)

limρ→0

ρa(ρ)

s+(ρ) + 1

= 0 for real ρ. (3.3.27)

4) The functions R±(x), defined by

R±(x) =1

2π ∞

−∞s±(ρ) exp(

±iρx) dρ, s−(ρ) :=

−s+(

−ρ)

a(−ρ)

a(ρ), (3.3.28)



194

are real and absolutely continuous, R±(x) ∈ L2(−∞, ∞), and for each fixed a > −∞,(3.2.19) holds.

Proof. The necessity part of Theorem 3.3.1 was proved above. We prove here thesufficiency. Let a set J + satisfying the hypothesis of Theorem 3.3.1 be given. According to

(3.3.25),

B(ρ) =1

2πi

∞−∞

θ(ξ )

ξ − ρdξ, ρ ∈ Ω+, θ(ξ ) := ln

1

1 − |s+(ξ )|2. (3.3.29)

For real ξ = 0, the function θ(ξ ) is continuous and

θ(ξ ) = θ(−ξ ) ≥ 0, (3.3.30)

θ(ξ ) = o 1

ξ 2 as ξ → ∞,

θ(ξ ) = O

ln1

ξ

as ξ → 0.

Furthermore, the function B(ρ) is analytic in Ω+, continuous in Ω+ \ 0 and for realρ = 0,

B(ρ) =1

2θ(ρ) +

1

2πi

∞−∞

θ(ξ )

ξ − ρdξ, (3.3.31)

where the integral in (3.3.31) is understood in the principal value sense: ∞−∞

:= limε→0

ρ−ε

−∞+ ∞

ρ+ε

.

It follows from assumption 3) of the theorem, (3.3.25), (3.3.30) and (3.3.31) that

a(ρ) = a(−ρ) for real ρ = 0. (3.3.32)

Moreover, for real ρ = 0, (3.3.25) and (3.3.31) imply

|a(ρ)|2 = | exp(B(ρ))|2 = exp(2Re B(ρ)) = exp(θ(ρ)),

and consequently,

1 − |s+(ρ)|2 =1

|a(ρ)|2for real ρ = 0. (3.3.33)

Furthermore, the function s−(ρ), defined by (3.3.28), is continuous for real ρ = 0, and byvirtue of (3.3.32),

s−(ρ) = s−(−ρ), |s−(ρ)| = |s+(ρ)|.Therefore, taking (3.3.23), (3.3.24) and (3.3.27) into account we obtain

s−(ρ) = o1

ρ

as |ρ| → ∞,

ρ2

1

− |s−(ρ)

|2

= O(1) as |ρ| → 0,



195

limρ→0

ρa(ρ)

s−(ρ) + 1

= 0 for real ρ.

Let us show thata2

1(ρk) < 0, k = 1, N , (3.3.34)

where a1(ρ) = ddρ a(ρ), ρk = iτ k. Indeed, it follows from (3.3.25) that

a1(ρk) =d

dρ

N j=1

ρ − iτ jρ + iτ j

|ρ=ρk

exp(B(ρk)).


B(ρk) =1

2πi

∞−∞

θ(ξ )

ξ − iτ kdξ

=

1

2πi ∞

−∞ξθ(ξ )

ξ 2 + τ 2k dξ +

τ k

2π ∞−∞ θ(ξ )

ξ 2 + τ 2k dξ =

τ k

2π ∞−∞ θ(ξ )

ξ 2 + τ 2k dξ.

Sinced

dρ

N j=1

ρ − iτ jρ + iτ j

|ρ=ρk

=1

2iτ k

N j=1j=k

τ k − τ jτ k + τ j

,

the numbers a1(ρk) are pure imaginary, and (3.3.34) is proved.Denote

α−k =

−1

a2

1(ρk)α+

k

, k = 1, N. (3.3.35)

According to (3.3.34)-(3.3.35),α−k > 0, k = 1, N .

Thus, we have the sets J ± = s±(ρ), λk, α±k ; ρ ∈ R, k = 1, N which satisfy Condition A.Therefore, Theorem 3.2.2 and Lemma 3.3.1 hold. Let A±(x, y) be the solutions of (3.2.1)-(3.2.2). We construct the functions e(x, ρ) and g(x, ρ) by (3.1.8) and the functions q ±(x)by (3.3.1). Then (3.3.2)-(3.3.4) are valid.


s+(ρ)e(x, ρ) + e(x, −ρ) =g(x, ρ)

a(ρ),

s−(ρ)g(x, ρ) + g(x, −ρ) =e(x, ρ)

a(ρ).

(3.3.36)

Proof. 1) Denote

Φ+

(x, y) = R+

(x + y) + ∞x A

+

(x, t)R+

(t + y) dt,

Φ−(x, y) = R−(x + y) + x

−∞A−(x, t)R−(t + y) dt.

For each fixed x, Φ±(x, y) ∈ L2(−∞, ∞), and ∞−∞

Φ+(x, y)exp(−iρy) dy = s+(ρ)e(x, ρ),

∞

−∞Φ−(x, y)exp(iρy) dy = s−(ρ)g(x, ρ).

(3.3.37)



196

Indeed, using (3.1.8) and (3.3.28), we calculate

s+(ρ)e(x, ρ) =

exp(iρx) + ∞

xA+(x, t) exp(iρt) dt

∞−∞

R+(ξ ) exp(−iρξ ) dξ

= ∞−∞ R+

(x + y)exp(−iρy) dy + ∞x A

+

(x, t) ∞

−∞ R+

(t + y)exp(−iρy) dy dt

= ∞−∞

R+(x + y) +

∞x

A+(x, t)R+(t + y) dt

exp(−iρy) dy = ∞−∞

Φ+(x, y) exp(−iρy) dy.

The second relation in (3.3.37) is proved similarly.On the other hand, it follows from (3.2.1)-(3.2.2) that

Φ+(x, y) = −A+(x, y) −N

k=1

α+k exp(−τ ky)e(x,iτ k), y > x,

Φ−(x, y) = −A−(x, y) −N

k=1

α−k exp(τ ky)g(x,iτ k), y < x.

Then ∞−∞

Φ+(x, y)exp(−iρy) dy = x

−∞Φ+(x, y)exp(−iρy) dy

− ∞

x A+(x, y) +N

k=1

α+k exp(

−τ ky)e(x,iτ k) exp(

−iρy) dy.

According to (3.1.8), ∞x

A+(x, y)exp(−iρy) dy = e(x, −ρ) − exp(−iρx),

and consequently,

∞

−∞Φ+(x, y)exp(

−iρy) dy =

x

−∞Φ+(x, y)exp(

−iρy) dy

+exp(−iρx) − e(x, −ρ) −N

k=1

α+k

τ k + iρexp(−iρx)exp(−τ kx)e(x,iτ k).

Similarly, ∞−∞

Φ−(x, y) exp(iρy) dy = ∞

xΦ+(x, y)exp(iρy) dy

+exp(iρx)−

g(x,−

ρ)−

N

k=1

α−kτ k + iρ

exp(iρx)exp(τ k

x)g(x,iτ k

).

Comparing with (3.3.37) we deduce

s+(ρ)e(x, ρ) + e(x, −ρ) =h−(x, ρ)

a(ρ),

s−(ρ)g(x, ρ) + g(x, −ρ) =h+(x, ρ)

a(ρ),

(3.3.38)



197

where

h−(x, ρ) := exp(−iρx)a(ρ)

1 + x

−∞Φ+(x, y)exp(iρ(x − y)) dy

−

N

k=1

α+k

τ k + iρexp(

−τ kx)e(x,iτ k),

h+(x, ρ) := exp(iρx)a(ρ)

1 + ∞

xΦ−(x, y) exp(iρ(y − x)) dy

−N

k=1

α−kτ k + iρ

exp(τ kx)g(x,iτ k)

.

(3.3.39)

2) Let us study the properties of the functions h±(x, ρ). By virtue of (3.3.38),

h−(x, ρ) = a(ρ)s+

(ρ)e(x, ρ) + e(x, −ρ),

h+(x, ρ) = a(ρ)

s−(ρ)g(x, ρ) + g(x, −ρ)

.

(3.3.40)

In particular, this yields that the functions h±(x, ρ) are continuous for real ρ = 0, and inview of (3.3.32), h±(x, ρ) = h±(x, −ρ). Since

limρ→0

ρa(ρ)

s±(ρ) + 1

= 0,

it follows from (3.3.40) thatlimρ→0

ρh±(x, ρ) = 0, (3.3.41)

and consequently the functions ρh±(x, ρ) are continuous for real ρ. By virtue of (3.3.39)the functions ρh±(x, ρ) are analytic in Ω+, continuous in Ω+, and (3.3.41) is valid forρ ∈ Ω+. Taking (3.3.40) and (3.1.7) into account we get

e(x, ρ), h−(x, ρ) = h+(x, ρ), g(x, ρ) = −2iρa(ρ). (3.3.42)

Since |s±(ρ)| < 1, it follows from (3.3.38) that for real ρ = 0,

supρ=0

1

a(ρ)h±(x, ρ)

< ∞. (3.3.43)


h+(x,iτ k) = ia1(iτ k)α−k g(x,iτ k), h−(x,iτ k) = ia1(iτ k)α+k e(x,iτ k), (3.3.44)

lim|ρ|→∞Im ρ≥0

h±(x, ρ) exp(∓

iρx) = 1, (3.3.45)

where a1(ρ) = ddρ

a(ρ).

3) It follows from (3.3.38) that

s+(ρ)e(x, ρ) + e(x, −ρ) =h−(x, ρ)

a(ρ),

e(x, ρ) + s+(−

ρ)e(x,−

ρ) =h−(x, −ρ)

a(−ρ).



198

Solving this linear algebraic system we obtain

e(x, ρ)

1 − s+(ρ)s+(−ρ)

=h−(x, −ρ)

a(−ρ)− s+(−ρ)

h−(x, ρ)

a(ρ).

By virtue of (3.3.32)-(3.3.33),

1 − s+(ρ)s+(−ρ) = 1 − |s+(ρ)|2 =1

|a(ρ)|2=

1

a(ρ)a(−ρ).

Therefore,e(x, ρ)

a(ρ)= s−(ρ)h−(x, ρ) + h−(x, −ρ). (3.3.46)

Using (3.3.46) and the second relation from (3.3.38) we calculate

h−(x, ρ)g(x, −ρ) − h−(x, −ρ)g(x, ρ) = G(ρ), (3.3.47)

where

G(ρ) :=1

a(ρ)

h+(x, ρ)h−(x, ρ) − e(x, ρ)g(x, ρ)

. (3.3.48)


h+(x,iτ k)h−(x,iτ k)

−e(x,iτ k)g(x,iτ k) = 0, k = 1, N,

and hence (after removing singularities) the function G(ρ) is analytic in Ω+ and continuousin Ω+ \ 0. By virtue of (3.3.45),

lim|ρ|→∞Im ρ≥0

h+(x, ρ)h−(x, ρ) = 1.

Since

a(ρ) = 1 + O1

ρas

|ρ

| → ∞, ρ

∈Ω+,

it follows from (3.3.48) thatlim|ρ|→∞Im ρ≥0

G(ρ) = 0.

By virtue of (3.3.47),G(−ρ) = −G(ρ) for real ρ = 0.

We continue G(ρ) to the lower half-plane by the formula

G(ρ) = −G(−ρ), Im ρ < 0. (3.3.49)

Then, the function G(ρ) is analytic in C \ 0, and

lim|ρ|→∞

G(ρ) = 0. (3.3.50)

Furthermore, it follows from (3.3.48), (3.3.26), (3.3.41) and (3.3.49) that

limρ→0 ρ

2

G(ρ) = 0,



199

i.e. the function ρG(ρ) is entire in ρ. On the other hand, using (3.3.48), (3.3.41) and(3.3.43) we obtain that for real ρ,

limρ→0

ρG(ρ) = 0,

and consequently, the function G(ρ) is entire in ρ. Together with (3.3.50), using Liouville’stheorem, this yields G(ρ) ≡ 0, i.e.

h+(x, ρ)h−(x, ρ) = e(x, ρ)g(x, ρ), ρ ∈ Ω+, (3.3.51)

h−(x, ρ)g(x, −ρ) = h−(x, −ρ)g(x, ρ) for real ρ = 0. (3.3.52)

4) Now we consider the function

p(x, ρ) :=h+(x, ρ)

e(x, ρ) .

Denote E = x : e(x, 0)e(x,iτ 1) . . . e(x,iτ N ) = 0. Since e(x, ρ) is a solution of thedifferential equation (3.3.3) and for ρ ∈ Ω+,

|e(x, ρ) exp(−iρx) − 1| ≤ ∞

x|A+(x, t)| dt → 0 as x → ∞, (3.3.53)

it follows that

E is a finite set. Take a fixed x /

∈ E . Let ρ∗

∈Ω+ be a zero of e(x, ρ), i.e.

e(x, ρ∗) = 0. Since x /∈ E , it follows that ρ∗ = 0, ρ∗ = iτ k, k = 1, N ; hence ρ∗a(ρ∗) = 0.In view of (3.3.42) this yields h−(x, ρ∗) = 0. According to (3.3.51), h+(x, ρ∗) = 0. Sinceall zeros of e(x, ρ) are simple (this fact is proved similarly to Theorem 2.3.3), we deducethat the function p(x, ρ) (after removing singularities) is analytic in Ω+ and continuous inΩ+ \ 0. It follows from (3.3.45) and (3.3.53) that

p(x, ρ) → 1 as |ρ| → ∞, ρ ∈ Ω+.

By virtue of (3.3.51)-(3.3.52),

p(x, ρ) = p(x, −ρ) for real ρ = 0.

We continue p(x, ρ) to the lower half-plane by the formula

p(x, ρ) = p(x, −ρ), Im ρ < 0.

Then, the function p(x, ρ) is analytic in C \ 0, and

lim|ρ|→∞ p(x, ρ) = 1. (3.3.54)

Since e(x, 0) = 0, it follows from (3.3.41) that

limρ→0

ρp(x, ρ) = 0,

and consequently, the function p(x, ρ) is entire in ρ. Together with (3.3.54) this yields p(x, ρ) ≡ 1, i.e.

h+

(x, ρ) ≡ e(x, ρ). (3.3.55)



200

Then, in view of (3.3.51),h−(x, ρ) ≡ g(x, ρ). (3.3.56)

From this, using (3.3.38), we arrive at (3.3.36). Lemma 3.3.2 is proved.

Let us return to the proof of Theorem 3.3.1. It follows from (3.3.36) and (3.3.3) that

q −(x) = q +(x) := q (x). (3.3.57)

Then (3.3.2) implies (3.1.2), and the functions e(x, ρ) and g(x, ρ) are the Jost solutionsfor the potential q defined by (3.3.57).

Denote by J ± = s+(ρ), λk, α+k ; ρ ∈ R, k = 1, N the scattering data for this potential

q, and let

a(ρ) :=

−

1

2iρe(x, ρ), g(x, ρ)

. (3.3.58)

Using (3.3.42) and (3.3.55)-(3.3.56) we get

e(x, ρ), g(x, ρ) = −2iρa(ρ).

Together with (3.3.58) this yields a(ρ) ≡ a(ρ), and consequently, N = N, λk = λk, k =1, N . Furthermore, it follows from (3.1.21) that

s+(ρ)e(x, ρ) + e(x,

−ρ) =

g(x, ρ)

a(ρ)

,

s−(ρ)g(x, ρ) + g(x, −ρ) =e(x, ρ)

a(ρ).

Comparing with (3.3.26) we infer s±(ρ) = s±(ρ), ρ ∈ R.By virtue of (3.1.26)

α+k =

dk

ia1(ρk), α−k =

1

idka1(ρk). (3.3.59)

On the other hand, it follows from (3.3.55)-(3.3.56) and (3.3.44) that

e(x,iτ k) = ia1(iτ k)α−k g(x,iτ k),

g(x,iτ k) = ia1(iτ k)α+k e(x,iτ k),

i.e.

dk = ia1(iτ k)α+k =

1

ia1(iτ k)α−k.

Comparing with (3.3.59) we get α± = α±, and Theorem 3.3.1 is proved. 2

Remark 3.3.1. There is a connection between the inverse scattering problem and theRiemann problem for analytic functions. Indeed, one can rewrite (3.1.27) in the form

Q−(x, ρ) = Q+(x, ρ)Q(ρ), (3.3.60)

where

Q−(x, ρ) = g(x, −ρ) e(x, −ρ)

g(x, −ρ) e(x, −ρ) , Q+(x, ρ) = e(x, ρ) g(x, ρ)

e(x, ρ) g(x, ρ) ,



201

Q(ρ) =

1

a(ρ)

b(−ρ)

a(ρ)

− b(ρ)

a(ρ)

1

a(ρ)

.

For each fixed x, the matrix-functions Q±(x, ρ) are analytic and bounded for

±Im ρ > 0.

By virtue of (3.1.26) and (3.1.53), the matrix-function Q(ρ) can be uniquely reconstructedfrom the given scattering data J + (or J − ). Thus, the inverse scattering problem is reducedto the Riemann problem (3.3.60). We note that the theory of the solution of the Riemannproblem can be found, for example, in [gak1]. Applying the Fourier transform to (3.3.60) asit was shown in Section 3.2, we arrive at the Gelfand-Levitan-Marchenko equations (3.2.1)-(3.2.2) or (3.2.21). We note that the use of the Riemann problem in the inverse problemtheory has only methodical interest and does not constitute an independent method, sinceit can be considered as a particular case of the contour integral method (see [bea1], [yur1]).

3.4. REFLECTIONLESS POTENTIALS. MODIFICATION OF THEDISCRETE SPECTRUM.

3.4.1. A potential q satisfying (3.1.2) is called reflectionless if b(ρ) ≡ 0. By virtue of (3.1.26) and (3.1.53) we have in this case

s±(ρ) ≡ 0, a(ρ) =

N

k=1

ρ

−iτ k

ρ + iτ k . (3.4.1)

Theorem 3.3.1 allows one to prove the existence of reflectionless potentials and to describeall of them. Namely, the following theorem is valid.

Theorem 3.4.1. Let arbitrary numbers λk = −τ 2k < 0, α+k > 0, k = 1, N be given.

Take s+(ρ) ≡ 0, ρ ∈ R, and consider the data J + = s+(ρ), λk, α+k ; ρ ∈ R, k = 1, N .

Then, there exists a unique reflectionless potential q satisfying (3.1.2) for which J + are the right scattering data.

Theorem 3.4.1 is an obvious corollary of Theorem 3.3.1 since for this set J + all conditionsof Theorem 3.3.1 are fulfilled and (3.4.1) holds.

For a reflectionless potential, the Gelfand-Levitan-Marchenko equation (3.2.1) takes theform

A+(x, y) +N

k=1

α+k exp(−τ k(x + y)) +

N k=1

α+k exp(−τ ky)

∞x

A+(x, t)exp(−τ kt) dt = 0. (3.4.2)

We seek a solution of (3.4.2) in the form:

A+(x, y) =N

k=1

P k(x) exp(−τ ky).

Substituting this into (3.4.2), we obtain the following linear algebraic system with respectto P k(x) :

P k(x) +N

j=1

α+ j

exp(−(τ k + τ j)x)

τ k + τ jP j(x) =

−α+

k exp(

−τ kx), k = 1, N . (3.4.3)



202

Solving (3.4.3) we infer P k(x) = ∆k(x)/∆(x), where

∆(x) = det

δ kl + α+

k

exp(−(τ k + τ l)x)

τ k + τ l

k,l=1,N

, (3.4.4)

and ∆k(x) is the determinant obtained from ∆(x) by means of the replacement of k-columnby the right-hand side column. Then

q (x) = −2d

dxA+(x, x) = −2

N k=1

∆k(x)

∆(x)exp(−τ kx),

and consequently one can show that

q (x

) =−

2d2

dx2ln∆(

x).

(3.4

.5)

Thus, (3.4.4) and (3.4.5) allow one to calculate reflectionless potentials from the given num-bers λk, α+

k k=1,N .

Example 3.4.1. Let N = 1, τ = τ 1, α = α+1 , and

∆(x) = 1 +α

2τ exp(−2τ x).

Then (3.4.5) gives

q (x) = − 4τ αexp(τ x) + α

2τ exp(−τ x)

2 .

Denote

β = − 1

2τ ln

2τ

α.

Then

q (x) = −2τ 2

cosh2(τ (x − β )) .

3.4.2. If q = 0, then s±(ρ) ≡ 0, N = 0, a(ρ) ≡ 1. Therefore, Theorem 3.4.1 showsthat all reflectionless potentials can be constructed from the zero potential and given dataλk, α+

k , k = 1, N . Below we briefly consider a more general case of changing the discretespectrum for an arbitrary potential q. Namely, the following theorem is valid.

Theorem 3.4.2. Let J + = s+(ρ), λk, α+k ; ρ ∈ R, k = 1, N be the right scattering

data for a certain real potential q satifying (3.1.2). Take arbitrary numbers λk =−

τ 2k

<

0, α+k > 0, k = 1, N , and consider the set J + = s+(ρ), λk, α+

k ; ρ ∈ R, k = 1, N with the same s+(ρ) as in J +. Then there exists a real potential q satisfying (3.1.2), for which J + represents the right scattering data.

Proof. Let us check the conditions of Theorem 3.3.1 for J +. According to (3.3.25) and(3.3.28) we construct the functions a(ρ) and s−(ρ) by the formulae

a(ρ) :=N

k=1

ρ − iτ k

ρ + iτ kexp −

1

2πi ∞

−∞

ln(1 − |s+(ξ )|2)

ξ − ρ

dξ , ρ

∈Ω+,



203

s−(ρ) := −s+(−ρ)a(−ρ)

a(ρ). (3.4.6)


a(ρ) = a(ρ)

N k=1

ρ−

iτ kρ + iτ k

N k=1

ρ + iτ kρ − iτ k . (3.4.7)


s−(ρ) = −s+(−ρ)a(−ρ)

a(ρ). (3.4.8)

Using (3.4.6)-(3.4.8) we gets−(ρ) = s−(ρ). (3.4.9)

Since the scattering data J + satisfy all conditions of Theorem 3.3.1 it follows from (3.4.7)and (3.4.9) that J + also satisfy all conditions of Theorem 3.3.1. Then, by Theorem 3.3.1there exists a real potential q satisfying (3.1.2) for which J + are the right scattering data.2



204

IV. APPLICATIONS OF THE INVERSE PROBLEM THEORY

In this chapter various applications of inverse spectral problems for Sturm-Liouville op-erators are considered. In Sections 4.1-4.2 we study the Korteweg-de Vries equation on the

line and on the half-line by the inverse problem method. Section 4.3 is devoted to a synthesisproblem of recovering parameters of a medium from incomplete spectral information. Sec-tion 4.4 deals with discontinuous inverse problems which are connected with discontinuousmaterial properties. Inverse problems appearing in elasticity theory are considered in Section4.5. In Section 4.6 boundary value problems with aftereffect are studied. Inverse problemsfor differential equations of the Orr-Sommerfeld type are investigated in Section 4.7. Section4.8 is devoted to inverse problems for differential equations with turning points.

4.1. SOLUTION OF THE KORTEWEG-DE VRIES EQUATIONON THE LINE

Inverse spectral problems play an important role for integrating some nonlinear evolutionequations in mathematical physics. In 1967, G.Gardner, G.Green, M.Kruskal and R.Miura[gar1] found a deep connection of the well-known (from XIX century) nonlinear Korteweg-deVries (KdV) equation

q t = 6qq x−

q xxx

with the spectral theory of Sturm-Liouville operators. They could manage to solve globallythe Cauchy problem for the KdV equation by means of reduction to the inverse spectralproblem. These investigations created a new branch in mathematical physics (for furtherdiscussions see [abl1], [lax1], [tak1], [zak1]). In this section we provide the solution of theCauchy problem for the KdV equation on the line. For this purpose we use ideas from [gar1],[lax1] and results of Chapter 3 on the inverse scattering problem for the Sturm-Liouvilleoperator on the line.

4.1.1. Consider the Cauchy problem for the KdV equation on the line:

q t = 6qq x − q xxx, −∞ < x < ∞, t > 0, (4.1.1)

q |t=0 = q 0(x), (4.1.2)

where q 0(x) is a real function satisfying (3.1.2). Denote by Q0 the set of real functionsq (x, t), −∞ < x < ∞, t ≥ 0, such that for each fixed T > 0,

max0≤t≤T

∞−∞

(1 + |x|)|q (x, t)| dt < ∞.

Let Q1 be the set of functions q (x, t) such that q, q, q , q , q ∈ Q0. Here and below, ”dot”denotes derivatives with respect to t, and ”prime” denotes derivatives with respect to x.We will seek the solution of the Cauchy problem (4.1.1)-(4.1.2) in the class Q1. First weprove the following uniqueness theorem.

Theorem 4.1.1. The Cauchy problem (4.1.1)-(4.1.2) has at most one solution.



205

Proof. Let q, q ∈ Q1 be solutions of the the Cauchy problem (4.1.1)-(4.1.2). Denotew := q − q. Then w ∈ Q1, w|t=0 = 0, and

wt = 6(qwx + wq x) − wxxx.

Multiplying this equality by w and integrating with respect to x, we get ∞−∞

wwt dx = 6 ∞−∞

w(qwx + wq x) dx − ∞−∞

wwxxx dx.

Integration by parts yields ∞−∞

wwxxx dx = − ∞−∞

wxwxx dx = ∞−∞

wxwxx dx,

and consequently ∞−∞

wwxxx dx = 0.

Since ∞−∞

qwwx dx = ∞−∞

q 1

2w2

x

dx = −1

2

∞−∞

q xw2 dx,

it follows that ∞−∞

wwt dx = ∞−∞

q x − 1

2q x

w2 dx.

Denote

E (t) = 12

∞−∞

w2 dx, m(t) = 12maxx∈R

q x − 12

q x.

Thend

dtE (t) ≤ m(t)E (t),

and consequently,

0 ≤ E (t) ≤ E (0) exp t

0m(ξ ) dξ

.

Since E (0) = 0 we deduce E (t) ≡ 0, i.e. w ≡ 0. 2

4.1.2. Our next goal is to construct the solution of the Cauchy problem (4.1.1)-(4.1.2)by reduction to the inverse scattering problem for the Sturm-Liouville equation on the line.

Let q (x, t) be the solution of (4.1.1)-(4.1.2). Consider the Sturm-Liouville equation

Ly := −y + q (x, t)y = λy (4.1.3)

with t as a parameter. Then the Jost-type of (4.1.3) solutions and the scattering data

depend on t. Let us show that equation (4.1.1) is equivalent to the equationL = [A, L], (4.1.4)

where in this sectionAy = −4y + 6qy + 3q y

is a linear differential operator, and [A, L] := AL − LA.Indeed, since Ly = −y + qy, we have

Ly = qy, ALy = −4(−y + qy) + 6q (−y + qy) + 3q (−y + qy),



206

LAy = −(−4y + 6qy + 3q y) + q (−4y + 6qy + 3q y),

and consequently (AL − LA)y = (6qq − q )y.Equation (4.1.4) is called the Lax equation or Lax representation, and the pair A, L is

called the Lax pair, corresponding to (4.1.3).

Lemma 4.1.1. Let q (x, t) be a solution of (4.1.1), and let y = y(x,t,λ) be a solution of (4.1.3). Then (L−λ)(y −Ay) = 0, i.e. the function y −Ay is also a solution of (4.1.3).

Indeed, differentiating (4.1.3) with respect to t, we get Ly + (L − λ)y = 0, or, in viewof (4.1.4), (L − λ)y = LAy − ALy = (L − λ)Ay. 2

Let e(x,t,ρ) and g(x,t,ρ) be the Jost-type solutions of (4.1.3) introduced in Section3.1. Denote e± = e(x,t, ±ρ), g± = g(x,t, ±ρ).


e+ = Ae+ − 4iρ3e+. (4.1.5)

Proof. By Lemma 4.1.1, the function e+ − Ae+ is a solution of (4.1.3). Since thefunctions e+, e− form a fundamental system of solutions of (4.1.3), we have

e+ − Ae+ = β 1e+ + β 2e−,

where β k = β k(t, ρ) , k = 1, 2 do not depend on x. As x→

+∞

,

e± ∼ exp(±iρx), e+ ∼ 0, Ae+ ∼ 4iρ3 exp(iρx),

consequently β 1 = −4iρ3, β 2 = 0, and (4.1.5) is proved. 2


a = 0, b = −8iρ3b, s+ = 8iρ3s+, (4.1.6)

λ j = 0, α+

j = 8κ3

j α+

j . (4.1.7)

Proof. According to (3.1.18),

e+ = ag+ + bg−. (4.1.8)

Differentiating (4.1.8) with respect to t, we get e+ = (ag+ + bg−) + (ag+ + bg−). Using(4.1.5) and (4.1.8) we calculate

a(Ag+ − 4iρ3

g+) + b(Ag− − 4iρ3

g−) = (ag+ + bg−) + (ag+ + bg−). (4.1.9)

Since g± ∼ exp(±iρx), g± ∼ 0, Ag± ∼ ±4iρ3 exp(±iρx) as x → −∞, then (4.1.9)yields −8iρ3 exp(−iρx) ∼ a exp(iρx) + b exp(−iρx), i.e. a = 0, b = −8iρ3b. Consequently,s+ = 8iρ3s+, and (4.1.6) is proved.

The eigenvalues λ j = ρ2 j = −κ2

j , κ j > 0, j = 1, N are the zeros of the function

a = a(ρ, t). Hence, according to a = 0, we have λ j = 0. Denote

e j = e(x,t,iκ j), g j = g(x,t,iκ j), j = 1, N .



207

By Theorem 3.1.3, g j = d je j, where d j = d j(t) do not depend on x. Differentiating therelation g j = d je j with respect to t and using (4.1.5), we infer

g j = d je j + d j e j = d j e j + d jAe j − 4κ3 j d je j

org j =

d j

d jg j + Ag j − 4κ3

j g j.

As x → −∞, g j ∼ exp(κ jx), g j ∼ 0, Ag j ∼ −4κ3 j exp(κ jx), and consequently d j =

8κ3 j d j, or, in view of (3.1.32), α+

j = 8κ3 j α+

j . 2

Thus, it follows from Lemma 4.1.3 that we have proved the following theorem.

Theorem 4.1.2. Let q (x, t) be the solution of the Cauchy problem (4.1.1)-(4.1.2), and

let J +

(t) = s+

(t, ρ), λ j(t), α

+

j (t), j = 1, N be the scattering data for q (x, t). Then

s+(t, ρ) = s+(0, ρ) exp(8iρ3t),

λ j(t) = λ j(0), α+ j (t) = α+

j (0) exp(8κ3 j t), j = 1, N (λ j = −κ2

j ).

(4.1.10)

The formulae (4.1.10) give us the evolution of the scattering data with respect to t, andwe obtain the following algorithm for the solution of the Cauchy problem (4.1.1)-(4.1.2).

Algorithm 4.1.1. Let the function q (x, 0) = q 0(x) be given. Then

1) Construct the scattering data s+(0, ρ), λ j(0), α+ j (0), j = 1, N .2) Calculate s+(t, ρ), λ j (t), α+

j (t), j = 1, N by (4.1.10).3) Find the function q (x, t) by solving the inverse scattering problem (see Section 3.3).

Thus, if the Cauchy problem (4.1.1)-(4.1.2) has a solution, then it can be found byAlgorithm 4.1.1. On the other hand, if q 0 ∈ Q1 then J +(t), constructed by (4.1.10),satisfies the conditions of Theorem 3.3.1 for each fixed t > 0, and consequently there existsq (x, t) for which J +(t) are the scattering data. It can be shown (see [mar1]) that q (x, t)

is the solution of (4.1.1)-(4.1.2).

We notice once more the main points for the solution of the Cauchy problem (4.1.1)-(4.1.2) by the inverse problem method:

(1) The presence of the Lax representation (4.1.4).(2) The evolution of the scattering data with respect to t.(3) The solution of the inverse problem.

4.1.3. Among the solutions of the KdV equation (4.1.1) there are very important

particular solutions of the form q (x, t) = f (x − ct). Such solutions are called solitons.Substituting q (x, t) = f (x−ct) into (4.1.1), we get f +6f f +cf = 0, or (f +3f 2+cf ) =0. Clearly, the function

f (x) = − c

2cosh2√

cx2

satisfies this equation. Hence, the function

q (x, t) =

−

c

2cosh2 √ c(x

−ct)

2 (4.1.11)



208

is a soliton.It is interesting to note that solitons correspond to reflectionless potentials (see Section

3.4). Consider the Cauchy problem (4.1.1)-(4.1.2) in the case when q 0(x) is a reflectionlesspotential, i.e. s+(0, ρ) = 0. Then, by virtue of (4.1.10), s+(t, ρ) = 0 for all t, i.e. thesolution q (x, t) of the Cauchy problem (4.1.1)-(4.1.2) is a reflectionless potential for all t.

Using (3.4.5) and (4.1.10) we derive

q (x, t) = −2d2

dx2∆(x, t),

∆(x, t) = det

δ kl + α+

k (0) exp(8κ3kt)

exp(−(κk+κl)x)

κk+κl

k,l=1,N

.

In particular, if N = 1, then α+1 (0) = 2κ1, and

q (x, t) = − 2κ21

cosh2(κ1(x − 4κ21t))

.

Taking c = 4κ21, we get (4.1.11).

4.2. KORTEWEG-DE VRIES EQUATION ON THE HALF-LINE.NONLINEAR REFLECTION

4.2.1. We consider the mixed problem for the KdV equation on the half-line:

q t = 6qq x − q xxx, x ≥ 0, t ≥ 0, (4.2.1)

q |t=0 = q 0(x), q |x=0 = q 1(t), q x|x=0 = q 2(t), q xx|x=0 = q 3(t). (4.2.2)

Here q 0(x) and q j(t), j = 1, 2, 3, are continuous complex-valued functions, and q 0(x) ∈L(0,

∞), q 0(0) = q 1(0). In this section the mixed problem (4.2.1)-(4.2.2) is solved by the

inverse problem method. For this purpose we use the results of Chapter 2 on recovering theSturm-Liouville operator on the half-line from the Weyl function. We provide the evolutionof the Weyl function with respect to t, and give an algorithm for the solution of the mixedproblem (4.2.1)-(4.2.2) along with necessary and sufficient conditions for its solvability.

Denote by Q the set of functions q (x, t) such that the functions q, q, q , q , q arecontinuous in D = (x, t) : x ≥ 0, t ≥ 0, and integrable on the half-line x ≥ 0, for eachfixed t. We will seek the solution of (4.2.1)-(4.2.2) in the class Q.

Consider the Sturm-Liouville equation

Ly := −y + q (x, t)y = λy. (4.2.3)

where q (x, t) ∈ L(0, ∞) for each fixed t ≥ 0. Let Φ(x,t,λ) be the solution of (4.2.3) underthe conditions Φ(0, t , λ) = 1, Φ(x,t,λ) = O(exp(iρx)), x → ∞ (for each fixed t ). DenoteM (t, λ) := Φ(0, t , λ). Then,

Φ(x,t,λ) =e(x,t,ρ)

e(0, t , ρ), M (t, λ) =

e(0, t , ρ)

e(0, t , ρ),



209

where e(x,t,ρ) is the Jost solution for q (x, t). The function M (t, λ) is the Weyl functionfor equation (4.2.3) with respect to the linear form U (y) := y(0) (see Remark 2.1.4). FromChapter 2 we know that the specification of the Weyl function uniquely determines thepotential q (x, t). Our goal is to obtain the evolution of the Weyl function M (t, λ) withrespect to t when q satisfies (4.2.1). For this purpose we consider two methods.

Method 1. Let us use the Lax equation (4.1.4), which is equivalent to the KdV equation.Let q (x, t) be a solution of the mixed problem (4.2.1)-(4.2.2). By Lemma 4.1.1, the functionΦ−AΦ is a solution of (4.2.3). Moreover, one can verify that Φ−AΦ = O(exp(iρx)). HenceΦ − AΦ = β Φ, where β = β (t, λ) does not depend on x. Taking x = 0, we calculateβ = −(AΦ)|x=0, and consequently

Φ = AΦ − (AΦ)|x=0Φ.

Differentiating this relation with respect to x and taking x = 0, we get

M = (A1Φ)|x=0 − M (AΦ)|x=0, (4.2.4)

where A1y := (Ay). If y is a solution of (4.2.3), then

Ay = −4y + 6qy + 3q y = F 11y + F 12y,

A1y =−

4y(4) + 6qy + 9q y + 3q y = F 21y + F 22y,

where F = F (x,t,λ) is defined by

F =

F 11 F 12

F 21 F 22

=

−q 4λ + 2q (4λ + 2q )(q − λ) − q q

. (4.2.5)

Hence

(AΦ)|x=0 = F 11(0, t , λ) + F 12(0, t , λ)M (t, λ), (A1Φ)|x=0 = F 21(0, t , λ)

−F 22(0, t , λ)M (t, λ).

Substituting this into (4.2.4), we obtain the nonlinear ordinary differential equation

M (t, λ) = F 21(0, t , λ) + (F 22(0, t , λ) − F 11(0, t , λ))M (t, λ) − F 12(0, t , λ)M 2(t, λ). (4.2.6)

The functions F jk (0, t , λ) depend on q j (t), j = 1, 2, 3, only, i.e. they are known functions.Equation (4.2.6) gives us the evolution of the Weyl function M (t, λ) with respect to t.This equation corresponds to the equations (4.1.6)-(4.1.7). But here the evolution equation(4.2.6) is nonlinear. This happens because of nonlinear reflection at the point x = 0. That

is the reason that the mixed problem on the half-space is more difficult than the Cauchyproblem.

Here is the algorithm for the solution of the mixed problem (4.2.1)-(4.2.2).

Algorithm 4.2.1. (1) Using the functions q 0, q 1, q 2 and q 3, construct the functions M (0, λ) and F jk (0, t , λ).

(2) Calculate M (t, λ) by solving equation (4.2.6) with the initial data M (0, λ).(3) Construct q (x, t) by solving the inverse problem.



210

Remark 4.2.1. Equation (4.2.6) is a (scalar) Riccati equation. Therefore it follows fromRadon’s lemma (see subsection 4.2.3 below) that the solution of (4.2.6) with given initialcondition M (0, λ) has the form

M (t, λ) =X 2(t, λ)

X 1(t, λ),

where X 1(t, λ) and X 2(t, λ) is the unique solution of the initial value problemX 1(t, λ)

X 2(t, λ)

= F (0, t , λ)

X 1(t, λ)X 2(t, λ)

,

X 1(0, λ)X 2(0, λ)

=

1

M (0, λ)

.

For convenience of the reader we present below in Subsection 4.2.3 a general version of Radon’s lemma.

Method 2. Instead of the Lax equation we will use here another representation of theKdV equation. Denote

G =

0 1

q − λ 0

, U = G − F + GF − F G,

where F is defined by (4.2.5). Then, it follows by elementary calculations that

U = 0 0u 0 ,

whereu = q t − 6qq x + q xxx.

Thus, the KdV equation is equivalent to the equation

U = 0. (4.2.7)

We shall see that Equation (4.2.7) is sometimes more convenient for the solution of the mixedproblem (4.2.1)-(4.2.2) than the Lax equation (4.1.4).

Let the matrix W (x,t,λ) and V (x,t,λ) be solutions of the Cauchy problems:

W = GW, W |x=0 = E, V = F V, V |t=0 = E, E =

1 00 1

, (4.2.8)

Denote by C (x,t,λ) and S (x,t,λ) the solutions of equation (4.2.3) under the initial con-ditions C

|x=0= S

|x=0= 1, S

|x=0= C

|x=0= 0. Clearly,

W =

C S C S

.

Lemma 4.2.1. Let q (x, t) be a solution of (4.2.1)-(4.2.2). Then

W (x,t,λ) = F (x,t,λ)W (x,t,λ) − W (x,t,λ)F (0, t , λ), (4.2.9)

W (x,t,λ) = V (x,t,λ)W (x, 0, λ)V −1

(0, t , λ). (4.2.10)



211

Proof. Using (4.2.8) we calculate

(W − F W ) − G(W − F W ) = UW.

Since U = 0, it follows that the matrix ξ = W −F W is a solution of the equation ξ = Gξ.

Moreover, ξ |x=0 = −F (0, t , λ). Hence, ξ (x,t,λ) = −W (x,t,λ)F (0, t , λ), i.e. (4.2.9) is valid.Denote Y = W (x,t,λ)V (0, t , λ), Z = V (x,t,λ)W (x, 0, λ). Then, using (4.2.8) and

(4.2.9), we calculate Y = F (x,t,λ)Y = Z, Y |t=0 = W (x, 0, λ) = Z |t=0. By virtue of theuniqueness of the solution of the Cauchy problem, we conclude that Y ≡ Z, i.e. (4.2.10) isvalid. 2

Consider the matrices

Ψ = Φ S

Φ S , N = 1 0

M 1 , N 1

:= N −1 = 1 0

−M 1 .

Since Φ = C + MS, we get Ψ = W N or W = ΨN 1. Then (4.2.9)-(4.2.10) take the form

Ψ(x,t,λ) = F (x,t,λ)Ψ(x,t,λ) − Ψ(x,t,λ)D0(t, λ), (4.2.11)

Ψ(x,t,λ) = V (x,t,λ)Ψ(x, 0, λ)B(t, λ), (4.2.12)

where

D0(t, λ) = (N 1(t, λ) + N 1(t, λ)F (0, t , λ))N (t, λ), D0 = d0

11 d012

d021 d0

22

, (4.2.13)

B(t, λ) = N 1(0, λ)V −1(0, t , λ)N (t, λ), B =

b11 b12

b21 b22

. (4.2.14)


˙B(t, λ) = −B(t, λ)D

0

(t, λ), B(0, λ) = E, (4.2.15)

d021(t, λ) ≡ 0, b21(t, λ) ≡ 0. (4.2.16)

Proof. Differentiating (4.2.12) with respect to t and using the equality V = F V, weget

Ψ = V (Ψ|t=0)B + V (Ψ|t=0)B = F Ψ + V (Ψ|t=0)B. (4.2.17)

On the other hand, it follows from (4.2.11) and (4.2.12) that

Ψ = F Ψ − V (Ψ|t=0)BD0.

Together with (4.2.17) this gives B = BD0. Moreover, according to (4.2.12), B(0, λ) = E.Rewriting (4.2.11) in coordinates, we obtain

Sd021 = −Φ + F 11Φ + F 12Φ − Φd0

11. (4.2.18)

Since

−˙

Φ + F 11Φ + F 12Φ − Φd

0

11 = O(exp(iρx)), x → ∞,



212

equality (4.2.18) is possible only if d021 = 0. Then (4.2.15) yields

b21 = −b21d011, b21 |t=0 = 0,

and consequently b21 ≡ 0. 2

Denote

R(t, λ) := V −1(0, t , λ), R =

R11 R12

R21 R22

.

Since R(V |x=0) = E, we calculate R(V |x=0) + R(V |x=0) = 0. Thus, the matrix R(t, λ) is asolution of the Cauchy problem

R(t, λ) = −R(t, λ)F (t, λ), R(0, λ) = E, (4.2.19)

where

F = F 11 F 12

F 21 F 22

= −q 2 4λ + 2q 1

(4λ + 2q 1)(q 1 − λ) − q 3 q 2

.

It is important that (4.2.19) is a linear system.


M (t, λ) = F 21(t, λ) + (F 22(t, λ) − F 11(t, λ))M (t, λ) − F 12(t, λ)M 2(t, λ), (4.2.20)

M (t, λ) =−

M 0(λ)R11(t, λ) − R21(t, λ)

M 0(λ)R12(t, λ) − R22(t, λ). (4.2.21)

where M 0(λ) = M (0, λ).

Proof. Rewriting (4.2.13) in coordinates, we get

d021(t, λ) = −M (t, λ) + F 21(0, t , λ) + (F 22(0, t , λ) − F 11(0, t , λ))M (t, λ) − F 12(0, t , λ)M 2(t, λ).

(4.2.22)By virtue of (4.2.16), d0

21 ≡ 0, i.e. (4.2.20) holds. Rewriting (4.2.14) in coordinates, wehave

b21(t, λ) = −M (0, λ)(R11(t, λ) + M (t, λ)R12(t, λ)) + (R21(t, λ) + M (t, λ)R22(t, λ)).

By virtue of (4.2.16), b21 ≡ 0, i.e. (4.2.21) holds. 2

Formula (4.2.21) gives us the solution of the nonlinear evolution equation (4.2.20). Thus,we obtain the following algorithm for the solution of the mixed problem (4.2.1)-(4.2.2) bythe inverse problem method.

Algorithm 4.2.2 (1) Using the functions q 0, q 1, q 2 and q 3, construct the functions M 0(λ) and F

jk(t, λ).

(2) Find R(t, λ) by solving the Cauchy problem (4.2.19).(3) Calculate M (t, λ) by (4.2.21).(4) Construct q (x, t) by solving the inverse problem (see Chapter 2).

4.2.2. In this subsection we study the existence of the solution of the mixed problem(4.2.1)-(4.2.2) and provide necessary and sufficient conditions for its solvability. The follow-ing theorem shows that the existence of the solution of (4.2.1)-(4.2.2) is equivalent to the

solvability of the corresponding inverse problem.



213

Theorem 4.2.1. Let for x ≥ 0, t ≥ 0 continuous complex-valued functions q 0(x), q 1(t),q 2(t), q 3(t) be given such that q 0(x) ∈ L(0, ∞), q 0(0) = q 1(0). Construct the function M (t, λ) according to steps (1)-(3) of Algorithm 4.2.2. Assume that there exists a function q (x, t) ∈ Q for which M (t, λ) is the Weyl function. Then q (x, t) is the solution of the mixed problem (4.2.1)-(4.2.2).

Thus, the problem (4.2.1)-(4.2.2) has a solution if and only if the solution of the corre-sponding inverse problem exists. Necessary and sufficient conditions for the solvability of the inverse problem for the Sturm-Liouville operator on the half-line from its Weyl functionand an algorithm for the solution are given in Chapter 2.

Proof. We will use the notations of Subsection 4.2.1. Since

(W − F W ) − G(W − F W ) = UW,

(W − F W )|x=0 = −F (0, t , λ),

we getW (x,t,λ) = F (x,t,λ)W (x,t,λ) − W (x,t,λ)χ(x,t,λ), (4.2.23)

whereχ(x,t,λ) = F (0, t , λ) −

x

0W −1(s,t,λ)U (s, t)W (s,t,λ) ds.

It follows from (4.2.23) and the relation W = ΨN 1 that

Ψ(x,t,λ) = F (x,t,λ)Ψ(x,t,λ) − Ψ(x,t,λ)D(x,t,λ), (4.2.24)

where

D(x,t,λ) = (N 1(t, λ) − N 1(t, λ)χ(x,t,λ))N (t, λ), D =

d11 d12

d21 d22

. (4.2.25)

Rewriting (4.2.24) and (4.2.25) in coordinates we obtain

d11(x,t,λ) = d011(t, λ) +

x

0u(s, t)Φ(s,t,λ)S (s,t,λ) ds, (4.2.26)

d21(x,t,λ) = d021(t, λ) −

x

0u(s, t)Φ2(s,t,λ) ds, (4.2.27)

Sd21 = −Φ + F 11Φ + F 12Φ − Φd11. (4.2.28)

Fix t ≥ 0 and λ (λ = ρ2, I m ρ > 0). It follows from (4.2.26) that d11 = O(1), x → ∞.From this and (4.2.28) we get d21 = O(exp(2iρx)), x → ∞, and consequently d21 → 0 as

x → ∞. Then, by virtue of (4.2.27),

d021(t, λ) =

∞0

u(x, t)Φ2(x,t,λ) dx.


M (t, λ) = F 21(0, t , λ) + (F 22(0, t , λ) − F 11(0, t , λ))M (t, λ) − F 12(0, t , λ)M 2(t, λ)

− ∞0 u(x, t)Φ

2

(x,t,λ) dx. (4.2.29)



214


M (0, λ) = M 0(λ). (4.2.30)

Since the specification of the Weyl function uniquely determines the potential we have from

(4.2.30) thatq (x, 0) = q 0(x). (4.2.31)

Differentiating (4.2.21) with respect to t we arrive at (4.2.20). Comparing (4.2.20) with(4.2.29) we obtain

F 21(t, λ) + (F 22(t, λ) − F 11(t, λ))M (t, λ) − F 12(t, λ)M 2(t, λ) − ∞

0u(x, t)Φ2(x,t,λ) dx,

where F jk (t, λ) = F jk (0, t , λ)−

F jk (t, λ). Hence

−

q xx(0, t) − q 3(t)

+ 2λ

q (0, t) − q 1(t)

+ 2

q 2(0, t) − q 21(t)

+ 2

q x(0, t) − q 2(t)

M (t, λ)

−2

q (0, t) − q 1(t)

M 2(t, λ) − ∞

0u(x, t)Φ2(x,t,λ) dx = 0. (4.2.32)

Since

Φ(x,t,λ) =e(x,t,ρ)

e(0, t , ρ), M (t, λ) =

e(0, t , ρ)

e(0, t , ρ),

we have for a fixed t,

M (t, λ) = (iρ) + O1

ρ

,

∞0

u(x, t)Φ2(x,t,λ) dx = O1

ρ

, |ρ| → ∞.

Then, (4.2.32) implies

q (0, t) = q 1(t), q x(0, t) = q 2(t), q xx(0, t) = q 3(t), (4.2.33)

∞0

u(x, t)Φ2(x,t,λ) dx = 0. (4.2.34)

In particular, (4.2.34) yields ∞0

u(x, t)e2(x,t,ρ) dx = 0. (4.2.35)


e(x,t,ρ) = exp(iρx) + ∞x A(x,t,ξ )exp(iρξ ) dξ,

and consequently,

e2(x,t,ρ) = exp(2iρx) + ∞

xE (x,t,ξ ) exp(2iρξ ) dξ, (4.2.36)

where

E (x,t,ξ ) = 4A(x,t, 2ξ − x) + 2ξ−x

xA(x, s)A(x, 2ξ − s) ds.



215

Substituting (4.2.36) into (4.2.35) we get ∞0

u(x, t) +

x

0E (ξ,t,x)u(ξ, t) dξ

exp(2iρx) dx = 0.

Hence, u(x, t) = 0, i.e. the function q (x, t) satisfies the KdV equation (4.2.1). Together

with (4.2.31) and (4.2.33) this yields that q (x, t) is a solution of the mixed problem (4.2.1)-(4.2.2). Theorem 4.2.1 is proved. 2

4.2.3. In this subsection we formulate and prove Radon’s lemma for the matrix Riccatiequation.

Let us consider the following Cauchy problem for the Riccati equation:

Z = Q21(t) + Q22(t)Z − ZQ11(t) − ZQ12(t)Z, t ∈ [t0, t1],

Z (t0) = Z 0, (4.2.37)

where Z (t), Z 0, Q11(t), Q12(t), Q21(t) and Q22(t) are complex matrix of dimensions m ×n, m × n, n × n, n × m, m × n, and m × m respectively, and Q jk (t) ∈ C [t0, t1].

Let the matrices (X (t), Y (t), t ∈ [t0, t1]), be the unique solution of the linear Cauchyproblem

X = Q11(t)X + Q12(t)Y, X (t0) = E,

Y = Q21(t)X + Q22(t)Y, Y (t0) = Z 0,

(4.2.38)

where E is the identity n × n matrix.

Radon’s Lemma.The Cauchy problem (4.2.37) has a unique solution Z (t) ∈ C 1[t0, t1],if and only if det X (t) = 0 for all t ∈ [t0, t1]. Moreover,

Z (t) = Y (t)(X (t))−1. (4.2.39)

Proof. 1) Let det X (t) = 0 for all t ∈ [t0, t1]. Define the matrix Z (t) via (4.2.39).

Then, using (4.2.38), we get

Z (t0) = Y (t0)(X (t0))−1 = Z 0,

Z = Y X −1 − Y X −1XX −1 = (Q21X + Q22Y )X −1 − Y X −1(Q11X + Q12Y )X −1

= Q21 + Q22Z − ZQ11 − ZQ12Z.

Thus, Z (t) is the solution of (4.2.37).2) Let Z (t)

∈C 1[t0, t1] be a solution of (4.2.37). Denote by X (t) the solution of the

Cauchy problem ˙X = (Q11(t) + Q12(t)Z (t))X, X (t0) = E. (4.2.40)

Clearly, det X (t) = 0 for all t ∈ [t0, t1]. Take

Y (t) := Z (t)X (t).

Then, it follows from (4.2.37) and (4.2.40) that

˙Y = Z X + Z

˙X = Q21X + Q22Y ,



216

˙X = Q11X + Q12Y ,

and X (t0) = E, Y (t0) = Z 0. By virtue of the uniqueness theorem, this yields

X (t) ≡ X (t), Y (t) ≡ Y (t),

and Radon’s lemma is proved. 2

4.3. CONSTRUCTING PARAMETERS OF A MEDIUM FROMINCOMPLETE SPECTRAL INFORMATION

In this section the inverse problem of synthesizing parameters of differential equationswith singularities from incomplete spectral information. We establish properties of the spec-

tral characteristics, obtain conditions for the solvability of such classes of inverse problemsand provide algorithms for constructing the solution.

4.3.1. Properties of spectral characteristics.

Let us consider the system

dy1

dx= iρR(x) y2,

dy2

dx= iρ

1

R (x)y1, x ∈ [0, T ] (4.3.1)

with the initial conditionsy1(0, ρ) = 1, y2(0, ρ) = −1.

Here ρ = k + iτ is the spectral parameter, and R(x) is a real function.For a wide class of problems describing the propagation of electromagnetic waves in a

stratified medium, Maxwell’s equations can be reduced to the canonical form (4.3.1), wherex is the variable in the direction of stratification, y1 and y2 are the components of theelectromagnetic field, R(x) is the wave resistance which describes the refractive properties

of the medium and ρ is the wave number in a vacuum. System (4.3.1) also appears in radioengineering for the design of directional couplers for heterogeneous electronic lines, whichconstitute one of the important classes of radiophysical synthesis problems [mes1].

Some aspects of synthesis problems for system (4.3.1) with a positive R(x) were studiedin [lit1], [mes1], [tik1]-[tik3], [zuy1] and other works. In this section we partially use theseresults. We study the inverse problem for system (4.3.1) from incomplete spectral infor-mation in the case when R(x) is nonnegative and can have zeros which are called turningpoints. More precisely, we shall consider two classes of functions R(x) . We shall say thatR(x)

∈B0 if R(x), R(x) are absolutely continuous on [0, T ], R(x)

∈L2(0, T ), R(x) >

0, R(0) = 1, R(0) = 0. We also consider the more general case when R(x) has zeros0 < x1 < . . . < x p < T, p ≥ 0 inside the interval (0, T ). We shall say that R(x) ∈ B+

0 if R(x) has the form

1

R(x)=

p j=1

R j

(x − x j )2+ R0(x), 0 < x1 < . . . < x p < T, R j > 0, 0 ≤ j ≤ p,

and R0(x), R0(x) are absolutely continuous on [0, T ], R

0(x) ∈ L2(0, T ), R(x) > 0 (x =x j), R(0) = 1, R(0) = 0. In particular, if here p = 0 then R(x) ∈ B0.



217

As the main spectral characteristics we introduce the amplitude reflection coefficient

r(ρ) =y1(T, ρ) + R0y2(T, ρ)

y1(T, ρ) − R0y2(T, ρ), R0 := R(T );

the power reflection coefficientσ(k) = |r(k)|2, k := Re ρ;

the transmission coefficients

f 1(ρ) =y1(T, ρ) − R0y2(T, ρ)

2√

R0, f 2(ρ) =

y1(T, ρ) + R0y2(T, ρ)

2√

R0

and the characteristic function

∆(ρ) =1√ R0

y1(T, ρ).

Clearly,r(ρ) = f 2(ρ)/f 1(ρ), (4.3.2)

∆(ρ) = f 1(ρ) + f 2(ρ). (4.3.3)

In many cases of practical interest, the phase is difficult or impossible to measure, while

the amplitude is easily accessible to measurement. Such cases lead us to the so-called in-complete inverse problems where only a part of the spectral information is available. Inthis paper we study one of the incomplete inverse problems, namely, the inverse problem of recovering the wave resistance from the power reflection coefficient:

Inverse Problem 4.3.1. Given σ(k) , construct R(x).

The lack of spectral information leads here to nonuniqueness of the solution of the inverseproblem. Let us briefly describe a scheme of the solution of Inverse Problem 4.3.1.

Denote α j (k) =

|f j(k)

|, j = 1, 2. Since α2

1(k)

−α2

2(k)

≡1 (see (4.3.19) below), we get

in view of (4.3.2),σ(k) = 1 − (α1(k))−2, 0 ≤ σ(k) < 1.

Firstly, from the given power reflection coefficient σ(k) we construct α j(k). Then, usinganalytic properties of the transmission coefficients f j (ρ) and, in particular, informationabout their zeros, we reconstruct the transmission coefficients from their amplitudes. Namelyon this stage we are faced with nonuniqueness. Problems concerning the reconstruction of analytic functions from their moduli often appear in applications and have been studied inmany works (see [hoe1], [hur1] and references therein). The last of our steps is to calculate

the characteristic function ∆(ρ) by (4.3.3) and to solve the inverse problem of recoveringR(x) from ∆(ρ) . Here we use the Gelfand-Levitan-Marchenko method (see [mar1], [lev2]and Chapters 1-2 of this book).

To realize this scheme, in Subsection 4.3.1 we study properties of the spectral character-istics. In Subsection 4.3.2 we solve the synthesis problem for R(x) from the characteristicfunction ∆(ρ) , obtain necessary and sufficient conditions for its solvability and providethree algorithms for constructing the solution. In Subsection 4.3.3 we study the problem of recovering the transmission coefficients from their moduli. In Subsection 4.3.4, the so-called

symmetrical case is considered, and in Subsection 4.3.5 we provide an algorithm for the



218

solution of Inverse Problem 1. We note that inverse problems for Sturm-Liouville equationswith turning points have been studied in [FY1] and [FY3]. Some aspects of the turning pointtheory and a number of its applications are described in [ebe4], [mch1] and [was1].

We shall say that R(x) ∈ B−0 if (R(x))−1 ∈ B+

0 . An investigation of the classes B+0

and B−0 is completely similar because the replacement R

→1/R is equivalent to the re-

placement (y1, y2) → (−y2, −y1). Below it will be more convenient for us to consider thecase R(x) ∈ B−

0 .

We transform (4.3.1) by means of the substitution

y1(x, ρ) =

R(x)u(x, ρ), y2(x, ρ) =1

R(x)

v(x, ρ) (4.3.4)

to the systemu + h(x)u = iρv, v − h(x)v = iρu, x ∈ [0, T ] (4.3.5)

with the initial conditions u(0, ρ) = 1, v(0, ρ) = −1, where

h(x) =R(x)

2R(x). (4.3.6)

Hence the function u(x, ρ) satisfies the equation

−u + q (x)u = λu, λ = ρ2 (4.3.7)

and the initial conditionsu(0, ρ) = 1, u(0, ρ) = −iρ,

whereq (x) = h2(x) − h(x) (4.3.8)

or

q (x) =3

4

R(x)

R(x)

2

− 1

2

R(x)

R(x).

Similarly,−v + g(x)v = λv, v(0, ρ) = −1, v(0, ρ) = iρ,

where g(x) = h2(x) + h(x).In view of (4.3.4), the transmission coefficients, the amplitude reflection coefficient and

the characteristic function take the form

f 1(ρ) =u(T, ρ) − v(T, ρ)

2, f 2(ρ) =

u(T, ρ) + v(T, ρ)

2,

r(ρ) =u(T, ρ) + v(T, ρ)

u(T, ρ) − v(T, ρ)=

f 2(ρ)

f 1(ρ), ∆(ρ) = u(T, ρ).

(4.3.9)

Since y1(x, 0) ≡ 1 , we have according to (4.3.4),

R(x)u(x, 0) ≡ 1. (4.3.10)



219

Lemma 4.3.1. R(x) ∈ B−0 if and only if q (x) ∈ L2(0, T ), ∆(0) = 0.

Proof. 1) Let R(x) ∈ B−0 , i.e.

R(x) = p

j=1

R j

(x−

x j)2+ R0(x), 0 < x1 < .. . < x p < T, R j > 0,

R0(x) ∈ W 22 (0, T ), R(x) > 0 (x = x j), R(0) = 1, R(0) = 0.

(4.3.11)

Using (4.3.6) and (4.3.11) we get for x → x j,

h(x) ∼ − 1

x − x j+ h∗ j (x), h∗ j (x) ∈ W 12 , h∗ j (x j) = 0,

and consequently q (x)

∈L2(0, T ). If follows from (4.3.10) that u(T, 0)

= 0, i.e. ∆(0)

= 0 .

2) Let now q (x) ∈ L2(0, T ), u(T, 0) = 0, and let 0 < x1 < . . . < x p < T be zeros of u(x, 0). It follows from (4.3.10) that

R(x) = (u(x, 0))−2.

Hence R(x) > 0 (x = x j), R(0) = 1, R(0) = 0, and

R(x) ∼ R j

(x − x j)2, x → x j, R j > 0.

Denote

R0(x) := R(x) − p

j=1

R j

(x − x j)2.

It is easy to see that R0(x) ∈ W 22 [0, T ]. 2

Remark 4.3.1. If R(x) ∈ B−0 , then the functions y1(x, ρ) and v(x, ρ) have singulari-

ties of order 1 at x = x j, and there exist finite limits limx→xj

(x−x j)v(x, ρ), limx→xj

(x−x j )y1(x, ρ).

In the other words, we continue solutions in the neighbourhoods of the singular points withthe help of generalized Bessel type solutions (see, [yur28]). It follows from (4.3.10) that

limx→xj

(x − x j )

R(x) = R j.

Example 4.3.1. Let R(x) =1

cos2 x. Then h(x) = tan x, q (x) = −1, u(x, 0) =

1 R(x)

= cos x, u(x, ρ) = cos µx−iρsin µx

µ, µ2 = ρ2+1. If T ∈

(2 p − 1)π

2,

(2 p + 1)π

2

,

then u(x, 0) has p zeros x j =(2 j

−1)π

2 , j = 1, p.

In the sequel, we denote by AC [a, b] the set of absolutely continuous functions on thesegment [a, b].

Theorem 4.3.1. Let R(x) ∈ B−0 . Then

(i) The characteristic function ∆(ρ) is entire in ρ, and the following representation holds

∆(ρ) = e−iρT + T

−T η(t)e−iρt dt, η(t)

∈AC [

−T, T ], η(t)

∈L

2(−

T, T ), η(−

T ) = 0, (4.3.12)



220

where η(t) is a real function, and

η(T ) =1

2

T

0q (t) dt = −h(T )

2+

1

2

T

0h2(t) dt.

(ii) For real ρ, δ (ρ) has no zeros. For I m ρ > 0, ∆(ρ) has at most a finite number of simple zeros of the form ρ j = iτ j, τ j > 0, j = 1,m, m ≥ 0.(iii) The function δ (ρ) has no zeros for Im ρ ≥ 0 if and only if R(x) ∈ B0.

Proof. 1) It is well-known (see [mar1] and Section 1.3 of this book) that u(x, ρ) hasthe form

u(x, ρ) = e−iρx + x

−xK (x, t)e−iρt dt,

where K (x, t) is a real, absolutely continuous function, K t(T, t) ∈ L2(−T, T ), K (x, −x) =

0, K (x, x) =1

2

x 0 q (t) dt. Moreover,

u(x, ρ) = u(x, −ρ), (4.3.13)

u(x, ρ), u(x, −ρ) ≡ 2iρ, (4.3.14)

where y, z := yz − yz. Hence we arrive at (4.3.12), where η(t) = K (T, t).2) If follows from (4.3.13) that for real ρ, ∆(ρ) = ∆(−ρ), and consequently, in view

of (4.3.14), for real ρ

= 0, ∆(ρ) has no zeros. By virtue of Lemma 4.3.1, ∆(0)

= 0.

Furthermore, denote Ω+ = ρ : Im ρ > 0. Let ρ = k + iτ ∈ Ω+ be a zero of ∆(ρ). Since

−u + q (x)u = ρ2u, −u + q (x)u = ρ2u,

we get

(ρ2 − ρ2) T

0|u|2dx =

T

0u, u = −i(ρ + ρ),

which is possible only if k := Re ρ = 0. Hence, all zeros of ∆(ρ) i n Ω+ are pure imaginary,

and by virtue of (4.3.12), the number of zeros in Ω+ is finite.3) Let us show that in Ω+ all zeros of ∆(ρ) are simple. Suppose that for a certain

ρ = iτ,τ > 0, ∆(ρ) = ∆(ρ) = 0, where ∆(ρ) =d

dρ∆(ρ) . Denote u1 = u . Then, by

virtue of (4.3.7),

−u1 + q (x)u1 = ρ2u1 + 2ρu, u1(0, ρ) = 0, u1(0, ρ) = −i.

Consequently,

2ρ T

0u2(x, ρ) dx =

T

0u(x, ρ)(−u1(x, ρ) + q (x)u1(x, ρ) − ρ2u1(x, ρ)) dx

= −T

0u, u1 = −i,

i.e.2τ

τ

0|u(x, ρ)|2 dx = −1,

which is impossible.



221

4) If R(x) ∈ B0 , then in view of (4.3.10) u(x, 0) > 0, x ∈ [0, T ]. Since u(0, iτ ) =1, τ ≥ 0 and u(x,iτ ) > 0, x ∈ [0, T ] for large τ , we get u(T,iτ ) > 0 for all τ ≥ 0 , i.e.∆(ρ) has no zeros in Ω+ . The inverse assertion is proved similarly. 2

In the analogous manner one can prove the following theorem.

Theorem 4.3.2.Let R(x) ∈ B−0 .(i) The functions f 1(ρ), f 2(ρ) are entire in ρ , and have the form

f 1(ρ) = e−iρT + T

−T g1(t)e−iρt dt, g1(t) ∈ AC [−T, T ], g1(t) ∈ L2(−T, T ), g1(−T ) = 0,

(4.3.15)

f 2(ρ) = T

−T g2(t)e−iρt dt, g2(t) ∈ AC [−T, T ], g2(t) ∈ L2(−T, T ), g2(−T ) = 0, (4.3.16)

where g j(t) are real, and

g1(T ) =1

2

T

0h2(t), g2(T ) = −h

2, h := h(T ). (4.3.17)

(ii)f 1(ρ)f 1(−ρ) − f 2(ρ)f 2(−ρ) ≡ 1. (4.3.18)

(iii) For real ρ, f 1(ρ) has no zeros. In Ω+, f 1(ρ) has a finite number of simple zeros of the form ρ∗ j = iτ ∗ j , τ ∗ j > 0, j = 1, m∗, m∗ ≥ 0 .

(iv) If R(x) ∈ B0 , then f 1(ρ) has no zeros in Ω+, i.e. m∗ = 0 .

We note that (4.3.18) follows from (4.3.14). Since f j(ρ) = f j(−ρ) , we obtain from(4.3.18) that

α21(k) − α2

2(k) ≡ 1. (4.3.19)


α21(k) = 1 +

h2

4k2

+ω(k)

k2

, α22(k) =

h2

4k2

+ω(k)

k2

,

σ(k) =h2

4k2 + h2+

ω1(k)

k2, |k| → ∞,

(4.3.20)

where ω(k), ω1(k) ∈ L2(−∞, ∞).

Lemma 4.3.2. Denote g∗ j (t) = g j(t − T ) . Then

g∗1(ξ ) +

ξ

0g∗1(t)g∗1(t + 2T − ξ ) dt =

ξ

0g∗2(t)g∗2(t + 2T − ξ ) dt, ξ ∈ [0, 2T ]. (4.3.21)

Proof. Indeed, using (4.3.15) and (4.3.16), we calculate

f 1(ρ)f 1(−ρ) = 1 + 2T

−2T G1(ξ )e−iρξ dξ, f 2(ρ)f 2(−ρ) =

2T

−2T G2(ξ )e−iρξ dξ, (4.3.22)

where G j(−ξ ) = G j(ξ ),

G1(ξ ) = g1(T −

ξ ) + T

ξ−T g1(t)g1(t

−ξ )dt, G2(ξ ) =

ξ

ξ−T g2(t)g2(t

−ξ )dt, ξ > 0.



222

Substituting (4.3.22) into (4.3.18) we arrive at (4.3.21). 2

Denote

p(x) =

q (T − x) for 0 ≤ x ≤ T,

0 for x > T,

and consider the equation −y + p(x)y = λy, x > 0 (4.3.23)

on the half-line. Let e(x, ρ) be the Jost solution of (4.3.23) such that e(x, ρ) ≡ eiρx forx ≥ T . Denote δ (ρ) = e(0, ρ) . Clearly,

e(x, ρ) = eiρT u(T − x, ρ), 0 ≤ x ≤ T, δ (ρ) = eiρT ∆(ρ).

Then

e(x, ρ) = eiρx + 2T −x

xG(x, t)eiρt dt, 0 ≤ x ≤ T, e(x, ρ) ≡ eiρx, x ≥ T, (4.3.24)

δ (ρ) = 1 + 2T

0θ(t)eiρt dt, θ(t) ∈ AC [0, 2T ], θ(t) ∈ L2(0, 2T ), θ(2T ) = 0, (4.3.25)

where G(x, t) = K (T − x, T − t), θ(t) = η(T − t) = G(0, t), G(x, 2T − x) = 0,

θ(0) =1

2

T

0 p(t) dt, G(x, x) =

1

2

T

x p(ξ ) dξ.

Denoteα j = (

∞0

e2(x, ρ j)dx)−1 > 0, ρ j = iτ j, τ j > 0, j = 1, m.

Lemma 4.3.3.

α j =iδ (−ρ j)

δ (ρ j), (4.3.26)

where δ (ρ) =d

dρ

δ (ρ).

Proof. Since −e + p(x)e = ρ2e and δ (ρ j) = 0 , we have

(ρ2 − ρ2 j ) ∞

0e(x, ρ)e(x, ρ j)dx =

∞0

e(x, ρ), e(x, ρ j) = −e(0, ρ j)δ (ρ).

If yields

α j = − 2ρ j

e(0, ρ j)δ (ρ j). (4.3.27)

Since e(x, ρ), e(x, −ρ) ≡ 2iρ , we get

δ (−ρ j)e(0, ρ j) = 2iρ j. (4.3.28)

Together with (4.3.27) it gives (4.3.26). 2

4.3.2. Synthesis of R(x) from the characteristic function ∆(ρ).

In this subsection we study the inverse problem of recovering the wave resistance R(x)

from the given ∆(ρ) in the class B−0 . This inverse problem has a unique solution. We



223

provide physical realization conditions, i.e. necessary and sufficient conditions on a function∆(ρ) to be the characteristic function for a certain R(x) ∈ B−

0 . We also obtain threealgorithms for the solution of the inverse problem.

Denote s(ρ) =δ (−ρ)

δ (ρ). For real ρ, s(ρ) is continuous and, by virtue of (4.3.25),

s(ρ) = 1 + O( 1ρ

), |ρ| → ∞ . Consequently, 1 − s(ρ) ∈ L2(−∞, ∞) . We introduce theFourier transform

F 0(t) =1

2π

∞−∞

(1 − s(ρ))eiρt dρ, (4.3.29)

1 − s(ρ) = ∞−∞

F 0(t)e−iρt dt. (4.3.30)

Substituting (4.3.25) and (4.3.30) into the relation δ (ρ) s(ρ) = δ (−ρ) , we obtain the con-

nection between F 0(t) and θ(t) :

θ(t) + F 0(t) + 2T

0θ(s)F 0(t + s) ds = 0, t > 0, (4.3.31)

where θ(t) ≡ 0 for t > 2T . Further, we consider the function

F (t) = F 0(t) + F 1(t), (4.3.32)

where

F 1(t) =m

j=1

α jeiρjt. (4.3.33)

Since

F 1(t) + 2T

0θ(s)F 1(t + s)ds =

m j=1

α jeiρjtδ (ρ j ) = 0,

we get by virtue of (4.3.31) that

θ(t) + F (t) + 2T

0θ(s)F (t + s) ds = 0, t > 0. (4.3.34)

Calculating the integral (4.3.29) by Jordan’s lemma for t > 2T and using Lemma 4.3.3we obtain

F 0(t) = im

j=1

Resρ=ρj

(1 − s(ρ))eiρt = −im

j=1

δ (−ρ j)

δ (ρ j )eiρjt = −

m j=1

α jeiρjt = −F 1(t).

Thus,F (t) ≡ 0 , t > 2T. (4.3.35)

Taking (4.3.35) into account we rewrite (4.3.34) as follows

θ(t) + F (t) + 2T

tθ(s − t)F (s) ds = 0, t ∈ (0, 2T ). (4.3.36)

In particular it yields F (t) ∈ AC [0, 2T ], F (t) ∈ L2(0, 2T ), F (2T ) = 0 .We note that the function F (t) can be constructed from equation (4.3.36) which is

usually better than to calculate F (t) directly by (4.3.32), (4.3.33), (4.3.29).



224

Acting in the same way as in [mar1] one can obtain that

G(x, t) + F (x, t) + 2T −t

xF (t + s)G(x, s) ds = 0, 0 ≤ x ≤ T, x < t < 2T − x. (4.3.37)

Equation (4.3.37) is called the Gelfand-Levitan-Marchenko equation.Now let us formulate physical realization conditions for the characteristic function ∆(ρ) .

Theorem 4.3.3. For a function ∆(ρ) of the form (4.3.12) to be the characteristic function for a certain R(x) ∈ B−

0 , it is necessary ans sufficient that all zeros of ∆(ρ) in Ω+ are simple, have the form ρ j = iτ j, τ j > 0, j = 1, m , m ≥ 0 , and

α j :=iδ (−ρ j )

δ (ρ j)> 0.

R(x) ∈ B0 if and only if ∆(ρ) has no zeros in Ω+ , i.e. m = 0 . The specification of the characteristic function ∆(ρ) uniquely determines R(x).

Proof. The necessity part of Theorem 4.3.3 was proved above. We prove the sufficiency.Put δ (ρ) = eiρT ∆(ρ) . Then (4.3.25) holds, where θ(t) = η(T − t) . Since for each fixedx ≥ 0 , the homogeneous integral equation

w(t) + ∞

xF (t + s)w(s) ds = 0, t > x (4.3.38)

has only the trivial solution (see [mar1]), then equation (4.3.37) has a unique solution G(x, t).

The function G(x, t) is absolutely continuous, G(x, 2T −x) = 0 andd

dxG(x, x) ∈ L2(0, T ) .

We construct the function e(x, ρ) by (4.3.24).Let us show that e(0, ρ) = δ (ρ) . Indeed, it follows from (4.3.24) and (4.3.25) that

e(0, ρ) − δ (ρ) = 2T

0(G(0, t) − θ(t))eiρt dt. (4.3.39)

By virtue of (4.3.36) and (4.3.37),

(G(0, t) − θ(t)) + 2T −t

0(G(0, s) − θ(s))F (t + s) ds = 0.

In view of (4.3.35), it gives us that the function w(t) := G(0, t) − θ(t) satisfies (4.3.38) forx = 0 . Hence G(0, t) = θ(t) , and according to (4.3.39), e(0, ρ) = δ (ρ) .

Put p(x) := −2dG(x, x)

dx, x ∈ [0, T ] , and p(x) ≡ 0, x > T . It is easily shown that

−e(x, ρ) + p(x)e(x, ρ) = ρ2e(x, ρ), x > 0.

We construct R(x) by the formula

R(x) =1

u2(x, 0), (4.3.40)

where u(x, ρ) = e−iρT e(T − x, ρ) . Since δ (0) = 0 , we have u(T, 0) = 0 . It follows from

(4.3.40) that R(x) ∈ B−0 , where 0 < x1 < . . . < x p < T are zeros of u(x, 0) . In particular,



225

if δ (ρ) has no zeros in Ω+ , then, by virtue of Theorem 1, R(x) ∈ B0 . The uniqueness of recovering R(x) from ∆(ρ) was proved for example in [yur5]. 2

Theorem 4.3.3 gives us the following algorithm for constructing R(x) from the charac-teristic function ∆(ρ) :

Algorithm 4.3.1. Let a function ∆(ρ) satisfying the hypothesis of Theorem 4.3.3 begiven. Then(1) Construct F (t), t ∈ (0, 2T ) from the integral equation (4.3.36), where θ(t) = η(T − t) ,or directly from (4.3.26), (4.3.29), (4.3.32), (4.3.33).(2) Find G(x, t) from the integral equation (4.3.37).(3) Calculate R(x) by

R(x) =1

e2(T − x)

, e(x) = 1 + 2T −x

x

G(x, t) dt. (4.3.41)

Next we provide two other algorithms for the synthesis of the wave resistance from thecharacteristic function, which sometimes may give advantage from the numerical point of view.

Let S (x, λ) be the solution of (4.3.23) under the conditions S (0, λ) = 0, S (0, λ) = 1.Then

S (x, λ) =sin ρx

ρ

+ x

0

Q(x, t)sin ρt

ρ

dt, (4.3.42)

where Q(x, t) is a real, absolutely continuous function, Q(x, 0) = 0, Q(x, x) =1

2

x

0 p(t)dt .

Since e(x, ρ), S (x, λ) ≡ δ (ρ) and δ (ρ j) = 0 we get

e(x, ρ j) = e(0, ρ j)S (x, λ j ), λ j = ρ2 j , (4.3.43)

and by virtue of (4.3.27) and (4.3.28),

β j := ∞

0S 2(x, λ j)dx

−1 = −2ρ je(0, ρ j)δ (ρ j)

= − 4iρ2

j

δ (−ρ j)δ (ρ j)> 0. (4.3.44)

Taking into account the relations

δ (ρ j) = eiρjT ∆(ρ j), δ (−ρ j) = e−iρjT ∆(−ρ j),

we obtain from (4.3.44)

β j = −4iρ2

j

∆(−ρ j )∆(ρ j) .

Consider the function

a(x) =m

j=1

β jcos ρ jx

ρ2 j

+2

π

∞0

cos ρx

1

|∆(ρ)|2− 1

dρ. (4.3.45)


ρ1

|∆(ρ)|2 −1 ∈

L2(0,∞

),



226

and consequently a(x) is absolutely continuous, and a(x) ∈ L2 . The kernel Q(x, t) from(4.3.42) satisfies the equation (see [lev2])

f (x, t) + Q(x, t) + x

0Q(x, s)f (s, t)ds = 0, x ≥ 0, 0 < t < x, (4.3.46)

wheref (x, t) =

1

2(a(x − t) − a(x + t)).

The function R(x) can be constructed by the following algorithm:

Algorithm 4.3.2. Let ∆(ρ) be given. Then(1) Construct a(x) by (4.3.45).(2) Find Q(x, t) from the integral equation (4.3.46).

(3) Calculate p(x) := 2d

dxQ(x, x) .

(4) Construct

R(x) :=1

e2(T − x),

where e(x) is the solution of the Cauchy problem y = p(x)y, y(T ) = 1, y(T ) = 0 .

Remark 4.3.2. We also can construct R(x) using (4.3.6) and (4.3.8), namely:

R(x) = exp(2

x

0h(t)dt),

where h(x) is the solution of the equation

h(x) = x

0h2(t)dt + 2Q(T − x, T − x) − 2Q(T, T ).

Remark 4.3.3. If R(x) ∈ B0 , then

a(x) =2

π ∞

0cos ρx

1

|∆(ρ)|2

−1 dρ,

and for constructing the solution of the inverse problem it is sufficient to specify |∆(k)| fork ≥ 0 .

Now we provide an algorithm for the solution of the inverse problem which uses discretespectral characteristics.

Let X k(x, λ), k = 1, 2 be solutions of (4.3.7) under the conditions X 1(0, λ) = X 2(0, λ) =1, X 1(0, λ) = X 2(0, λ) = 0 . Denote ∆k(λ) = X k(T, λ) . Clearly,

u(x, ρ) = X 1(x, λ) − iρX 2(x, λ), ∆(ρ) = ∆1(λ) − iρ ∆2(Λ),

and consequently

∆1(λ) =∆(ρ) + ∆(−ρ)

2, ∆2(λ) =

∆(ρ) − ∆(−ρ)

2iρ.

Let µnn≥1 be the zeros of ∆2(λ) , and

γ n := T

0 X

2

2 (x, µn)dx > 0.



227

It is easily shown that

γ n = (∆1(µn))−1(d

dλ∆2(λ))|λ=µn ,

and

√ µn = πnT

+ 12πn

T

0q (ξ )dξ + ωn

n, γ n = T

3

2n2π2 + ωn1n3 , ωn, ωn1 ∈ 2. (4.3.47)

We introduce the function

A(x) =2

T

∞n=1

T cos

√ µnx

2γ nµn

− cosπnx

T

. (4.3.48)

By virtue of (4.3.47), A(x) is absolutely continuous, and A(x) ∈ L2. It is known (see[mar1]) that

X 2(x, λ) =sin ρx

ρ+ x

0D(x, t)

sin ρtρ

dt,

where D(x, t) is a real, absolutely continuous function, and D(x, 0 ) = 0, D(x, x) =12

x 0

q (t)dt . The function D(x, t) satisfies the equation (see [lev2])

D(x, t) + B(x, t) + x

0D(x, s)B(s, t)ds = 0, 0 ≤ x ≤ T , 0 < t < x, (4.3.49)

where B(x, t) = 12

(A(x − t) − A(x + t)) . The function R(x) can be constructed from the

discrete data µn, γ nn≥1 by the the following algorithm.

Algorithm 4.3.3. Let µn, γ nn≥1 be given. Then(1) Construct A(x) by (4.3.48).(2) Find D(x, t) from the integral equation (4.3.49).

(3) Calculate q (x) := 2dD(x, x)

dx.

(4) Construct R(x) := 1u2(x)

, where u(x) is the solution of the Cauchy problem

u = q (x)u, u(0) = 1, u(0) = 0,

or by

R(x) = exp(2 x

0h(t)dt), h(x) =

x

0h2(t)dt − 2D(x, x).

4.3.3. Reconstruction of the transmission coefficients from their moduli.

Lemma 4.3.4.Suppose that a function γ (ρ) is regular in Ω+ , has no zeros in Ω+ ,and for |ρ| → ∞, ρ ∈ Ω+, γ (ρ) = 1 + O( 1

ρ) . Let γ (k) = |γ (k)|e−iβ (k), k = Re ρ . Then

β (k) =1

π

∞−∞

ln |γ (ξ )|ξ − k

dξ. (4.3.50)

In (4.3.50) (and everywhere below, where necessary) the integral is understood in the principal

value sense.



228

Proof. First we suppose that γ (k) = 0 for real k . By Cauchy’s theorem, taking intoaccount the hypothesis of the lemma, we obtain

1

2πi

C r,

ln γ (ξ )

ξ

−k

dξ = 0, (4.3.51)

where C r, is the closed contour (with counterclockwise circuit) consisting of the semicirclesC r = ξ : ξ = reiϕ, ϕ ∈ [0, π], Γ = ξ : ξ − k = eiϕ, ϕ ∈ [0, π] and the intervalsξ ∈ [−r, r] \ [k − , k + ] of the real axis. Since

lim→0

1

2πi

Γ

ln γ (ξ )

ξ − kdξ = −1

2ln γ (k),

limr→∞

1

2πi C r

ln γ (ξ )

ξ − kdξ = 0,

we get from (4.3.51) that

ln γ (k) =1

πi

∞−∞

ln γ (ξ )

ξ − kdξ.

Separating here real and imaginary parts, we arrive at (4.3.50).Suppose now that for real ρ , the function γ (ρ) has one zero ρ0 = 0 of multiplicity s

(the general case is treated in the same way). Denote

γ (ρ) = γ (ρ)ρ + i

ρ

s, > 0; γ (k) = |γ (k)|e−iβ (k).

Thenβ (k) = β (k) + s arctg

k. (4.3.52)

For the function γ (ρ) , (4.3.50) has been proved. Hence, (4.3.52) takes the form

β (k) =1

π ∞−∞ ln

|γ (ξ )

|ξ − k dξ +s

2π ∞−infty

ln(1 + 2

ξ2)

ξ − k dξ + s arctg

k .

When → 0 , it gives us (4.3.50). 2

Lemma 4.3.5. Suppose that a function γ (ρ) is regular in Ω+, γ (−ρ) = γ (ρ) , and for |ρ| → ∞, ρ ∈ Ω+, γ (ρ) = 1 + O( 1

ρ) . Let γ (k) = |γ (k)|e−iβ (k) , and let ρ j = k j + iτ j, τ j >

0, j = 1, s be zeros of γ (ρ) in Ω+ . Then

β (k) = 1π ∞−∞ ln |γ (ξ )|ξ − k dξ + 2

s j=1

arctg τ jk − k j. (4.3.53)

Proof. Since γ (−ρ) = γ (ρ) , the zeros of γ (ρ) i n Ω+ are symmetrical with respect tothe imaginary axis. If ρ j = iτ j, τ j > 0 , then

argk + ρ j

k − ρ j= 2arctg

τ jk

. (4.3.54)



229

If ρ j = k j + iτ j , k j > 0, τ j > 0 , then

arg(k + ρ j)(k − ρ j)

(k − ρ j)(k + ρ j)= 2arctg

τ jk − k j

+ 2arctgτ j

k + k j. (4.3.55)

Denoteγ (ρ) = γ (ρ)

s j=1

ρ + ρ j

ρ − ρ j.

The function γ (ρ) satisfies the hypothesis of Lemma 4.3.4. Then using (4.3.50), (4.3.54)and (4.3.55) we arrive at (4.3.53). 2

Using Lemma 5 one can construct the transmission coefficients from their moduli andinformation about their zeros in Ω+ . For definiteness we confine ourselves to the caseh

= 0 .

Theorem 4.3.4.Let f 1(k) = α1(k)e−iδ1(k), (4.3.56)

and let ρ∗ j = iτ ∗ j , τ ∗ j > 0, j = 1, m∗

be zeros of f 1(ρ) in Ω+ . Then

δ 1(k) = kT +1

π

∞−∞

ln α1(ξ )

ξ − kdξ + 2

m∗ j=1

arctg τ ∗ jk

. (4.3.57)

In particular, if R(x)∈

B0

, then

δ 1(k) = kT +1

π

∞−∞

ln α1(ξ )

ξ − kdξ. (4.3.58)

Proof. Denote γ (ρ) = eiρT f 1(ρ) . If follows from (4.3.15) that

γ (ρ) = 1 + T

−T g1(t)eiρ(T −t)dt,

and consequently, γ (−ρ) = γ (ρ) , and for |ρ| → ∞, ρ ∈ Ω+ , γ (ρ) = 1 + O 1

ρ

. Thus,

the function γ (ρ) satisfies the hypothesis of Lemma 4.3.5. Using (4.3.53) and the relations|γ (k)| = α1(k), δ 1(k) = β (k) + kT , we arrive at (4.3.57). 2

Similarly we prove the following theorem.

Theorem 4.3.5. Let f 2(k) = α2(k)e−iδ2(k), (4.3.59)

and let ρ0 j = k0

j + iτ 0 j , τ 0 j > 0, j = 1, m0 be zeros of f 2(ρ) in Ω+ . Then

δ 2(k) =π

2sign

k

h

+ kT +

1

π

∞−∞

ln |2ξh α2(ξ )|

ξ − kdξ + 2

m0 j=1

arctg τ 0 j

k − k0 j

. (4.3.60)

In particular, if f 2(ρ) has no zeros in Ω+ , then

δ 2(k) =π

2sign

k

h + kT +1

π ∞

−∞

ln |2ξh

α2(ξ )|ξ − k

dξ. (4.3.61)



230

Thus, the specification of α j(k) uniquely determines the transmission coefficients onlywhen they have no zeros in Ω+ . In particular, for the class B0 , f 1(ρ) is uniquely deter-mined by its modulus, and all possible transmission coefficients f 2(ρ) can be constructedby means of solving the integral equation (4.3.21).

4.3.4. Symmetrical case.

The wave resistance is called symmetrical if R(T −x) = R(x) . For the symmetrical casethe transmission coefficient f 2(k) is uniquely determined (up to the sign) from its modulus.

Theorem 4.3.6. For the wave resistance to be symmetrical it is necessary and sufficient that Re f 2(k) = 0 . Moreover, f 2(k) = −f 2(−k), g2(t) = −g2(−t) , and (see (4.3.16))

f 2(k) = −2i T

0g2(t)sin ktdt.

Proof. According to (4.3.9) and (4.3.5),

f 2(ρ) =1

2iρ(u(T, ρ) + iρu(T, ρ) + hu(T, ρ)).

Since u(x, ρ) = X 1(x, λ) − iρX 2(x, λ) , we calculate

Re f 2(k) =1

2(X 1(T, λ) − X 2(T, λ) − hX 2(T, λ)). (4.3.62)

If R(x) = R(T −x) , then it follows from (4.3.6) and (4.3.8) that h(x) = −h(T −x), q (x) =q (T − x) , and consequently h = 0 , X 1(T, λ) ≡ X 2(T, λ) (see [yur2]), i.e. Re f 2(k) = 0 .

Conversely, if Re f 2(k) = 0, then (4.3.62) gives us

X 1(T, λ) − X 2(T, λ) − hX 2(T, λ) ≡ 0. (4.3.63)

Since for k → ∞ ,

X 2(T, k2) =sin kT

k+ O

1

k2

, X 1(T, λ) − X

2(T, λ) = O

1

k2

,

we obtain from (4.3.63) that h = 0, X 1(T, λ) ≡ X 2(T, λ). Consequently, q (x) = q (T − x)(see [yur2]). Similarly one can prove that g(x) = g(T − x) . Hence h(x) = −h(T − x) , andR(x) = R(T − x) .


Re f 2(k) = T

−T g2(t)cos kt dt, Im f 2(k) = −

T

−T g2(t)sin ktdt.

For the symmetrical case Re f 2(k) = 0 , and consequently g2(t) = −g2(−t) . Then

f 2(k) = i I m f 2(k) = −2i T

0g2(t)sin kt dt, f 2(−k) = −f 2(k).

2



231

Thus, f 2(k) can be constructed (up to the sign) by the formulas

Re f 2(k) = 0, |Im f 2(k)| = α2(k), f 2(k) = −f 2(−k).

4.3.5. Synthesis of the wave resistance from the power reflection coefficient.

In this subsection, using results obtained above, we provide a procedure for constructingR(x) from the given power reflection coefficient σ(k) . For definiteness we confine ourselvesto the case h = 0 . Let σ(k) (0 ≤ σ(k) < 1, σ(k) = σ(−k)) be given. Our scheme of calculation is:

Step 1. Calculate α1(k) and α2(k) by the formula

α21(k) = 1 + α2

2(k) =1

1

−σ(k)

. (4.3.64)

Step 2 . Construct f 1(k) by (4.3.56), (4.3.57) or for R(x) ∈ B0 by (4.3.56), (4.3.58).Find g1(t) from the relation T

−T g1(t)e−ikt dt = f 1(k) − eikT .

Step 3 . Construct f 2(k) by (4.3.59), (4.3.60) or, if f 2(ρ) has no zeros in Ω+ , by(4.3.59), (4.3.61). Find g2(t) from the relation

T

−T g2(t)e−ikt dt = f 2(k).

We note that g2(t) can be constructed also from the integral equation (4.3.21). In this casef 2(k) is calculated by (4.3.16).

Step 4. Calculate η(t) = g1(t) + g2(t) and ∆(ρ) by (4.3.12).Step 5 . Construct R(x) using one of the Algorithms 4.3.1 - 4.3.3.

Remark 4.3.4. In some concrete algorithms it is not necessary to make all calculationsmentioned above. For example, let R(x)

∈B0 , and let us use Algorithm 4.3.2. Then it is

not necessary to calculate g1(t), g2(t) and η(t) , since we need only |∆(k)|.Now we consider a concrete algorithm which realizes this scheme. For simplicity, we

consider the case R(x) ∈ B0 .For t ∈ [0, T ] we consider the functions

ϕ j(t) = g j(t) + g j (−t), ψ j (t) = g j(t) − g j(−t), j = 1, 2, (4.3.65)

ϕ(t) = η(t) + η(−t), ψ(t) = η(t) − η(−t). (4.3.66)

Since η(t) = g1(t) + g2(t), we get

ψ(t) = ϕ1(t) + ϕ2(t), ψ(t) = ψ1(t) + ψ2(t). (4.3.67)

Solving (4.3.65) and (4.3.66) with respect to g j(t) and η(t) we obtain

g j(t) =

1

2(ϕ j(t) + ψ j(t)) , t > 0,

1

2(ϕ j(−t) − ψ j(−t)) , t < 0,

η(t) =

1

2(ϕ(t) + ψ(t)) , t > 0

1

2(ϕ(−t) + ψ(−t)) , t < 0

(4.3.68)



232


ϕ1(T ) = ψ1(T ) = −w1, ϕ2(T ) = ψ2(T ) = −h

2, ψ1(0) = ψ2(0) = 0, (4.3.69)

where

w1 = −12

T

0h2(ξ ) dξ.

By virtue of Lemma 4.3.3,

f 1(k) = (cos kT + C 1(k)) − i(sin kT + S 1(k)),

f 2(k) = C 2(k) − iS 2(k),

∆(k) = ((cos kT + C (k))

−i(sin kT + S (k)),

(4.3.70)

where

C j(k) = T

0ϕ j(t)cos kt dt, S j(k) =

T

0ψ j(t)sin ktdt, (4.3.71)

C (k) = T

0ϕ(t)cos kt dt, S (k) =

T

0ψ(t)sin ktdt. (4.3.72)

Using (4.3.69) and (4.3.71) we obtain the following asymptotic formulae for C j(k) andS j(k) as k → ∞ :

C 1(k) = −w1sin kT

k+

ω(k)

k, C 2(k) = −h

2

sin kT

k+

ω(k)

k,

S 1(k) = w1cos kT

k+

ω(k)

k, S 2(k) =

h

2

cos kT

k+

ω(k)

k.

(4.3.73)

Here and below, one and the same symbol ω(k) denotes various functions from L2(−∞, ∞) .Comparing (4.3.70) with the relations f j(k) = α j(k) e−iδj(k) , we derive

C 1(k) = α1(k)cos δ 1(k) − cos kT, S 1(k) = α1(k)sin δ 1(k) − sin kT , (4.3.74)

C 2(k) = α2(k)cos δ 2(k), S 2(k) = α2(k)sin δ 2(k). (4.3.75)

For calculating the arguments of the transmission coefficients we will use (4.3.58) and(4.3.61), i.e.

δ 1(k) = kT + δ 1(k), δ 2(k) =π

2w + kT + δ 2(k), k > 0, (4.3.76)

where w = sign h ,

δ 1(k) =

1

π ∞−∞ln α1(ξ )

ξ − k dξ, (4.3.77)

δ 2(k) =1

π

∞−∞

ln α2(ξ )

ξ − kdξ, α2(ξ ) :=

2ξ

hα2(ξ )

. (4.3.78)

It follows from (4.3.73)-(4.3.75) and (4.3.20) that for k → +∞,

δ 1(k) = kT +w1

k+

ω(k)

k, δ 2(k) =

π

2w + kT + ω(k). (4.3.79)



233

Substituting (4.3.76) into (4.3.74) and (4.3.75), we calculate for k > 0,

C 1(k) = cos kT (α1(k)cos δ 1(k) − 1) − sin kT (α1(k)sin δ 1(k)),

S 1(k) = sin kT (α1(k)cos δ 1(k) − 1) + cos kT (α1(k)sin δ 1(k)),

(4.3.80)

C 2(k) = −α2(k)w(sin kT cos δ 2(k) + cos kT sin δ 2(k)),

S 2(k) = α2(k)w(cos kT cos δ 2(k) − sin kT sin δ 2(k)).

(4.3.81)

Furthermore, consider the functions

ϕ∗1(t) = ϕ1(t) + w1, ψ∗1(t) = ψ1(t) + w1

tT

,

ϕ∗2(t) = ϕ2(t) + h2 , ψ∗

2(t) = ψ2(t) + h2

· tT .

(4.3.82)

Then in view of (4.3.69),

ϕ∗ j (T ) = ψ∗ j (T ) = ψ∗

j (0) = 0, j = 1, 2. (4.3.83)

Denote

C ∗ j (k) = T

0ϕ∗ j (t)cos kt dt, S ∗ j (k) =

T

0ψ∗

j (t)sin ktdt. (4.3.84)

Integrating the integrals in (4.3.84) by parts and taking (4.3.83) into account, we obtain

C ∗ j (k) = − T

0 ϕ∗

j (t)

sin kt

k dt, S ∗ j (k) = T

0 ψ∗

j (t)

cos kt

k dt. (4.3.85)

Clearly,

C 1(k) = C ∗1 (k) − w1sin kT

k, S 1(k) = S ∗1 (k) + w1

cos kT

k− sin kT

T k2

, (4.3.86)

C 2(k) = C ∗2 (k) − h

2

sin kT

k, S 2(k) = S ∗2 (k) +

h

2

cos kT

k− sin kT

T k2

. (4.3.87)

Now let the power reflection coefficient σ(k) (0 ≤ σ(k) < 1, σ(k) = σ(−k)) be given for|k| ≤ B , and put

σ(k) =h2

4k2 + h2, |k| > B, (4.3.88)

where B > |h| is chosen sufficiently large, such that (see (4.3.20)) σ(k) is sufficientlyaccurate for |k| > B.

Using (4.3.64) we calculate α1(k) and α2(k) . Then

α

2

1(k) − α

2

2(k) = 1, α j(k) = α j(−k) > 0,and

α21(k) = 1 +

h2

4k2, α2

2(k) =h2

4k2, |k| > B. (4.3.89)

In order to calculate δ 1(k) we use (4.3.77). First let |k| > B + χ, χ > 0 . In view of (4.3.89), equality (4.3.77) takes the form

δ 1(k) =1

π B

−B

ln α1(ξ )

ξ − kdξ +

1

2π |ξ|>B

ln(1 + h2

4ξ2)

ξ − kdξ.



234

Since

1

2π

|ξ|>B

ln

1 + h2

4ξ2

ξ − k

dξ =1

2π

∞ j=1

(−1) j−1

j

h2

4

j |ξ|>B

dξ

ξ 2 j(ξ − k)

=

1

2π

∞

j=1

(

−1) j−1

j h2

4 j

|ξ|>B 1

k2 j 1

ξ − k −1

ξ −2 j−1

s=1

1

k2 j−sξ s+1 dξ,

we get

δ 1(k) =1

π

B

−B

ln α1(ξ )

ξ − kdξ +

1

2πln

1 +2B

k − B

ln

1 +

h2

4k2

+1

π

∞ j=1

(−1) j

j

h2

4

j j−1µ=0

1

(2 j − 2µ − 1)B2 j−2µ−1k2µ+1.

(4.3.90)

Using the relation

1

π

B

−B

ln α1(ξ )

ξ − kdξ = − 1

kπ

B

−Bln α1(ξ ) dξ +

1

π

B

−B

ξ ln α1(ξ )

k(ξ − k)dξ,

we separate in (4.3.90) the terms of order 1/k , i.e.

δ 1(k) =w1

k+ δ ∗1(k), (4.3.91)

whereδ ∗1(k) =

1

π

B

−B

ξ ln α1(ξ )

k(ξ − k)dξ +

1

2πln

1 +2B

k − B

ln

1 +

h2

4k2

+1

π

∞ j=1

(−1) j

j

h2

4

j j−1µ=1

1

(2 j − 2µ − 1)B2 j−2µ−1k2µ+1, |k| > B + χ, (4.3.92)

w1 = − 1

π

B

−Bln α1(ξ ) dξ +

1

π

∞ j=1

(−1) j

j

h2

4B2

j

· B

2 j − 1< 0. (4.3.93)

It is easy to show that

|δ ∗1(k)| ≤ Ω∗1

k2,

where

Ω∗1 =

B + χ

πχ

B

0ξ ln α1(ξ )dξ +

h2

8πln

1 +

2B

χ

+

h2

2π.

Now let |k| ≤ B + χ . In this case we rewrite (4.3.77) in the form

δ 1(k) = 1π ∞

0ln α1(ξ + k) − ln α1(ξ − k)

ξ dξ. (4.3.94)

Take r ≥ 2B + χ . Since ln

1 +h2

4ξ 2

− h2

4ξ 2

≤ h4

32ξ 4,

we get

1

π ∞

r

ln α1(ξ + k) − ln α1(ξ − k)

ξ dξ =

h2

8π ∞

r

1

ξ 1

(ξ + k)2 −1

(ξ − k)2 dξ + (k),



235

where

|(k)| ≤ h4

96πr(r − B − χ)3.

Calculating this integral and substituting into (4.3.94), we obtain

δ 1(k) = 1π

r

0

ln α1(ξ + k) − ln α1(ξ − k)ξ

dξ + h2

8π

1k2

ln r + kr − k

− 1k

1R + k

+ 1R − k

+(k), δ 1(0) = 0, |k| ≤ B + χ, r ≥ 2B + χ. (4.3.95)

In order to calculate δ 2(k) we use (4.3.78). Let |k| > B + χ, χ > 0. According to(4.3.89) we have α2(k) = 1 for |k| > B . Hence

δ 2(k) =1

π B

−B

ln α2(ξ )

ξ − k

dξ,

|ξ

|> B + χ. (4.3.96)

For |k| ≤ B + χ it is more convenient to use another formula:

δ 2(k) =1

π

r

0

ln α2(ξ + k) − ln α2(ξ − k)

ξ dξ, |k| ≤ B + χ, r = 2B + χ. (4.3.97)

Let us calculate ψ∗ j (t), ϕ∗ j (t) . By virtue of (4.3.84),

ψ∗ j (t) =2

T

∞n=1 S ∗ j nπ

T sin

nπ

T t,

ϕ∗ j (t) = 1T

C ∗ j (0) + 2T

∞n=1

C ∗ j

nπ

T

cos

nπ

T t.

(4.3.98)

According to (4.3.85), the series in (4.3.98) converge absolutely and uniformly. By virtue of

(4.3.80), (4.3.81), (4.3.86), (4.3.87), the coefficients S ∗ j

nπT

and C ∗ j

nπT

can be calculated

via the formulae

S ∗1 (k) = (−1)n

(α1(k)sin˜δ 1(k) −

w1

k ), k =nπ

T , n ≥ 1,

S ∗2 (k) = (−1)n(α2(k)w cos δ 2(k) − h2k ), k = nπ

T , n ≥ 1,

C ∗1 (k) = (−1)n(α1(k)cos δ 1(k) − 1) + δ n0w1T, k = nπT

, n ≥ 0,

C ∗2 (k) = (−1)n+1w sin δ 2(k) + δ n0hT

2, k = nπ

T , n ≥ 0,

(4.3.99)

where δ nj is the Kronecker delta.

Using the formulae obtained above we arrive at the following numerical algorithm for thesolution of Inverse Problem 4.3.1.

Algorithm 4.3.4. Let the power reflection coefficient σ(k) be given for |k| ≤ B , andtake σ(k) according to (88) for |k| > B . Then

(1) Construct α1(k) and α2(k) by (4.3.64).

(2) Find the number w1 by (4.3.93).

(3) Calculate δ 1(k), δ 2(k) by (4.3.91), (4.3.92), (4.3.96) for

|k

|> B + χ , and by (4.3.95),

(4.3.97) for |k| ≤ B + χ .



236

(4) Construct ϕ∗ j (t), ψ∗ j (t) by (4.3.98), where S ∗ j

nπT

, C ∗ j

nπT

is defined by (4.3.99).

(5) Find ϕ j(t), ψ j(t) using (4.3.82).

(6) Calculate η(t), t ∈ [−T, T ] by (4.3.67), (4.3.68).

(7) Find F (t), 0 < t < 2T from the integral equation (4.3.36), where θ(t) = η(T

−t) .

(8) Calculate G(x, t) from the integral equation (4.3.37).

(9) Construct R(x), x ∈ [0, T ] via (4.3.41).

Remark 4.3.5. This algorithm is one of several possible numerical algorithms forthe solution of Inverse Problem 4.3.1. Using the obtained results one can construct variousalgorithms for synthesizing R(x) from spectral characteristics.

4.4. DISCONTINUOUS INVERSE PROBLEMS

This section deals with Sturm-Liouville boundary value problems in which the eigenfunc-tions have a discontinuity in an interior point. More precisely, we consider the boundaryvalue problem L for the equation:

y := −y + q (x)y = λy (4.4.1)

on the interval 0 < x < T with the boundary conditions

U (y) := y(0) − hy(0) = 0, V (y) := y(T ) + Hy(T ) = 0, (4.4.2)

and with the jump conditions

y(a + 0) = a1y(a − 0), y(a + 0) = a−11 y(a − 0) + a2y(a − 0). (4.4.3)

Here λ is the spectral parameter; q (x), h , H , a1 and a2 are real; a∈

(0, T ), a1 > 0, q (x)∈L2(0, T ). Attention will be focused on the inverse problem of recovering L from its spectral

characteristics.Boundary value problems with discontinuities inside the interval often appear in math-

ematics, mechanics, physics, geophysics and other branches of natural sciences. As a rule,such problems are connected with discontinuous material porperties. The inverse problemof reconstructing the material properties of a medium from data collected outside of themedium is of central importance in disciplines ranging from engineering to the geoscienes.For example, discontinuos inverse problems appear in electronics for constructing parameters

of heterogeneous electronic lines with desirable technical characteristics ([lit1], [mes1]). Afterreducing the corresponding mathematical model we come to the boundary value problemL where q (x) must be constructed from the given spectral information which describesdesirable amplitude and phase characteristics. Spectral information can be used to recon-struct the permittivity and conductivity profiles of a one-dimensional discontinuous medium([kru1], [she1]). Boundary value problems with discontinuities in an interior point also ap-pear in geophysical models for oscillations of the Earth ([And1], [lap1]). Here the maindiscontinuity is cased by reflection of the shear waves at the base of the crust. Further,

it is known that inverse spectral problems play an important role for investigating some



237

nonlinear evolution equations of mathematical physics. Discontinuous inverse problems helpto study the blow-up behavior of solutions for such nonlinear equations. We also note thatinverse problems considered here appear in mathematics for investigating spectral propertiesof some classes of differential, integrodifferential and integral operators.

Direct and inverse spectral problems for differential operators without discontinuities

have been studied in Chapter 1. The presence of discontinuities produces essential qualita-tive modifications in the investigation of the operators. Some aspects of direct and inverseproblems for discontinuous boundary value problems in various formulations have been con-sidered in [she1], [hal1], [akt2], [ebe4] and other works. In particular, it was shown in [hal1]that if q (x) is known a priori on [0, T /2], then q (x) is uniquely determined on [T/2, T ]by the eigenvalues. In [she1] the discontinuous inverse problem is considered on the half-line.

Here in Subsection 4.4.1 we study properties of the spectral characteristics of the bound-ary value problem L. Subsections 4.4.2-4.4.3 are devoted to the inverse problem of recovering

L from its spectral characteristics. In Subsection 4.4.2 the uniqueness theorems are proved,and in Subsection 4.4.3 we provide necessary and sufficient conditions for the solvability of the inverse problem and also obtain a procedure for the solution of the inverse problem.

In order to study the inverse problem for the boundary value problem L we use themethod of spectral mappings described in Section 1.6. We will omit the proofs (or provideshort versions of the proofs) for assertions which are similar to those given in Section 1.6.

4.4.1. Properties of the spectral characteristics. Let y(x) and z (x) be continu-ously differentiable functions on [0, a] and [a, T ]. Denote

y, z

:= yz

−yz. If y(x) and

z (x) satisfy the jump conditions (4.4.3), then

y, z |x=a+0 = y, z |x=a−0, (4.4.4)

i.e. the function y, z is continuous on [0, T ]. If y(x, λ) and z (x, µ) are solutions of theequations y = λy and z = µz respectively, then

d

dxy, z = (λ − µ)yz. (4.4.5)

Let ϕ(x, λ), ψ(x, λ), C (x, λ), S (x, λ) be solutions of (4.4.1) under the initial conditionsC (0, λ) = ϕ(0, λ) = S (0, λ) = ψ(T, λ) = 1, C (0, λ) = S (0, λ) = 0, ϕ(0, λ) = h, ψ(T, λ) =−H, and under the jump conditions (4.4.3). Then U (ϕ) = V (ψ) = 0. Denote

∆(λ) = ϕ(x, λ), ψ(x, λ).

By virtue of (4.4.4) and Liouville’s formula for the Wronskian [cod1, p.83], ∆(λ) does notdepend on x. The function ∆(λ) is called the characteristic function of L. Clearly,

∆(λ) = −V (ϕ) = U (ψ). (4.4.6)

Theorem 4.4.1. 1) The eigenvalues λnn≥0 of the boundary value problem L coincide with zeros of the characteristic function ∆(λ). The functions ϕ(x, λn) and ψ(x, λn) are eigenfunctions, and

ψ(x, λn) = β nϕ(x, λn), β n = 0. (4.4.7)

2) Denote

αn = T

0 ϕ

2

(x, λn) dx. (4.4.8)



238

Then β nαn = ∆1(λn), (4.4.9)

where ∆1(λ) =d

dλ∆(λ). The data λn, αnn≥0 are called the spectral data of L.

3) The eigenvalues

λn

and the eigenfunctions ϕ(x, λn), ψ(x, λn) are real. All zeros

of ∆(λ) are simple, i.e. ∆1(λn) = 0. Eigenfunctions related to different eigenvalues are orthogonal in L2(0, T ).

We omit the proof of Theorem 4.4.1, since the arguments here are the same as for theclassical Sturm-Liouville problem (see Section 1.1).

Let C 0(x, λ) and S 0(x, λ) be smooth solutions of (4.4.1) on the interval [0, T ] underthe initial conditions C 0(0, λ) = S 0(0, λ) = 1, C 0(0, λ) = S 0(0, λ) = 0. Then, using the jumpcondition (4.4.3) we get

C (x, λ) = C 0(x, λ), S (x, λ) = S 0(x, λ), x < a, (4.4.10)

C (x, λ) = A1C 0(x, λ) + B1S 0(x, λ), S (x, λ) = A2C 0(x, λ) + B2S 0(x, λ), x > a, (4.4.11)

where

A1 = a1C 0(a, λ)S 0(a, λ) − a−11 C 0(a, λ)S 0(a, λ) − a2C 0(a, λ)S 0(a, λ),

B1 = (a−11 − a1)C 0(a, λ)C 0(a, λ) + a2C 20 (a, λ),

A2 = (a1 − a−11 )S 0(a, λ)S 0(a, λ) − a2S 20 (a, λ),B2 = a−1

1 C 0(a, λ)S 0(a, λ) − a1C 0(a, λ)S 0(a, λ) + a2C 0(a, λ)S 0(a, λ).

(4.4.12)

Let λ = ρ2, ρ = σ + iτ. It was shown in Section 1.1 that the function C 0(x, λ) satisfiesthe following integral equation:

C 0(x, λ) = cos ρx + x

0

sin ρ(x − t)

ρq (t)C 0(t, λ) dt, (4.4.13)

and for |ρ| → ∞,

C 0(x, λ) = cos ρx + O1

ρexp(|τ |x)

.

Then (4.4.13) implies

C 0(x, λ) = cos ρx+sin ρx

2ρ

x

0q (t) dt+

1

2ρ

x

0q (t)sin ρ(x−2t) dt+ O

1

ρ2exp(|τ |x)

, (4.4.14)

C 0(x, λ) = −ρ sin ρx+ cos ρx2

x

0q (t) dt+ 1

2 x

0q (t)cos ρ(x−2t) dt+O1

ρexp(|τ |x), (4.4.15)

Analogously,

S 0(x, λ) =sin ρx

ρ− cos ρx

2ρ2

x

0q (t) dt+

1

2ρ2

x

0q (t)cos ρ(x−2t) dt+O

1

ρ3exp(|τ |x)

, (4.4.16)

S 0(x, λ) = cos ρx +sin ρx

2ρ x

0q (t) dt− 1

2ρ x

0q (t)sin ρ(x−2t) dt+ O

1

ρ2exp(|τ |x)

. (4.4.17)



239

By virtue of (4.4.12) and (4.4.14)-(4.4.17),

A1 = b1 + b2 cos2ρa +

b2

a

0q (t) dt − a2

2

sin2ρa

ρ+ O

1

ρ2

,

B1 = b2ρ sin2ρa − cos2ρa a

0 q (t) dt − a

0 q (t)cos2ρ(a − t) dt+

a2

2

1 + cos 2ρa

+ O

1

ρ

, A2 = b2

sin2ρa

ρ+ O

1

ρ2

, B2 = b1 + b2 cos2ρa + O

1

ρ

,

where b1 = (a1 +a−11 )/2, b2 = (a1−a−1

1 )/2. Since ϕ(x, λ) = C (x, λ)+hS (x, λ), we calculateusing (4.4.10)-(4.4.12), (4.4.14)-(4.4.17):

ϕ(x, λ) = cos ρx + h +1

2 x

0

q (t) dtsin ρx

ρ

+ o1

ρ

exp(

|τ

|x), x < a, (4.4.18)

ϕ(x, λ) = (b1 cos ρx + b2 cos ρ(2a − x))+

f 1(x)sin ρx

ρ+ f 2(x)

sin ρ(2a − x)

ρ+ o

1

ρexp(|τ |x)

, x > a, (4.4.19)

ϕ(x, λ) = −ρ sin ρx +

h +1

2

x

0q (t) dt

cos ρx + o(exp(|τ |x)), x < a, (4.4.20)

ϕ(x, λ) = ρ(

−b1 sin ρx + b2 sin ρ(2a

−x))+

f 1(x)cos ρx − f 2(x)cos ρ(2a − x) + o(exp(|τ |x)), x > a, (4.4.21)

where

f 1(x) = b1

h +

1

2

x

0q (t) dt

+

a2

2, f 2(x) = b2

h − 1

2

x

0q (t) dt +

a

0q (t) dt

− a2

2.

In particular, (4.4.18)-(4.4.21) yield

ϕ(ν )(x, λ) = O(|ρ|

ν exp(|τ |x)), 0

≤x

≤π. (4.4.22)

Similarly,ψ(ν )(x, λ) = O(|ρ|ν exp(|τ |(T − x))), 0 ≤ x ≤ π. (4.4.23)

It follows from (4.4.6),(4.4.19) and (4.4.21) that

∆(λ) = ρ(b1 sin ρT −b2 sin ρ(2a−T ))−ω1 cos ρT −ω2 cos ρ(2a−T ) + o(exp(|τ |T )), (4.4.24)

where

ω1 = b1

H +h+

1

2

T

0q (t) dt

+

a2

2, ω2 = b2

H −h+

1

2

T

0q (t) dt−

a

0q (t) dt

− a2

2. (4.4.25)

Using (4.4.24), by the well-known methods (see, for example, [bel1]) one can obtain thefollowing properties of the characteristic function ∆(λ) and the eigenvalues λn = ρ2

n of theboundary value problem L :

1) For |ρ| → ∞, ∆(λ) = O(|ρ| exp(|τ |T )).2) There exist h > 0, C h > 0 such that |∆(λ)| ≥ C h|ρ| exp(|τ |T ) for |Im ρ| ≥ h.

Hence, the eigenvalues λn lie in the domain |Im ρ| < h.



240

3) The number N a of zeros of ∆(λ) in the rectangle Ra = ρ : |τ | ≤ h, σ ∈ [a, a + 1]is bounded with respect to a.

4) Denote Gδ = ρ : |ρ − ρn| ≥ δ . Then

|∆(λ)| ≥ C δ|ρ| exp(|τ |T ), ρ ∈ Gδ. (4.4.26)

5) There exist numbers RN → ∞ such that for sufficiently small δ > 0, the circles|ρ| = RN lie in Gδ for all N.

6) Let λ0n = (ρ0

n)2 be zeros of the function

∆0(λ) = ρ(b1 sin ρT − b2 sin ρ(2a − T )). (4.4.27)

Thenρn = ρ0

n + o(1), n → ∞. (4.4.28)

Substituting (4.4.18),(4.4.19) and (4.4.28) into (4.4.8) we get

αn = α0n + o(1), n → ∞, (4.4.29)

where

α0n =

a

2+b2

1 + b22

2+ b1b2 cos2ρ0

na

(T − a). (4.4.30)


|αn| C, (4.4.31)

which means that αn = O(1) and (αn)−1 = O(1). By virtue of (4.4.7),

β n = ψ(0, λn) =1

ϕ(T, λn). (4.4.32)

Taking (4.4.22) and (4.4.23) into account, we deduce |β n| C. Together with (4.4.9) and(4.4.31) this yields

|∆1(λn)| C. (4.4.33)We need more precise asymptotics for ρn and αn than (4.4.28) and (4.4.29). Substi-

tuting (4.4.24) into the relation ∆(λn) = 0, we get

b1 sin ρnT − b2 sin ρn(2a − T ) = O 1

ρ0n

. (4.4.34)

According to (4.4.28), ρn = ρ0n + εn, εn → 0. Since

∆0

(λ0n) = ρ

0n(b1 sin ρ

0nT − b2 sin ρ

0n(2a − T )) = 0,


εn(b1T cos ρ0nT − b2(2a − T )cos ρ0

n(2a − T )) = O 1

ρ0n

+ O(ε2

n).

In view of (4.4.27),

∆

0

1(λ

0

n) := d

dλ∆

0

(λ)|λ=λ0n = 2−1b1T cos ρ

0

nT − b2(2a − T )cos ρ

0

n(2a − T ).



241

Hence

εn∆01(λ0

n) = O 1

ρ0n

+ O(ε2

n).

By virtue of (4.4.33), |∆01(λ0

n)| C, and we conclude that

εn = O 1ρ0

n

. (4.4.35)

Substituting (4.4.24) into the relation ∆(λn) = 0 again and using (4.4.35) we arrive at

ρn = ρ0n +

θn

ρ0n

+κn

ρ0n

, (4.4.36)

where κn = o(1), and

θn = (ω1 cos ρ0nT + ω2 cos ρ0

n(2a − T ))(2∆01(λ0

n))−1.

Here ω1 and ω2 are defined by (4.4.25). At last, using (4.4.8), (4.4.18), (4.4.19) and(4.4.36), one can calculate

αn = α0n +

θn1

ρ0n

+κn1

ρ0n

, (4.4.37)

where κn1 = o(1), and

θn1 = b0 sin2ρ0na + b

2

1

4sin2ρ0

nT − b2

2

4sin2ρ0

n(2a − T ) + b1b2

2sin2ρ0

n(T − a),

b0 = −2aθn − b21

4+

b22

4+ (T − a)

b1b2(2h +

a

0q (t) dt) +

a2

2(b2 − b1)

.

Furthermore, one can show that κn, κn1 ∈ 2. If q (x) is a smooth function one canobtain more precise asymptotics for the spectral data.

4.4.2. Formulation of the inverse problem. Uniqueness theorems. Let Φ(x, λ)

be the solution of (4.4.1) under the conditions U (Φ) = 1, V (Φ) = 0 and under the jumpconditions (4.4.3). We set M (λ) := Φ(0, λ). The functions Φ(x, λ) and M (λ) are calledthe Weyl solution and the Weyl function for the boundary value problem L, respectively.Clearly,

Φ(x, λ) =ψ(x, λ)

∆(λ)= S (x, λ) + M (λ)ϕ(x, λ), (4.4.38)

ϕ(x, λ), Φ(x, λ) ≡ 1, (4.4.39)

M (λ) =δ (λ)

∆(λ) , (4.4.40)

where δ (λ) = ψ(0, λ) = V (S ) is the characteristic function of the boundary value problemL1 for equation (4.4.1) with the boundary conditions U (y) = 0, y(T ) = 0 and with the

jump conditions (4.4.3). Let µnn≥0 be zeros of δ (λ), i.e. the eigenvalues of L1.

In this section we study the following inverse problems of recovering L from its spectralcharacteristics:

(i) from the Weyl function M (λ);

(ii) from the spectral data λn, αnn≥0;



242

(iii) from two spectra λn, µnn≥0.These inverse problems are a generalization of the inverse problems for Sturm-Liouville

equations without discontinuities (see Section 1.4).

First, let us prove the uniqueness theorems for the solutions of the problems (i)-(iii).

For this purpose we agree that together with L we consider a boundary value problem˜Lof the same form but with different coefficients q (x), h, H, a, ak. If a certain symbol γ

denotes an object related to L, then γ will denote the analogous object related to L, andγ = γ − γ.

Theorem 4.4.2. If M (λ) = M (λ), then L = L. Thus, the specification of the Weyl function uniquely determines the operator.

Proof. It follows from (4.4.23), (4.4.26) and (4.4.38) that

|Φ(ν )

(x, λ)| ≤ C δ|ρ|ν −

1

exp(−|τ |x), ρ ∈ Gδ. (4.4.41)

Let us define the matrix P (x, λ) = [P jk (x, λ)] j,k=1,2 by the formula

P (x, λ)

ϕ(x, λ) Φ(x, λ)

ϕ(x, λ) Φ(x, λ)

=


.

Taking (4.4.39) into account we calculate

P j1(x, λ) = ϕ( j−1)(x, λ)Φ(x, λ) − Φ( j−1)(x, λ)ϕ(x, λ),

P j2(x, λ) = Φ( j−1)(x, λ)ϕ(x, λ) − ϕ( j−1)(x, λ)Φ(x, λ),

(4.4.42)

ϕ(x, λ) = P 11(x, λ)ϕ(x, λ) + P 12(x, λ)ϕ(x, λ),

Φ(x, λ) = P 11(x, λ)Φ(x, λ) + P 12(x, λ)Φ(x, λ).

(4.4.43)

According to (4.4.38) and (4.4.42), for each fixed x, the functions P jk (x, λ) are meromor-phic in λ with simple poles in the points λn and λn. Denote G0

δ = Gδ

∩Gδ. By virtue of

(4.4.22), (4.4.41) and (4.4.42) we get

|P 12(x, λ)| ≤ C δ|ρ|−1, |P 11(x, λ)| ≤ C δ, ρ ∈ G0δ. (4.4.44)


P 11(x, λ) = ϕ(x, λ)S (x, λ) − S (x, λ)ϕ(x, λ) + (M (λ) − M (λ))ϕ(x, λ)ϕ(x, λ),

P 12(x, λ) = S (x, λ)ϕ(x, λ)−

ϕ(x, λ)S (x, λ) + (M (λ)−

M (λ))ϕ(x, λ)ϕ(x, λ).

Thus, if M (λ) ≡ M (λ) , then for each fixed x, the functions P 1k(x, λ) are entire in λ.Together with (4.4.44) this yields P 12(x, λ) ≡ 0, P 11(x, λ) ≡ A(x). Using (4.4.43) we derive

ϕ(x, λ) ≡ A(x)ϕ(x, λ), Φ(x, λ) ≡ A(x)Φ(x, λ). (4.4.45)

It follows from (4.4.18) and (4.4.19) that for |ρ| → ∞, arg ρ ∈ [ε, π − ε], ε > 0,

ϕ(x, λ) = 2−1b exp(

−iρx)(1 + O(ρ−1)),



243

where b = 1 for x < a, and b = b1 for x > a. Similarly, one can calculate

Φ(x, λ) = (iρb)−1 exp(iρx)(1 + O(ρ−1)).

Together with (4.4.39) and (4.4.45) this gives b1 = b1, A(x) ≡ 1, i.e. ϕ(x, λ) ≡ ϕ(x, λ),

Φ(x, λ) ≡ Φ(x, λ) for all x and λ. Consequently, L = L.2

Theorem 4.4.3. If λn = λn, αn = αn, n ≥ 0, then L = L. Thus, the specification of the spectral data λn, αnn≥0 uniquely determines the operator.

Proof. It follows from (4.4.40) that the Weyl function M (λ) is meromorphic withsimple poles λn. Using (4.4.40), (4.4.32) and (4.4.9) we calculate

Resλ=λn

M (λ) =δ (λn)

∆1(λn)=

β n∆1(λn)

=1

αn. (4.4.46)

Let Γ = λ = u + iv : u = (2h2)−2v2 − h2 be the image of the set Im ρ = ±h underthe mapping λ = ρ2. Denote Γn = Γ ∩ λ : |λ| ≤ rn, and consider the closed contoursΓn0 = Γn ∪ λ : |λ| = rn, λ /∈ int Γ, Γn1 = Γn ∪ λ : |λ| = rn, λ ∈ int Γ (withcounterclockwise circuit). Since the Weyl function M (λ) is regular for λ ∈ int Γn0, weget by Cauchy’s theorem that

M (λ) =1

2πi Γn0M (µ)

µ−

λdµ, λ ∈ int Γn0.

Furthermore, since δ (λ) = ψ(0, λ), estimate (4.4.23) gives

δ (λ) = O(exp(|τ |T )). (4.4.47)

Using (4.4.40), (4.4.26) and (4.4.47) we obtain

|M (λ)| ≤ C δ|ρ|−1, ρ ∈ Gδ. (4.4.48)

Hencelim

n→∞1

2πi

|µ|=rn

M (µ)

µ − λdµ = 0,

i.e.

M (λ) = limn→∞

1

2πi

Γn1

M (µ)

λ − µdµ.

Calculating this integral by the residue theorem and taking (4.4.46) into account we arriveat

M (λ) =

∞k=0

1

αk(λ − λk) , (4.4.49)

where the series converges ”with brackets”:∞

k=0

= limn→∞

|λk|<rn

. Moreover, it is obvious that

the series converges absolutely. Under the hypothesis of the theorem we get, in view of (4.4.49), that M (λ) ≡ M (λ), and consequently by Theorem 4.4.2, L = L. 2

Theorem 4.4.4. If λn = λn, µn = µn, n ≥ 0, then q (x) = q (x) a.e. on (0, T ), h =

h, H = H, a = a, a1 = a1 and a2 = a2.



244

Proof. It follows from (4.4.24) that the function ∆(λ) is entire in λ of order 1/2, andconsequently ∆(λ) is uniquely determined up to a multiplicative constant by its zeros:

∆(λ) = C ∞

n=0

1 − λ

λn

(4.4.50)

(the case when ∆(0) = 0 requires minor modifications). According to (4.4.27),

∆0(λ) = Ω0λ∞

n=1

1 − λ

λ0n

, Ω0 = T b1 − (2a − T )b2.

Then∆(λ)

∆0(λ)= C

λ0 − λ

λ0Ω0λ

∞n=1

λ0n

λn

∞n=1

1 +

λn − λ0n

λ0n − λ

.

Since

limλ→−∞

∆(λ)

∆0(λ)= 1, lim

λ→−∞

∞n=1

1 +

λn − λ0n

λ0n − λ

= 1,

we get

C = −λ0Ω0

∞n=1

λn

λ0n

.

Substituting into (4.4.50) we arrive at

∆(λ) = Ω0(λ − λ0)∞

n=1

λn − λλ0

n

. (4.4.51)

It follows from (4.4.51) that the specification of the spectrum λnn≥0 uniquely determinesthe characteristic function ∆(λ). Analogously, the function δ (λ) is uniquely determined byits zeros µnn≥0. Let now λn = λn, µn = µn, n ≥ 0. Then ∆(λ) ≡ ∆(λ), δ (λ) ≡ δ (λ),and consequently by (4.4.40), M (λ) ≡ M (λ). From this, in view of Theorem 4.4.2, weobtain q (x) = q (x) a.e. on (0, T ), h = h, H = H, a = a, a1 = a1 and a2 = a2. 2

4.4.3 Solution of the inverse problem. For definiteness we consider the inverseproblem of recovering L from the spectral data λn, αnn≥0. Let boundary value problemsL and L be such that

a = a,∞

n=0

ξ n|ρn| < ∞, (4.4.52)

where ξ n := |ρn − ρn| + |αn − αn|. Denote


Qkj (x, λ) =ϕ(x, λ), ϕkj (x)

αkj (λ − λkj )=

1

αkj

x

0ϕ(t, λ)ϕkj (t) dt, Qni,kj(x, λ) = Qkj (x, λni).

Here we lean on (4.4.4) and (4.4.5). It follows from (4.4.18)-(4.4.21) and (4.4.31) that

|ϕ(ν )nj (x)| ≤ C (|ρ0

n| + 1)ν , |ϕ(ν )nj (x)| ≤ C (|ρ0

n| + 1)ν , (4.4.53)

|Qni,kj(x)

| ≤C

|ρ0n − ρ0k| + 1,

|Q

(ν +1)

ni,kj

(x)| ≤

C (|ρ0

n|+

|ρ0

k|+ 1)ν . (4.4.54)



245


ϕ(x, λ) = ϕ(x, λ) +∞

k=0

(Qk0(x, λ)ϕk0(x) − Qk1(x, λ)ϕk1(x)). (4.4.55)

Proof. By virtue of (4.4.52) we have

a = a, a1 = a1. (4.4.56)

Then it follows from (4.4.18)-(4.4.21) that

|ϕ(ν )(x, λ) − ϕ(ν )(x, λ)| ≤ C |ρ|ν −1 exp(|τ |x). (4.4.57)

Similarly,

|ψ(ν )

(x, λ) − ψ(ν )

(x, λ)| ≤ C |ρ|ν −

1

exp(|τ |(T − x)). (4.4.58)Denote G0

δ = Gδ ∩ Gδ. In view of (4.4.38), (4.4.23), (4.4.26) and (4.4.58) we obtain

|Φ(ν )(x, λ) − Φ(ν )(x, λ)| ≤ C δ|ρ|ν −2 exp(−|τ |x), ρ ∈ G0δ. (4.4.59)

Let P (x, λ) be the matrix defined in Subsection 4.4.2. Since for each fixed x, the functionsP 1k(x, λ) are meromorphic in λ with simple poles λn and λn, we get by Cauchy‘s theorem

P 1k(x, λ) − δ 1k =

1

2πi Γn0P 1k(x, ξ )

−δ 1k

ξ − λ dξ, k = 1, 2, (4.4.60)

where λ ∈ int Γn0, and δ jk is the Kronecker delta.Further, (4.4.39) and (4.4.42) imply

P 11(x, λ) = 1 + (ϕ(x, λ) − ϕ(x, λ))Φ(x, λ) − (Φ(x, λ) − Φ(x, λ))ϕ(x, λ). (4.4.61)

Using (4.4.22), (4.4.23), (4.4.41), (4.4.42), (4.4.57), (4.4.59) and (4.4.61) we infer

|P 1k(x, λ) − δ 1k| ≤ C δ|ρ|−1, ρ ∈ G0δ. (4.4.62)


limn→∞

1

2πi

|ξ|=rn

P 1k(x, ξ ) − δ 1k

ξ − λdξ = 0,

and consequently, (4.4.60) yields

P 1k(x, λ)

−δ 1k = lim

n

→∞

1

2πi Γn1P 1k(x, ξ )

ξ −

λdξ.

Substituting into (4.4.43) we obtain

ϕ(x, λ) = ϕ(x, λ) + limn→∞

1

2πi

Γn1

ϕ(x, λ)P 11(x, ξ ) + ϕ(x, λ)P 12(x, ξ )

λ − ξ dξ.

Taking (4.4.42) into account we calculate

ϕ(x, λ) = ϕ(x, λ) + limn→∞

1

2πi Γn1(ϕ(x, λ)(ϕ(x, ξ )Φ(x, ξ )

−Φ(x, ξ )ϕ

(x, ξ ))+



246

ϕ(x, λ)(Φ(x, ξ )ϕ(x, ξ ) − ϕ(x, ξ )Φ(x, ξ ))dξ

λ − ξ ,

or, in view of (4.4.38),

ϕ(x, λ) = ϕ(x, λ) + limn→∞1

2πi Γn1 ϕ(x, λ), ϕ(x, ξ )

λ − ξ ˆ

M (ξ )ϕ(x, ξ ) dξ. (4.4.63)


Resξ=λkj

ϕ(x, λ), ϕ(x, ξ )λ − ξ

M (ξ )ϕ(x, ξ ) = Qkj (x, λ)ϕkj (x).

Calculating the integral in (4.4.63) by the residue theorem we arrive at (4.4.55). 2

Analogously one can obtain the following relation

Φ(x, λ) = Φ(x, λ) +∞

k=0

(Dk0(x, λ)ϕk0(x) − Dk1(x, λ)ϕk1(x)), (4.4.64)

where

Dkj (x, λ) :=Φ(x, λ), ϕkj (x)

αkj (λ − λkj ).



k=0

(Qni,k0(x)ϕk0(x) − Qni,k1(x)ϕk1(x)). (4.4.65)

Let V be a set of indices u = (n, i), n ≥ 0, i = 0, 1. For each fixed x, we define thevectors

ψ(x) = [ψu(x)]u∈V =

ψn0(x)ψn1(x)

n≥0

, ψ(x) = [ψu(x)]u∈V =

ψn0(x)

ψn1(x)

n≥0

by the formulasψn0(x) = (ϕn0(x) − ϕn1(x))ξ −1

n , ψn1(x) = ϕn1(x),

ψn0(x) = (ϕn0(x) − ϕn1(x))ξ −1n , ψn1(x) = ϕn1(x)

(if ξ n = 0 for a certain n, then we put ψn0(x) = ψn0(x) = 0 ). Similarly we define theblock matrix

H (x) = [H u,v(x)]u,v∈V = H n0,k0(x) H no,k1(x)˜

H n1,k0(x)˜

H n1,k1(x) n,k≥0

,

where u = (n, i), v = (k, j),

H n0,k0(x) = (Qn0,k0(x) − Qn1,k0(x))ξ kξ −1n , H n1,k1(x) = Qn1,k0(x) − Qn1,k1(x),

H n1,k0(x) = Qn1,k0(x)ξ k, H n0,k1(x) = (Qn0,k0(x) − Qn1,k0(x) − Qn0,k1(x) + Qn1,k1(x))ξ −1n .

It follows from (4.4.18)-(4.4.21), (4.4.31), (4.4.36), (4.4.33), (4.4.53), (4.4.54) and Schwarz‘slemma that

|ψ(ν )

nj (x)| ≤ C (|ρ0n| + 1)

ν

, |ψ(ν )

nj (x)| ≤ C (|ρ0n| + 1)

ν

, (4.4.66)



247

|H ni,kj(x)| ≤ Cξ k|ρ0

n − ρ0k| + 1

, |H (ν +1)ni,kj (x)| ≤ C (|ρ0

n| + |ρ0k| + 1)ν ξ k. (4.4.67)

Let us consider the Banach space m of bounded sequences α = [αu]u∈V with the normαm = sup

u∈V |αu|. It follows from (4.4.67) that for each fixed x, the operator E + H (x)

(here E is the identity operator), acting from m to m , is a linear bounded operator, and

H (x) ≤ C supn

k

ξ k|ρ0

n − ρ0k| + 1

< ∞.

Taking into account our notations, we can rewrite (4.4.65) in the form

ψni(x) = ψni(x) +∞

k=0

(H ni,k0(x)ψk0(x) + H ni,k1(x)ψk1(x)),

orψ(x) = (E + H (x))ψ(x). (4.4.68)

Thus, for each fixed x, the vector ψ(x) ∈ m is a solution of equation (4.4.68) in the Banachspace m. Equation (4.4.68) is called the main equation of the inverse problem.

Now let us denote

ε0(x) =∞

k=0 1

αk0

ϕk0(x)ϕk0(x)

−1

αk1

ϕk1(x)ϕk1(x), ε(x) =

−2ε0(x). (4.4.69)

By virtue of (4.4.52), the series converges absolutely and uniformly on [0, a] and [a, T ]; ε0(x)is absolutely continuous on [0, a] and [a, T ], and ε(x) ∈ L2(0, T ).

Lemma 4.4.2 The following relations hold

q (x) = q (x) + ε(x), (4.4.70)

a2

= a2

+ (a−1

1 −a3

1)ε

0(a

−0), (4.4.71)

h = h − ε0(0), H = H + ε0(T ). (4.4.72)

Proof. Differentiating (4.4.55) twice with respect to x and using (4.4.5) we get

ϕ(x, λ) − ε0(x)ϕ(x, λ) = ϕ(x, λ) +∞

k=0

(Qk0(x, λ)ϕk0(x) − Qk1(x, λ)ϕk1(x)), (4.4.73)

ϕ(x, λ) = ϕ(x, λ) +

∞k=0

(Qk0(x, λ)ϕk0(x) − Qk1(x, λ)ϕk1(x))+

2ϕ(x, λ)∞

k=0

1


αk1ϕk1(x)ϕk1(x)

+

∞k=0

1

αk0(ϕ(x, λ)ϕk0(x))ϕk0(x) − 1


.

We replace here the second derivatives, using equation (4.4.1), and then we replace ϕ(x, λ),

using (4.4.55). After canceling terms with ϕ(x, λ) we arrive at (4.4.70).



248

By virtue of (4.4.3), it follows from (4.4.69) that

ε0(a + 0) = a21ε0(a − 0). (4.4.74)

In (4.4.73) we set x = a + 0. Since the functions Qkj (x, λ) are continuous with respect to

x ∈ [0, T ], we calculate using (4.4.3), (4.4.56) and (4.4.74):

a−11 ϕ(a − 0, λ) + a2ϕ(a − 0, λ) − a3

1ε0(a − 0)ϕ(a − 0, λ) = a−11 ϕ(a − 0, λ) + a2ϕ(a − 0, λ)+

a−11

∞k=0

(Qk0(a, λ)ϕk0(a − 0) − Qk1(a, λ)ϕk1(a − 0))+

a2

∞k=0

(Qk0(a, λ)ϕk0(a − 0) − Qk1(a, λ)ϕk1(a − 0)).

Replacing here ϕ(a − 0, λ) and ϕ(a − 0, λ) from (4.4.55) and (4.4.73) respectively, we getafter simple calculations

(a2 − a31ε0(a − 0) + a−1

1 ε0(a − 0) − a2)ϕ(a − 0, λ) = 0,

i.e. (4.4.71) holds. Denote h0 = −h, hT = H, U 0 = U, U T = V. In (4.4.55) and (4.4.73)we put x = 0 and x = T. Then

ϕ(d, λ) + (hd

−ε0(d))ϕ(d, λ) = U d(ϕ(x, λ))+

∞k=0

(Qk0(d, λ)U d(ϕk0) − Qk0(d, λ)U d(ϕk0)), d = 0, T. (4.4.75)

Let d = 0. Since U 0(ϕ(x, λ)) = 0, ϕ(0, λ) = 1, ϕ(0, λ) = −h0, we get h0 − h0 − ε0(0) = 0,i.e. h = h − ε0(0). Let d = T. Since

U T (ϕ(x, λ)) = −∆(λ), U T (ϕk0) = 0, U T (ϕk1) = −∆(λk1),

ϕ(x, λ), ϕ(x, µ) = ϕ(T, µ)∆(λ) − ϕ(T, λ)∆(µ),it follows from (4.4.75) that

ϕ(T, λ) + (hT − ε0(T ))ϕ(T, λ) = −∆(λ) +∞

k=0

ϕk1(T )∆(λ)

αk1(λ − λk1)∆(λk1).

For λ = λn1 this yields

ϕn1(T ) + (hT

−ε0(T ))ϕn1(T ) = ∆(λn1)−

1 + (αn1)−1ϕn1(T )∆1(λn1).

By virtue of (4.4.7) and (4.4.9),

(αn1)−1ϕn1(T )∆1(λn1) = 1,

i.e.ϕn1(T ) + (hT − ε0(T ))ϕn1(T ) = 0.

On the other hand,

ϕn1(T ) + hT ϕn1(T ) = −∆(λn1) = 0.



249

Then (hT − ε0(T ) − hT )ϕn1(T ) = 0, i.e. hT − hT = ε0(T ). 2

Now let us formulate necessary and sufficient conditions for the solvability of the inverseproblem.

Theorem 4.4.5. For real numbers

λn, αn

n≥0 to be the spectral data for a certain

boundary value problem L of the form (4.4.1)-(4.4.3), it is necessary and sufficient that αn > 0, λn = λm (n = m), and there exists L such that (4.4.52) holds. The boundary value problem L can be constructed by (4.4.56), (4.4.70)-(4.4.72).

Proof. We give only a short sketch of the proof of Theorem 4.4.5, since it is similar tothe proof of Theorem 1.6.2 for the classical Sturm-Liouville boundary value problems.

The necessity part of Theorem 4.4.5 was proved above. We prove the sufficiency. Letnumbers λn, αnn≥0 satisfying the hypothesesis of Theorem 4.4.5 be given. We constructψ(x) and H (x), and consider the main equation (4.4.68).

Lemma 4.4.3 For each fixed x ∈ [0, T ], the operator E + H (x), acting from m tom, has a bounded inverse operator, and the main equation (4.4.68) has a unique solution ψ(x) ∈ m.

We omit the proof of Lemma 4.4.3, since the arguments here are the same as in the proof of Lemma 1.6.6.

Let ψ(x) = [ψu(x)]u∈V be the solution of the main equation (4.4.68). Then the functions

ψ(ν )ni (x), ν = 0, 1 are absolutely continuous on [0, a] a n d [a, T ], and (4.4.66) holds. We

define the functions ϕni(x) by the formulae

ϕn1(x) = ψn1(x), ϕn0(x) = ψn0(x)ξ n + ψn1(x).

Then (4.4.65) and (4.4.53) are valid. Furthermore, we construct the functions ϕ(x, λ) andΦ(x, λ) via (4.4.55) and (4.4.64), and the boundary value problem L via (4.4.56), (4.4.69)-(4.4.72). Clearly, ϕni(x) = ϕ(x, λni).

Using (4.4.55), (4.4.64) and (4.4.65) we get analogously to the proof of Lemma 1.6.9,

ϕni(x) = λniϕni(x), ϕ(x, λ) = λϕ(x, λ), Φ(x, λ) = λΦ(x, λ).

Furthermore, from (4.4.55) by differentiation we get (4.4.73). For λ = λni, (4.4.73) implies

ϕni(x) = ϕni(x) + ε0(x)ϕni(x) +∞

k=0

Qni,k0(x)ϕk0(x) − Qni,k1(x)ϕk1(x)

. (4.4.76)

Let us show that the functions ϕ(x, λ) and Φ(x, λ) satisfy the jump conditions (4.4.3).

For this purpose we put we put x = a + 0 in (4.4.65). Using the jump conditions for ϕni(x),we get

a1ϕni(a − 0) = ϕni(a + 0) +∞

k=0

(Qni,k0(a)ϕk0(a + 0) − Qni,k1(a)ϕk1(a + 0)).

Comparing with (4.4.65) for x = a − 0, we infer

gni +∞

k=0

(Qni,k0(a)gk0

−Qni,k1(a)gk1) = 0, gni := ϕni(a + 0)

−a1ϕni(a

−0).



250

According to Lemma 4.4.3 this yields gni = 0, i.e.

ϕni(a + 0) = a1ϕni(a − 0). (4.4.77)

Taking x = a + 0 in (4.4.76) and using (4.4.77) and the jump conditions for ϕni(x), we get

a−11 ϕni(a − 0) + a2ϕni(a − 0) = ϕni(a + 0) + a3

1ε0(a − 0)ϕni(a + 0)+

∞k=0

(Qni,k0(a)ϕk0(a + 0) − Qni,k1(a)ϕk1(a + 0)).

Replacing here ϕni(a − 0) and ϕni(a − 0) from (4.4.65) and (4.4.76) and taking (4.4.71)into account, we obtain

Gni +

∞k=0

(Qni,k0(a)Gk0 − Qni,k1(a)Gk1) = 0, Gni := ϕni(a + 0) − a−1

1 ϕni(a − 0) − a2ϕni(a − 0).

By Lemma 4.4.3 this yields Gni = 0, i.e.

ϕni(a + 0) = a−11 ϕni(a − 0) + a2ϕni(a − 0). (4.4.78)

It follows from (4.4.55), (4.4.71), (4.4.73), (4.4.77), (4.4.78) and the jump conditions forϕ(x, λ) that

ϕ(a + 0, λ) = a1ϕ(a − 0, λ), ϕ(a + 0, λ) = a−11 ϕ(a − 0, λ) + a2ϕ(a − 0, λ).

Similarly one can calculate

Φ(a + 0, λ) = a1Φ(a − 0, λ), Φ(a + 0, λ) = a−11 Φ(a − 0, λ) + a2Φ(a − 0, λ).

Thus, the functions ϕ(x, λ) and Φ(x, λ) satisfy the jump conditions (4.4.3).Let us now show that

ϕ(0, λ) = 1, ϕ(0, λ) = h, U (Φ) = 1, V (Φ) = 0. (4.4.79)

Indeed, using (4.4.55) and (4.4.64) and acting in the same way as in the proof of Lemma4.4.2, we get

U d(ϕ(x, λ)) = U d(ϕ(x, λ)) +∞

k=0

Qk0(d, λ)U d(ϕk0) − Qk1(d, λ)U d(ϕk1)

, (4.4.80)

U d(Φ(x, λ)) = U d(Φ(x, λ)) + ∞k=0

Dk0(d, λ)U d(ϕk0) − Dk1(d, λ)U d(ϕk1), (4.4.81)

where d = 0, T ; U 0 = U, U T = V. Since ϕ(x, λ), ϕkj (x)|x=0 = 0, it follows from (4.4.55)

and (4.4.80) that ϕ(0, λ) = 1, U 0(ϕ) = 0, and consequently ϕ(0, λ) = h. In (4.4.81) weput d = 0. Since U 0(ϕ) = 0 we get U 0(Φ) = 1.

Denote ∆(λ) := −V (ϕ). It follows from (4.4.80) and (4.4.81) for d = T that

∆(λ) = ∆(λ) +∞

k=0 Qk0(T, λ)∆(λk0)

−Qk1(T, λ)∆(λk1), (4.4.82)



251

0 = V (Φ) −∞

k=0

Dk0(T, λ)∆(λk0) − Dk1(T, λ)∆(λk1)

. (4.4.83)

In (4.4.82) we set λ = λn1 :

∆(λn1) +

∞k=0

Qn1,k0(T )∆(λk0) − Qn1,k1(T )∆(λk1) = 0.

Since Qn1,k1(T ) = δ nk we get

∞k=0

Qn1,k0(T )∆(λk0) = 0.

Then, by virtue of Lemma 4.4.3, ∆(λk0) = 0, k ≥ 0. Substituting this into (4.4.83) andusing the relation

Φ(x, λ), ϕ

k1(x)

|x=T = 0, we obtain V (Φ) = 0, and (4.4.79) is proved.

We additionally proved that ∆(λn) = 0, i.e. the numbers λnn≥0 are eigenvalues of L. It follows from (4.4.64) for x = 0 that

M (λ) = M (λ) +∞

k=0

1

αk0(λ − λk0)− 1

αk1(λ − λk1)

.

But according to (4.4.49),

˜M (λ) =

∞

k=0

1

αk1(λ − λk1) .

Hence

M (λ) =∞

k=0

1

αk(λ − λk).

Thus, the numbers λn, αnn≥0 are the spectral data for the constructed boundary valueproblem L. Theorem 4.4.5 is proved. 2

Example 4.4.1. Take L such that q (x) = 0, h = H = a2 = 0. Let λn, αnn≥0 be

the spectral data of L. Clearly,

λ0 = 0, α0 = a + (T − a)a1, ϕ00(x) = 1 (x < a), ϕ00(x) = a1 (x > a).

Let λn = λn (n ≥ 0), αn = αn (n ≥ 1), and let α0 > 0 be an arbitrary positive

number. Denote A :=1

α0

− 1

α0. Then, (4.4.65) and (4.4.69) yield

ϕ00(x) = ϕ00(x)1 + A x

0ϕ2

00(t) dt, ε0(x) = Aϕ00(x)ϕ00(x),

and consequently,

ϕ00(x) =

(1 + Ax)−1, x < a,a1(B + Aa2

1x)−1, x > a,ε0(x) =

A(1 + Ax)−1, x < a,Aa2

1(B + Aa21x)−1, x > a,

where B = 1 + Aa − Aa21a. Using (4.4.70)-(4.4.72) we calculate

q (x) = 2A2(1 + Ax)−2, x < a,

2A2

a41(B + Aa

21x)−

2

, x > a,



252

h = −A, H = Aa21(B + Aa2

1T )−1, a2 = (a−11 − a3

1)A(1 + Aa)−1.

Remarks. 1) Analogous results are also valid for boundary value problems having anarbitrary number of discontinuities.

2) In Theorem 4.4.5, (4.4.52) is a condition on the asymptotic behaviour of the spectral

data. It follows from the proof that Theorem 4.4.5 is also valid if instead of (4.4.52) we takeξ nρn ∈ l2.3) In order to choose a model boundary value problem L, one can use the asymptotics

of the spectral data. We note that in applications a model boundary value problem often isgiven a priori. It describes a known object, and its perturbations are studied.

4) For boundary value problems without discontinuities (i.e. a1 = 1, a2 = 0 ), Theorem4.4.5 gives the following assertion as a corollary:

For real numbers λn, αnn≥0 to be the spectral data for a certain boundary value

problem L of the form (4.4.1)-(4.4.2), it is necessary and sufficient that αn > 0, λn =λm (n = m), and

ρn =πn

T +

ω

n+

κn

n, αn =

T

2+

κn1

n, κn, κn1 ∈ 2.

5) This method also works for nonselfadjoint boundary value problems and for moregeneral jump conditions.

4.5. INVERSE PROBLEMS IN ELASTICITY THEORY

The problem of determining the dimensions of the transverse cross-sections of a beamfrom the given frequencies of its natural vibrations is examined. Frequency spectra areindicated that determine the dimensions of the transverse cross-sections of the beam uniquely,an effective procedure is presented for solving the inverse problem, and a uniqueness theoremis proved.

4.5.1. Consider the differential equation describing beam vibrations in the form(hµ(x)y) = λh(x)y, 0 ≤ x ≤ T. (4.5.1)

Here h(x) is a function characterizing the beam transverse section, and µ = 1, 2, 3 isa fixed number. We will assume that the function h(x) is absolutely continuous in thesegment [0, T ], and h(x) > 0, h(0) = 1. The inverse problem for (4.5.1) in the case µ = 2(similar transverse sections) was investigated in [ain1] in determining small changes in thebeam transverse sections from given small changes in a finite number of its natural vibration

frequencies.

Let λkjk≥1, j = 1, 2 be the eigenvalues of the boundary value problems Q j for (4.5.1)with the boundary conditions

y(0) = y( j)(0) = y(T ) = y(T ) = 0.

The inverse problem is formulated as follows.

Inverse Problem 4.5.1. Find the function h(x), x ∈ [0, T ] from the given spectra

λkjk≥1, j=1,2.



253

Let us show that this inverse problem can be reduced to the inverse problem of recoveringthe differential equation (4.5.1) from the corresponding Weyl function. Let Φ(x, λ) be asolution of (4.5.1) under the conditions

Φ(0, λ) = Φ(T, λ) = Φ(T, λ) = 0, Φ(0, λ) = 1.

We set M (λ) : = Φ(0, λ). The function M (λ) is called the Weyl function for (4.5.1).Let the functions C ν (x, λ), ν = 0, 3 be solutions of (4.5.1) under the initial conditionsC (µ)

ν (0, λ) = δ νµ , ν,µ = 0, 3. Denote

∆ j(λ) = C 3− j (T, λ)C 3(T, λ) − C 3(T, λ)C 3− j(T, λ), j = 1, 2.

Then

Φ(x, λ) =1

∆1(λ)

det[C ν (x, λ), C ν (T, λ), C ν (T, λ)]ν =1,2,3,

and hence

M (λ) = −∆2(λ)

∆1(λ).

Denoteγ (x) =

x

0(h(t))(1−µ)/4 dt, τ = γ (T ).

Let λ = ρ4. By the well-known method (see, for example [nai1, Ch.1]) one can obtain thefollowing asymptotic formulae

λkj =kπ

τ

41 +

A j1

k+ O

1

k2

, k → ∞, (4.5.2)

∆ j(λ) = O(ρ j−5 exp(C |ρ|)), |λ| → ∞. (4.5.3)

For definiteness, let ρ ∈ S := ρ : arg ρ ∈ [0, π/4]. Then, for |λ| → ∞, ρ ∈ S, arg λ =ϕ = 0,

∆ j(λ) = ρ j−5A j2 exp(ρ(1 − i)τ )1 + O1

ρ, (4.5.4)

Φ(ν )(x, λ) = ρν −12

ξ=1

(Rξγ (x))ν gξ(x)exp(ρRξγ (x))

1 + O1

ρ

, x ∈ [0, T ], (4.5.5)

where R1 = −1, R2 = i; the functions gξ(x) are absolutely continuous, and gξ(x) =0, g1(0) = −g2(0) = (−1 − i)−1. The numbers A js in (4.5.2) and (4.5.4) depend on τ. Inparticular, (4.5.5) yields

M (λ) = ρ(−1 + i)

1 + O

1

ρ.

Lemma 4.5.1. The Weyl function M (λ) is uniquely defined by giving the spectra λkjk≥1, j=1,2 .

Proof. The eigenvalues λkjk≥1, j=1,2 of the boundary value problems Q j coincidewith the zeros of the entire functions ∆ j(λ). Indeed, let λ0 be an eigenvalue and ψ(x) aneigenfunction of the boundary value problem Q j. Then

ψ(x) =3

µ=0

β µC µ(x, λ0),



254

where3

µ=0

β µC (k)µ (0, λ0) = 0,

3µ=0

β µC (s)µ (T, λ0), k = 0, j s = 0, 1.

Since ψ = 0, this linear homogeneous algebraic system has non-zero solutions and, therefore,

its determinant equals zero, i.e. ∆ j(λ

0

) = 0. Repeating all arguments in reverse order, weobtain that if ∆ j(λ0) = 0, then λ0 is an eigenvalue of the boundary value problem Q j.It follows from (4.5.3) that the order of the entire functions ∆ j(λ) is less than 1, and

therefore, according to Hadamard’s factorization theorem [con1, p.289]

∆ j(λ) = B j

∞k=1

1 − λ

λkj

, (4.5.6)

with some constants B j. Let us consider a positive function h(x) ∈ AC [0, T ] such that

h(0) = 1. As before we agree that if a certain symbol α denotes an object related to h,then the corresponding symbol α with tilde denotes the analogous object related to h, andα := α − α.

Let τ = τ . We have from (4.5.6)

∆ j(λ)

∆ j(λ)=

B j

B j

∞k=1

λkj

λkj

∞k=1

1 − λkj − λkj

λkj − λ

.

By virtue of (4.5.2) and (4.5.4), we have for

|λ

| → ∞, arg λ = ϕ

= 0, ρ

∈S,

∆ j(λ)

∆ j(λ)→ 1,

∞k=1

1 − λkj − λkj

λkj − λ

→ 1.

This yieldsB j

B j

∞k=1

λkj

λkj= 1,

and therefore,

M (λ) = B ∞k=1

λk1

λk2

· λk2 − λλk1 − λ

, B = −B2

B1

.

Hence the specification of the spectra λkjk≥1, j=1,2 uniquely determines the Weyl functionM (λ). 2

Thus, Inverse Problem 4.5.1 is reduced to the following inverse problem.

Inverse Problem 4.5.2. Given the Weyl function M (λ), find h(x), x ∈ [0, T ].

4.5.2. We shall solve Inverse Problem 4.5.2 by the method of standard models. Firstwe prove several auxiliary assertions.

Lemma 4.5.2. Let p(x) := hµ(x). The following relationship holds T

0

h(x)λΦ(x, λ)Φ(x, λ) − ˆ p(x)Φ(x, λ)Φ(x, λ)

dx = M (λ). (4.5.7)

Proof. Denote

λy = ( p(x)y) − λh(x)y, L(y, z ) = ( p(x)y)z − p(x)yz + p(x)yz − y( p(x)z ).



255

Then T

0λy(x)z (x) dx = L(y(x), z (x))

T

0+ T

0y(x)λz (x) dx. (4.5.8)

Using (4.5.8) and the equality λΦ(x, λ) = λΦ(x, λ) = 0, we obtain

T

0(λ − λ)Φ(x, λ)Φ(x, λ) dx = − T

0λΦ(x, λ)Φ(x, λ) dx

= −L(Φ(x, λ), Φ(x, λ))T

0− T

0Φ(x, λ)λΦ(x, λ) dx

= Φ(0, λ)Φ(0, λ) − Φ(0, λ)Φ(0, λ) = −M (λ).

On the other hand, integrating by parts we have

T

0 (λ − λ)Φ(x, λ)Φ(x, λ) dx = T

0 ((ˆ p(x)Φ(x, λ)) − λh(x)Φ(x, λ))Φ(x, λ) dx

= ((ˆ p(x)Φ(x, λ))Φ(x, λ) − ˆ p(x)Φ(x, λ)Φ(x, λ))T

0

+ T

0(ˆ p(x)Φ(x, λ)Φ(x, λ) − λh(x)Φ(x, λ))Φ(x, λ)) dx.

Since the substitution vanishes, we hence obtain (4.5.7). 2

Lemma 4.5.3. Consider the integral

J (z ) = T

0f (x)H (x, z ) dx, (4.5.9)

where

f (x) ∈ C [0, T ], f (x) ∼ f αxα

α!(x → +0), H (x, z ) = exp(−za(x))

1 +

ξ (x, z )

z

,

a(x) ∈ C 1[0, T ], 0 < a(x1) < a(x2) (0 < x1 < x2),

a(ν )(x) ∼ a0x1−ν (x → +0, ν = 0, 1), a(x) > 0,

the function ξ (x, z ) is continuous and bounded for x ∈ [0, T ],

z ∈ Q :=

z : arg z ∈− π

2+ δ 0,

π

2− δ 0

, δ 0 > 0

.

Then as z → ∞, z ∈ Q,J (z ) ∼ f α(a0z )−α−1. (4.5.10)

Proof. The function t = a(x) has the inverse x = b(t), where b(t) ∈ C 1

[0, T 1], T 1 =a(T ), b(t) > 0 for t > 0, and b(ν )(t) ∼ (a0)−1t1−ν , ν = 0, 1 as t → +0. Let us make thechange of variable t = a(x) in the integral in (4.5.9). We obtain

J (z ) = T

0g(t)exp(−zt)(1 + z −1ξ (b(t), z )) dt, (4.5.11)

where g(t) = f (b(t))b(t). It is clear that for t → +0,

g(t) ∼ (a0)−1

−α

f α

tα

α! .



256

Applying Lemma 1.7.1 to (4.5.11), we obtain (4.5.10). 2

Denote

Aα =1

(R1

−R2)2

2

ξ,s=1

(−1)ξ+s(1 − µR2ξ R2

s)

(Rξ + Rs)α+1, α ≥ 1.

Let us show that Aα = 0 for all α ≥ 1. For definiteness, we put arg ρ ∈ (0, π4

), i.e.R1 = −1, R2 = i. Then

Aα = − 1

(R1 − R2)2· 1

(−2)α+1aα,

whereaα = (µ − 1)(1 + iα+1) + 2(µ + 1)(1 + i)α+1.

Since |1 + iα+1| ≤ 2 and |1 + i|α+1 = (√

2)α+1, it follows that aα = 0 for all α ≥ 1. Hence

Aα = 0 for all α ≥ 1. 2

Lemma 4.5.3. As x → 0 let h(x) ∼ hα(α!)−1xα. Then as |λ| → ∞, arg λ = ϕ = 0,there exists a finite limit I α = lim ρα−1M (λ), and

Aαhα = I α. (4.5.12)

Proof. Since p(x) = hµ(x), then by virtue of the conditions of the lemma we have

ˆ p(x) ∼ µhα xα

α!, x → +0.

Using the asymptotic formulae (4.5.5) and Lemma 4.5.3 we find as λ → ∞, arg λ = ϕ =0, ρ ∈ S :

T

0h(x)λΦ(x, λ)Φ(x, λ) dx ∼ 1

ρα−1· hα

(R1 − R2)2

2ξ,s=1

(−1)ξ+s

(Rξ + Rs)α+1,

T

0ˆ p(x)Φ(x, λ)Φ(x, λ) dx ∼ 1

ρα−1· µhα

(R1 − R2)2

2ξ,s=1

(−1)ξ+sR2ξ R2s

(Rξ + Rs)α+1.

Substituting the expressions obtained in (4.5.7), we obtain the assertion of the lemma. 2

Let A be the set of analytic functions on [0, T ]. From the facts presented above wehave the following theorem.

Theorem 4.5.1. Inverse Problem 4.5.2 has a unique solution in the class h(x) ∈ A.This solution can be found according to the following algorithm:

1) We calculate hα = h(α)(0), α ≥ 0, h0 = 1. For this purpose we successively perform the following operations for α = 1, 2, . . . : we construct the function h(x) ∈ A, h(x) > 0such that h(ν )(0) = hν , ν = 0, α − 1, and arbitrary in the rest, and we calculate hα by (4.5.12).

We construct the function h(x) by the formula

h(x) =∞

α=0

hαxα

α!

, 0 < x < R,



257

where

R =

limα→∞

α

|hα|α!

−1.

If R < T, then for R < x < T the function h(x) is constructed by analytic continuation.

We note that the inverse problem in the class of piecewise-analytic function can be solvedin an analogous manner.

4.6. BOUNDARY VALUE PROBLEMS WITH AFTEREFFECT

In this section, the perturbation of the Sturm-Liouville operator by a Volterra integraloperator is considered. The presence of an ”aftereffect” in a mathematical model producesqualitative changes in the study of the inverse problem. The main results of the section areexpressed by Theorems 4.6.1-4.6.3. We use here several methods which were described inChapter 1. In order to prove the uniqueness theorem (Theorem 4.6.1) the transformationoperator method is applied. With the help of the method of standard models we get analgorithm for the solution of the inverse problem (Theorem 4.6.2), and the Borg method isused for the local solvability of the inverse problem and for the study the stability of thesolution (Theorem 4.6.3).

4.6.1. Let

λn

n≥1 be the eigenvalues of a boundary value problem L = L(q, M ) of

the form

y(x) ≡ −y(x) + q (x)y(x) + x

0M (x − t)y(t) dt = λy(x) = ρ2y(x), (4.6.1)

y(0) = y(π) = 0. (4.6.2)

Considered the following problem:

Inverse Problem 4.6.1. Given the function q (x) and the spectrum λnn≥1, find the function M (x) .

Put

M 0(x) = (π − x)M (x), M 1(x) = x

0M (t) dt, Q(x) = M 0(x) − M 1(x).

We shall assume that q (x), Q(x) ∈ L2(0, π), M k(x) ∈ L(0, π), k = 0, 1.Let S (x, λ) be the solution of (4.6.1) under the initial conditions S (0, λ) = 0, S (0, λ) =

1.

Lemma 4.6.1. The representation

S (x, λ) =sin ρx

ρ+ x

0K (x, t)

sin ρt

ρdt, (4.6.3)

holds, where K (x, t) it is a continuous function, and K (x, 0) = 0.

Proof. The function S (x, λ) is the solution of the integral equation

S (x, λ) =sin ρx

ρ+

x

0

sin ρ(x − τ )

ρ q (τ )S (τ, λ) + τ

0

M (τ −

s)S (s, λ) ds dτ. (4.6.4)



258

Since x

0

sin ρ(x − τ )

ρf (τ ) dτ =

x

0

t

0f (τ )cos ρ(t − τ ) dτ

dt,

then (4.6.4) is transformable to the form

S (x, λ) = sin ρxρ

+ x

0

t

0

q (τ )S (τ, λ) +

τ

0M (τ − s)S (s, λ) ds

cos ρ(t − τ ) dτdt. (4.6.5)

Apply the method of successive approximations to solve the equation (4.6.5):

S 0(x, λ) =sin ρx

ρ,

S n+1(x, λ) = x

0 t

0 q (τ )S n(τ, λ) + τ

0

M (τ

−s)S n(s, λ) ds cos ρ(t

−τ ) dτdt.

Transform S 1(x, λ) :

S 1(x, λ) =1

2ρ

x

0sin ρt

t

0q (τ ) dτ

dt +

1

2ρ

x

0

t

0q (τ )sin ρ(2τ − t) dτ

dt

+1

2ρ

x

0

t

0

τ

0M (τ −s)sin ρ(s+t−τ ) dsdτ dt+

1

2ρ

x

0

t

0

τ

0M (τ −s)sin ρ(s−t+τ ) ds dτ dt.

Carrying out the change of variables ξ = 2τ

−t, ξ = s + t

−τ, ξ = s

−t + τ, respectively,

in the last three integrals and reversing the order of integration, we obtain

S 1(x, λ) = x

0K 1(x, ξ )

sin ρξ

ρdξ,

where

K 1(x, ξ ) =1

2

ξ

0q (τ ) dτ +

1

4

x

ξ

q ξ + t

2

− q

t − ξ

2

dt

+

1

2 x

ξξM (t − ξ ) + t

ξ+t2

M (2τ − ξ − t) dτ − t−ξ

t−ξ2

M (2τ + ξ − t) dτ dt. (4.6.6)

Clearly K 1(x, 0) = 0 . In an analogous manner we compute

S n(x, λ) = x

0K n(x, ξ )

sin ρξ

ρdξ,

where the functions K n(x, ξ ) are determined by the recurrence formula

K n+1(x, ξ ) =1

2 x

ξ t

t−ξ q (τ )K n(τ, ξ + τ − t) dτ + t

ξ+t2

q (τ )K n(τ, ξ + t − τ ) dτ

− t−ξ

t−ξ2

q (τ )K n(τ, −ξ + t − τ ) dτ + t

t−ξ

τ

ξ−t+τ M (τ − s)K n(s, ξ − t + τ ) ds

dτ

+ ξ

t+ξ2

τ

ξ+t−τ M (τ − s)K n(s, ξ + t − τ ) ds

dτ

− t−ξ

t−ξ2 τ

−ξ+t−τ

M (τ −

s)K n(s,−

ξ + t−

τ ) ds dτ dt. (4.6.7)



259

Clearly, K n(x, 0) = 0 . From (4.6.6) and (4.6.7) we obtain by induction the estimate

|K n(x, ξ )| ≤ 1

n!(Cx)n, 0 ≤ ξ ≤ x ≤ π.

Thus,

S (x, λ) =∞

n=0

S n(x, λ) = sin ρxρ

+ x

0K (x, ξ )sin ρξ

ρdξ,

where

K (x, ξ ) =∞

n=1

K n(x, ξ ), (4.6.8)

and the series (4.6.8) converges absolutely and uniformly for 0 ≤ ξ ≤ x ≤ π . Lemma 4.6.1is proved. 2

Denote ∆(λ) = S (π, λ). The eigenvalues λnn≥1 of the boundary value problem Lcoincide with the zeros of the function ∆(λ), and like in the proof of Theorem 1.1.3 we getfor n → ∞,

ρn =

λn = n +A1

n+

κn

n, κn ∈ 2, A1 =

1

2π

π

0q (t) dt. (4.6.9)

The following assertion can be proved like Theorem 1.1.4.

Lemma 4.6.2. The function ∆(λ) is uniquely determined by its zeros. Moreover,

∆(λ) = π∞

n=1

λn − λ

n2. (4.6.10)

We shall now prove the uniqueness theorem for the solution of Inverse Problem 4.6.1.Let λnn≥1 be the eigenvalues of the boundary value problem L = L(q, M ) .

Theorem 4.6.1. If λn = λn, n ≥ 1, then M (x) = M (x), x ∈ (0, π).

Proof. Let the function S ∗(x, λ) be the solution of the equation

∗z (x) := −z (x) + q (x)z (x) + π

xM (t − x)z (t) dt = λz (x) (4.6.11)

under the conditions S ∗(π, λ) = 0, S ∗(π, λ) = −1. Put ∆∗(λ) = S ∗(0, λ), and denoteM (x) := M (x)

−M (x). Then π

0S ∗(x, λ)

x

0M (x − t)S (t, λ) dtdx

= π

0S ∗(x, λ)S (x, λ) dx −

π

0S ∗(x, λ)S (x, λ) dx

= π

0∗S ∗(x, λ)S (x, λ) dx −

π

0S ∗(x, λ)S (x, λ) dx

+ ˜S (x, λ)S ∗(x, λ) −

˜S (x, λ)S ∗(x, λ)

π

0 = ∆∗(λ) −˜∆(λ).



260

For = we have ∆∗(λ) ≡ ∆(λ), and consequently π

0S ∗(x, λ)

x

0M (x − t)S (t, λ) dtdx = ∆(λ). (4.6.12)

Transform (4.6.12) into π

0M (x)

π

xS ∗(t, λ)S (t − x, λ) dt

dx = ∆(λ). (4.6.13)

Denote w(x, λ) = S ∗(π − x, λ), N (x) = M (π − x),

ϕ(x, λ) = x

0w(t, λ)S (t − x, λ) dt. (4.6.14)

Then (4.6.13) takes the form π

0N (x)ϕ(x, λ) dx = ∆(λ). (4.6.15)

Lemma 4.6.3. The representation

ϕ(x, λ) =1

2ρ2− x cos ρx +

x

0V (x, t)cos ρtdt

(4.6.16)

holds, where V (x, t) is a continuous function.

Proof. Since w(x, λ) = S ∗(π −x, λ), the function w(x, λ) is the solution of the Cauchyproblem

−w(x, λ) + q (π − x)w(x, λ) + x

0M (x − t)w(t, λ) dt = λw(x, λ),

w(0, λ) = 0, w(0, λ) = 1.

Therefore, by Lemma 4.6.1, the representation

w(x, λ) =sin ρx

ρ+ x

0K 0(x, t)

sin ρt

ρdt (4.6.17)

holds, where K 0(x, t) is a continuous function. Substituting (4.6.3) and (4.6.17) into(4.6.14), we obtain

ϕ(x, λ) = ϕ1(x, λ) + ϕ2(x, λ) + ϕ3(x, λ) + ϕ4(x, λ),

whereϕ1(x, λ) =

1

ρ2

x

0sin ρt sin ρ(x − t) dt,

ϕ2(x, λ) =1

ρ2

x

0sin ρ(x − t)

t

0K 0(t, ξ )sin ρξ dξ

dt,

ϕ3(x, λ) =1

ρ2

x

0sin ρt

x−t

0K (x − t, η)sin ρη dη

dt,

ϕ4(x, λ) =1

ρ2 x

0 t

0

K 0(t, ξ )sin ρξ x−t

0

K (x−

t, η)sin ρη dη dξ dt.



261

For ϕ1(x, λ) we have

ϕ1(x, λ) =1

2ρ2

x

0

cos ρ(x − 2t) − cos ρx

dt =

1

2ρ2

− x cos ρx +

x

0cos ρtdt

.

Transform ϕ2(x, λ) :

ϕ2(x, λ) =1

2ρ2

x

0

t

0K 0(t, ξ )

cos ρ(x − t − ξ ) − cos ρ(x − t + ξ )

dξ dt

=1

2ρ2

x

0

x−t

x−2tK 0(t, x − t − s)cos ρsds −

x

x−tK 0(t, s + t − x)cos ρsds

dt.

Changing the order of integration, we obtain

ϕ2(x, λ) =1

2ρ2

x

0V 2(x, t)cos ρtdt,

where

V 2(x, t) = x−t

x−t2

K 0(s, x − t − s) ds + x

x+t2

K 0(s, x + t − s) ds − x

x−tK 0(s, t + s − x) ds.

The integrals ϕ3(x, λ) and ϕ4(x, λ) are transformable in an analogous manner. Lemma4.6.3 is proved. 2

Let us return to proving Theorem 4.6.1. Since λn = λn, n ≥ 1, we have by Lemma4.6.2

∆(λ) ≡ ∆(λ).

Then, substituting (4.6.16) into (4.6.15), we obtain

π

0cos ρx

−xN (x) +

π

xV (t, x)N (t) dt

dx ≡ 0,

and consequently,

−xN (x) + π

xV (t, x)N (t) dt = 0. (4.6.18)

For each fixed ε > 0, (4.6.18) is a homogeneous Volterra integral equation of the secondkind in the interval (ε, π). Consequently, N (x) = 0 a.e. in (ε, π) and, since ε is arbitrary,this holds in the whole interval (0, π). Thus, M (x) = M (x) a.e. in (0, π). 2

4.6.2. Relation (4.6.15) also makes it possible to obtain an algorithm for solving Inverse

Problem 4.6.1 in the case when M (x) ∈ P A[0, π] (i.e. M is piecewise analytic in [0, π] ).Indeed, consider the boundary value problems L(q, M ) and L(q, M ), and assume thatq (x) ∈ L2(0, π); M (x), M (x) ∈ P A[0, π]. Let for some fixed a > 0

N (x) = 0, x ∈ (a, π); N (x) ∼ N aα(α!)−1(a − x)α, x → a − 0. (4.6.19)

It follows from (4.6.16) that as |ρ| → ∞, arg ρ ∈ [δ, π − δ ], x ∈ [ε, π], δ > 0, ε > 0, theasymptotic formula

ϕ(x, λ) =−

x

4ρ2

exp(−

iρx)1 + O1

ρ (4.6.20)



262

holds. Furthermore, we infer from (4.6.16) that

|ϕ(x, λ)| < C |ρ−2 exp(−iρx)|, x ∈ [0, π], Im ρ ≥ 0. (4.6.21)

With (4.6.21) we obtain the estimate

ε

0N (x)ϕ(x, λ) dx

< C |ρ−2 exp(−iρε)|, Im ρ ≥ 0. (4.6.22)

Using (4.6.19), (4.6.20) and Lemma 1.7.1, we get for |ρ| → ∞, a

εN (x)ϕ(x, λ) dx =

a

4(−iρ)α+3exp(−iρa)(N aα + o(1)). (4.6.23)

Since N (x) = 0 for x ∈ (a, π), it follows from (4.6.15), (4.6.22) and (4.6.23) that for

|ρ| → ∞, arg ρ ∈ [δ, π − δ ],

∆(λ) =a

4(−iρ)−α−3 exp(−iρa)(N aα + o(1)),

and consequently

N aα =4

alim ∆(λ)(−iρ)α+3 exp(iρa), |ρ| → ∞, arg ρ ∈ [δ, π − δ ]. (4.6.24)

Thus we have proved the following theorem.

Theorem 4.6.2. Let λnn≥1 be the eigenvalues of L(q, M ), where q (x) ∈ L2(0, π) ,M (x) ∈ P A[0, π]. Then the solution of Inverse Problem 4.6.1 can be found by the following algorithm:1) From λnn≥1 construct the function ∆(λ) by (4.6.10).2) Take a = π.3) For α = 0, 1, 2, . . . carry out successively the following operations: construct a function M (x) ∈ P A[0, π] such that N (x) = 0, x ∈ (a, π); N (k)(a − 0) = 0, k = 0, α − 1 and find N a

α= (

−1)αN (α)(a

−0) from (4.6.24).

4) Construct N (x) for x ∈ (a+, a) by the formula

N (x) =∞

α=0

N aα(a − x)α

α!.

5) If a+ > 0 set a := a+ and go to step 3.

4.6.3. We shall now investigate local solvability of Inverse Problem 4.6.1 and itsstability. First let us prove an auxiliary assertion.

Lemma 4.6.4. Consider in a Banach space B the nonlinear equation

r = f +∞

j=2

ψ j(r), (4.6.25)

where

ψ j (r) ≤ (C r) j

, ψ j(r) − ψ j(r∗) ≤ r − r∗(C max(r, r∗)) j−

1

.



263

There exists δ > 0 such that if f < δ, then in the ball r < 2δ equation (4.6.25) has a unique solution r ∈ B, for which r ≤ 2f .

Proof. Assume that C ≥ 1. Put

ψ(r) =

∞ j=2 ψ j(r), C 0 = 2C

2

, δ =

1

4C 0 .

If r, r∗ ≤ (2C 0)−1, then

ψ(r) ≤∞

j=2

(C r) j ≤ C 0r2 ≤ 1

2r,

ψ(r) − ψ(r∗) ≤ r − r∗∞

j+2

(C (2C 0)−1) j−1 ≤ 1

2r − r ∗ .

(4.6.26)

Let f ≤ δ ; construct

r0 = f, rk+1 = f + ψ(rk), k ≥ 0.

By induction, using (4.6.26), we obtain the estimates

rk ≤ 2f , rk+1 − rk ≤ 1

2k+1f , k ≥ 0.

Concequently, the series

r = r0 +∞

k=0

(rk+1 − rk)

converges to the solution of (4.6.25), and r ≤ 2f . Lemma 4.6.4 is proved. 2

Theorem 4.6.3. For the boundary value problem L = L(q, M ) with the spectrum λnn≥1 there exists δ > 0 (which depends on L ) such that if numbers λnn≥1 satisfy the condition

Λ := ∞

n=1

|λn − λn|21/2

< δ,

then there exists a unique L = L(q, M ) for which the numbers λnn≥1 are the eigenvalues,and, furthermore,

Q(x) − Q(x)L2(0,π) ≤ C Λ,

M k(x) − M k(x)L(0,π) ≤ C Λ, k = 0, 1.

Here and below, C denotes various constants depending on L.

Proof. For brevity, we confine ourselves to the case when all eigenvalues are simple.The Cauchy problem y(x) − λy(x) + f (x) = 0, y(0) = y(0) = 0 has a unique solution

y(x) = x

0g(x,t,λ)f (t) dt,

where g(x,t,λ) is the Green function satisfying the relations

−gxx(x,t,λ) + q (x)g(x,t,λ) − λg(x,t,λ) + x

t M (x − τ )g(τ, t , λ) dτ = 0, x > t,



265

Due to (4.6.30), the functions ψn0(x)n≥1 constitute a Riesz basis in L2(0, π). Denote

ψ∗∗n0(x) = sn(x) +

π

xK (t, x)sn(t) dt. (4.6.31)

It follows from (4.6.29)-(4.6.31) that π

0ψn0(x)ψ∗∗

m0(x) dx = π

0ψn0

sm(x) +

π

xK (t, x)sm(t) dt

dx

= π

0sm(x)

ψn0(x) +

x

0K (x, t)ψn0(t) dt

dx =

π

0sn(x)sm(x) dx = δ nmαn. (4.6.32)

Furthermore, we compute

ψn(x) =n

x x

0w(t, λn)S (x

−t, λn) dt. (4.6.33)

SinceS (x, λ) = cos ρx +

x

0K 1(x, t)cos ρt dt, (4.6.34)

we obtain, substituting (4.6.34) and (4.6.17) into (4.6.33) like in the proof of Lemma 4.6.3,

ψn(x) = ψn0(x) + x

0V 0(x, t)ψn0(t) dt, (4.6.35)

where V 0(x, t) is a continuous function with V 0(x, 0) = 0. Solving the integral equation(4.6.35), we find

ψn0(x) = ψn(x) + x

0V 1(x, t)ψn(t) dt, V 1(x, 0) = 0. (4.6.36)

Consider the functions

ψ∗∗n (x) = ψ∗∗

n0(x) + π

xV 1(t, x)ψ∗∗

n0(t) dt. (4.6.37)

It follows from (4.6.32), (4.6.36) and (4.6.37) that π

0ψn(x)ψ∗∗

m (x) dx = δ nmαn. (4.6.38)

By virtue of (4.6.35) and (4.6.38), the functions ψn(x)n≥1 constitute a Riesz basis inL2(0, π), and the biorthogonal basis ψ∗

n(x)n≥1 has the form ψ∗n(x) = α−1

n ψ∗∗n (x). Substi-

tuting (4.6.31) into (4.6.37), we have

ψ∗∗n (x) = sn(x) + π

x V

0

1 (t, x)sn(t) dt, V

0

1 (t, 0) = 0.

Hence we obtain the required properties of the biorthogonal basis. 2

Since ηn(x) = ψn(π − x), Lemma 4.6.5 implies

Corollary 4.6.1. The function ηn(x)n≥1 constitute a Riesz basis in L2(0, π), and the biorthogonal basis χn(x)n≥1 has the properties:1) χ∗n(x) ∈ W 120;

2) |χn(x)| ≤ C, n ≥ 1, x ∈ [0, π];



266

3) for any θn ∈ 2

θ(x) =∞

n=1

θn

nχn(x) ∈ W 120, θ(x)W 1

2≤ C

∞n=1

θn2

1/2

.

Let us return to proving Theorem 4.6.3. Put

ε(x) =∞

n=1

εn

nχn(x). (4.6.39)

Using Lemma 4.6.1, the relations ∆(λ) = S (π, λ), ∆(λn) = 0, and formulae (4.6.9),(4.6.27), we obtain the estimate

|εn

|= n2

|∆(λn)

−∆(λn

| ≤C

|λn

−λn

|.

Now by Corollary 4.6.1 we have

ε(x) ∈ W 120, ε(x)W 12

≤ C Λ.

Consider in W 120 the nonlinear equation

r = ε +∞

j=2

ψ j(r), (4.6.40)

where ε(x) is defined by (4.6.39), and the operators z j = ψ j(r) act from W 120 to W 120

according to the formula

z j(x) = −∞

n=1

π

0. . .

π

0

j

r(t1) . . . r(t j )Bnj(t1, . . . , t j) dt1 . . . d t j

χn(x),

where

r(x) ∈ W 120, Bnj (t1, . . . , t j) = n(π − t1) . . . (π − t j)

× π

0. . .

π

0 j

vn(t1, s1)Gn(s1, t2, s2) . . . Gn(s j−1, t j, s j)yn(s j) ds1 . . . d s j

andψ j(r)W 1

2≤ (C rW 1

2) j,

ψ j(r) − ψ j(r∗)W 1

2 ≤ r − r∗W 1

2 (C max(rW 1

2 , r∗W 1

2 ))

j

−1

.By Lemma 4.6.4, there exists δ > 0 such that for Λ < δ equation (4.6.40) has a solutionr(x) ∈ W 120, r(x)W 1

20≤ C Λ. Put M (x) = M (x) − ((π − x)−1r(x)), and consider the

boundary value problem L = L(q, M ). Clearly

Q(x) = Q(x) − r(x) ∈ L2(0, π), Q(x) − Q(x)L2(0,π) ≤ C Λ.

Sinceˆ

M 1(x) = −1

π − x π

x

ˆQ(t) dt,

ˆM 0(x) =

ˆQ(x) +

ˆM 1(x),



267

we haveM k(x) − M k(x)L2(0,π) ≤ C Λ, k = 0, 1.

It remains to show that the numbers λn≥1 are the eigenvalues of the problem L. To dothis, consider the functions yn(x) which are solutions of the integral equations

yn(x) = yn(x) + π

0M 1(t)

π

0Gn(x,t,s)yn(s) ds

dt (4.6.41)

or, which is the same,

yn(x) = yn(x) + x

0M 1(t)

x−t

0G(x, s + t, λn)yn(s) ds

dt. (4.6.42)

After integration by parts, (4.6.42) takes the form

yn(x) = yn(x) − x

0M 1(t)

x

tg(x,s, λn)yn(s − t) dsdt.

Reverse the integration order:

yn(x) = yn(x) − x

0g(x,t, λn)

t

0M 1(s)yn(t − s) dsdt.

Integrate by parts:

yn(x) = yn(x) − x

0g(x,t, λn)

t

0M (t − s)yn(s) dsdt. (4.6.43)


(yn(x) − yn(x)) = x

0M (t − s)yn(s) ds = ( − )yn(x),

and consequently,yn(x)

−λnyn(x), yn(0) = (0), yn(0) = 1.

Since the solution of the Cauchy problem is unique, we have yn(x) = S (x, λn).Write (4.6.13) in the form

π

0M (x)

π−x

0w(π − x − t, λ)S (t, λ) dt

dx = ∆(λ).

Integrating by parts, we obtain for λ = λn

π

0 M 1(x) π

0 vn(x, t)yn(t) dt dx = ∆(λn). (4.6.44)

Solving (4.6.41) by the method of successive approximations, we have

yn(x) = yn(x) + Y n(x), (4.6.45)

where

Y n(x) =∞

j=1

π

0. . .

π

0 j

M 1(t1) . . . M 1(t j) π

0. . .

π

0

j

Gn(x, t1, s1)



268

×Gn(s1, t2, s2) . . . Gn(s j−1, t j, s j)yn(s j) ds1 . . . d s j

dt1 . . . d t j.

Furthermore, multiplying (4.6.40) by ηn(x) and integrating from 0 to π, we obtain

π

0 r(x)ηn(x) dx +

∞ j=2

π

0 . . . π

0 j

r(t1) . . . r(t j)Bnj(t1) . . . t j ) dt1 . . . d t j =

εn

n . (4.6.46)

Since r(x) = (π − x)M 1(x), ηn(x) = n(π − x)−1ξ n(x), we can tramsform (4.6.46) to theform π

0M 1(x)ξ n(x) dx +

∞ j=2

π

0. . .

π

0

j

M 1(t1) . . . M 1(t j ) π

0. . .

π

0

j

vn(t1, s1)

×Gn(s1, t2, s2) . . . Gn(s j−1, t j, s j)yn(s j) ds1 . . . d s j

dt1 . . . d t j = εn

n2.

Hence, taking (4.6.45) into account, we obtain π

0M 1(x)

π

0vn(x, t)yn(t) dt

dx = ∆(λn). (4.6.47)

Comparing (4.6.44) with (4.6.47), we find that ∆(λn) = 0. Hence the number λnn≥1 arethe eigenvalues of the boundary value problem L. Theorem 4.6.3 is proved. 2

4.7. DIFFERENTIAL OPERATORS OF THE ORR-SOMMERFELD TYPE

In this section we study the inverse problem of recovering differential operators of theOrr-Sommerfeld type (or Kamke type) from the Weyl matrix. Properties of the Weyl matrixare investigated, and an uniqueness theorem for the solution of the inverse problem is proved.

4.7.1. Let us consider the differential equation

Ly = zy (4.7.1)

where

Ly = y(n) +n−2k=0

pk(x)y(k), y = y(m) +m−1 j=0

q j(x)y( j), n > m ≥ 1, x ∈ (0, T ).

Here pk and q j are complex-valued integrable functions.

Differential equations of the form (4.7.1) play an important role in various areas of math-ematics as well as in applications. The Orr-Sommerfeld equation from the theory of hydro-dynamic stability is a typical example for equation (4.7.1). Spectral properties of boundaryvalue problems for the Orr-Sommerfeld equation and the more general equation (4.7.1) havebeen studied in many works (see [ebe5], [tre1], [shk1] and the references given therein). How-ever at present inverse spectral problems for such classes of operators have not been studiedyet because of their complexity.

In this paper we study the inverse problem of recovering the coefficients of equation

(4.7.1) from the so-called Weyl matrix. We introduce and investigate the Weyl matrix and



269

prove an uniqueness theorem for the solution of the inverse problem. We note that forthe case of m = 0, i.e. for the equation Ly = zy, inverse problems of recovering Lfrom various spectral characteristics have been studied in [bea1], [bea2], [kha3], [lei1], [sak1],[yur1], [yur16], [yur22] and other works. In particular, for such operators the theory of thesolution of the inverse problem by means of the Weyl matrix (and its applications) has been

constructed in [yur1], [yur12], [yur15], [yur16], [yur22]. The Weyl matrix introduced in thispaper is a generalization of the Weyl matrix for the equation Ly = zy introduced in [yur16].

4.7.2. Let the functions Φk(x, z ), k = 1, n, be solutions of equation (4.7.1) under the

conditions Φ( j−1)k (0, z ) = δ kj , j = 1, k, Φ

(s−1)k (T, z ) = 0, s = 1, n − k. Here δ kj is the

Kronecker delta. Denote M kj (z ) = Φ( j−1)k (0, z ), j = k + 1, n. The functions Φk(x, z ) are

called the Weyl solutions, and the functions M kj (z ) are called the Weyl functions. Thematrix M (z ) = [M kj (z )]k,j=1,n , M kj (z ) = δ kj , j = 1, k is called the Weyl matrix for

equation (4.7.1). The inverse problem studied in this section is formulated as follows:Given the Weyl matrix, construct L and .

Let us consider the fundamental system of solutions of equation (4.7.1) C k(x, z ), k = 1, n,

which satisfies the conditions C ( j−1)k (0, z ) = δ kj , j = 1, n. Then

Φk(x, z ) = C k(x, z ) +n

j=k+1

M kj (z )C j (x, z ), (4.7.2)

det[Φ( j−1)k (x, z )]k,j=1,n = det[C

( j−1)k (x, z )]k,j=1,n =

1, m < n − 1,exp(zx), m = n − 1.

(4.7.3)

Let us denote

∆kj (z ) = (−1)k+ j det[C (ξ−1)ν (T, z )]ξ=1,n−k; ν =k.n,ν = j.

The function ∆kj (z ) is entire in z of order less than 1, and its zeros coincide withthe eigenvalues of the boundary-value problem for equation (4.7.1) under the conditions

y(s−1)(0) = y( p−1)(T ) = 0; s = 1, k − 1, j; p = 1, n − k. The function ∆kj (z ) is uniquellydetermined by its zeros.

Using the boundary conditions on the Weyl solutions we calculate

Φk(x, z ) = (∆kk(z ))−1 det[C j(x, z ), C j(T, z ), . . . , C (n−k−1)

j (T, z )] j=k,n, (4.7.4)

and consequentlyM kj (z ) = (∆kk(z ))−1∆kj (z ). (4.7.5)

Let hk(x), k = 1, m, be the solutions of the differential equation y = 0 under theinitial conditions h

(ν −1)k (0) = δ kν , ν = 1, m. Denote

Bs(x) = det[hk(x), hk(T ), hk(T ), . . . , h(m−s−1)k (T )]k=s,m , s = 1, m − 1, Bm(x) = hm(x),

αs = det[h(ν )k (T )]k=s,m; ν =0,m−s, s = 1, m, αm+1 = 1.

Clearly,

α1 = exp(− T

0 q m−1(t) dt).



270

For definiteness, we shall assume that αs = 0, s = 2, m. Other cases require separatecalculations. We shall also suppose the coefficients pk and q j to be sufficiently smoothsuch that the asymptotic formulae (4.7.12)-(4.7.13) hold (see [ebe5], [tre1]).

By virtue of (4.7.5), the Weyl functions M kj (z ) are meromorphic. It follows from abovethat the specification of the Weyl matrix M (z ) is equivalent to the specification of the

spectra of the corresponding boundary-value problems for equation (4.7.1) or corresponds tothe specification of poles and residues of the Weyl functions. Therefore the inverse problemof recovering equation (4.7.1) from the Weyl matrix is a generalization of the well-knowninverse problems for the Sturm-Liouville operators from discrete spectral characteristics (see,for example, [lev2] and Chapter 1).

Let bs(x), s = 1, m be the solutions of the equation y = 0 with the conditionsb(ν −1)

s (0) = δ sν , ν = 1, s; b(ξ−1)s (T ) = 0, ξ = 1, m − s. In other words, the functions bs(x)

are the Weyl solutions for the equation y = 0. As in (4.7.3) and (4.7.4), we see that

bs(x) = (αs+1)−1Bs(x)

anddet[b(ν −1)

s (x)]s,ν =1,m = det[h(ν −1)s (x)]s,ν =1,m = exp(−

x

0q m−1(t) dt).

Together with L and we consider operators L and of the same form but withdifferent coefficients. We agree that if a symbol α denotes an object related to L and ,then α will denote the analogous object related to L and .

The main result of this section is the following uniqueness theorem for the solution of the inverse problem by means of the Weyl matrix.

Theorem 4.7.1. If M (z ) = M (z ) then L = L and = .

Proof. Let

C (x, z ) = [C ( j−1)k (x, z )] j,k=1,n , Φ(x, z ) = [Φ

( j−1)k (x, z )] j,k=1,n .

Then (4.7.2) takes the form

Φ(x, z ) = C (x, z )M t(z ), (4.7.6)where t denotes transposition. We define the matrix Q(x, z ) = [Q jk (x, z )] j,k=1,n from the

relation Q(x, z )Φ(x, z ) = Φ(x, z ). Let r := n − m.


Q jk (x, z ) = Q j+1,k(x, z )−Q j,k−1(x, z )+(−1)n−k+1Q j,n(x, z )πk−1(x, z ), j ,k = 1, n − 1, r ≥ 1,

(4.7.7)Q

jn(x, z ) = Q j+1,n(x, z )−

Q j,n−

1(x, z )−

zQ j,n(x, z ), j = 1, n−

1, r = 1, (4.7.8)

Q jn (x, z ) = Q j+1,n(x, z ) − Q j,n−1(x, z ), j = 1, n − 1, r > 1, (4.7.9)

where πs(x, z ) = ˜ ps(x) − z q s(x).

Proof of Lemma 1. Let us denote W (x, z ) = det Φ(x, z ). In virtue of (4.7.3), W (x, z ) =0, and consequently one can write Q(x, z ) = Φ(x, z )(Φ(x, z ))−1 or in the coordinates

Q jk (x, z ) = (W (x, z ))−1 det[Φs(x, z ), Φs(x, z ) . . . Φ(k−2)

s (x, z ), Φ( j−1)s (x, z ), Φ(k)

s (x, z ),

Φ(k+1)s (x, z ) . . . Φ

(n−

2)s (x, z ), Φ

(n−

1)s (x, z )]s=1,n. (4.7.10)



272

whereω jk = det[Rν

ξ ]ξ= j,k; ν =o,k− j,

σs = (s(s − 1) + (n − s − 1)(n − s) − (m − 1)m)/2, s = 1, µ,

σs+µ = ((µ+s

−1)(µ+s)

−s(s

−1)+(n

−µ

−s

−1)(n

−µ

−s)

−(m

−s)(m

−s

−1))/2, s = 0, m,

σs+µ+m = ((µ − s − 1)(µ − s) + (m + µ + s − 1)(m + µ + s) − (m − 1)m)/2, s = 0, µ.

Moreover|δ s(ρ)| ≤ C |ρ|β exp(α|ρ|), |ρ| → ∞, ρ ∈ S, s = 1, n,

for some α > 0 and β > 0.Using the boundary conditions on the Weyl solutions we calculate

Φk(x, z ) = (δ k(ρ))−1 det[Y s(0, ρ), . . . , Y (k−2)s (0, ρ),

Y s(x, ρ), Y s(T, ρ), . . . , Y (n−k−1)s (T, ρ)]s=1,n. (4.7.17)

Substituting the asymptotic formulas (4.7.12)-(4.7.16) into (4.7.17) we obtain for |ρ| →∞, ρ ∈ S, arg z = const = 0, π,

Φ(ν )s (x, z ) = ω1,s−1(ω1s)−1a(x)ρ1−s(ρRs)ν exp(ρRsx)(1 + O(ρ−1)), s = 1, µ, (4.7.18)

Φ(ν )µ+s(x, z ) = (−1)µ(R1 · · · Rµ)−1b(ν )

s (x)ρ−µ(1 + O(ρ−1)), s = 1, m, (4.7.19)

Φ(ν )µ+m+s(x, z ) = ω1,µ+s

−1(ω1,µ+s)−1R−m

µ+sa(x)ρ1−s−m−µ(ρRµ+s)ν exp(ρRµ+sx)×

(1 + O(ρ−1)), s = 1, µ, (4.7.20)

where ω10 = 1.Now we transform the matrix Q(x, z ). For this we use (4.7.6). Since M (z ) = M (z ) by

the assumption of Theorem 4.7.1, we get

Q(x, z ) = Φ(x, z )(Φ(x, z ))−1 = C (x, z )M t(z )(M t(z ))−1(C (x, z ))−1 = C (x, z )(C (x, z ))−1.

Using (4.7.3) we conclude that for each fixed x the functions Q jk (x, z ) are entire in z

of order 1/r. Substituting (4.7.18)-(4.7.20) into (4.7.10) we calculate for z → ∞, arg z =const = 0, π,

Q jn(x, z ) = O(ρ−1), j = 1, n − 1.

Then the theorems of Phragmen-Lindelof and Liouville give us

Q jn (x, z ) ≡ 0, j = 1, n − 1.

By virtue of Lemma 4.7.2, this yields

Q jk (x, z ) ≡ 0, j < k.

Since Q(x, z )Φ(x, z ) = Φ(x, z ), we obtain

Q11(x, z )Φk(x, z ) ≡ Φk(x, z ), k = 1, n.

This implies

Φ(ν )k (x, z ) = Q11(x, z )Φ

(ν )k (x, z ) +

ν −1

s=0

C sν Q(ν −s)11 (x, z )Φ

(s)k (x, z ),



273

and hence Qn11(x, z )W (x, z ) ≡ W (x, z ). In view of (4.7.3), W (x, z ) ≡ W (x, z ), i.e. Qn

11(x, z )1. Since Q11(0, z ) ≡ 1, we have Q11(x, z ) ≡ 1. Consequently, Φk(x, z ) ≡ Φk(x, z ), k =1,n, L = L, = . Theorem 4.7.1 is proved. 2

Using this method one can also obtain a constructive procedure for the solution of this

inverse problem along with necessary and sufficient conditions of its solvability.

4.8. DIFFERENTIAL EQUATIONS WITH TURNING POINTS

4.8.1. We consider in this section boundary value problems L of the form

y := −y + q (x)y = λR2(x)y, x ∈ [0, 1], (4.8.1)

U (y) := y(0) − hy(0) = 0, (4.8.2)

V (y) := y(1) + h1y(1) = 0. (4.8.3)

Here λ = ρ2 is the spectral parameter; R2 and q are real functions, and h, h1 are realnumbers. We suppose that

R2(x) =m

ν =1

(x − xν )νR0(x),

where 0 < x1 < x2 < .. . < xm < 1, ν

∈N, R0(x) > 0 for x

∈I := [0, 1] , and R0 is twice

continuously differentiable on I . In the other words, R2 has in I m zeros xν , ν = 1, mof order ν . The zeros xν of R2 are called turning points. We also assume that q isbounded and integrable on I .

In this section we study the following three inverse problems of recovering L from itsspectral characteristics, namely

(i) from the Weyl function,

(ii) from two spectra and

(iii) from the so-called spectral data.

Differential equations with turning points play an important role in various areas of mathe-matics as well as in applications (see [ebe1]-[ebe4], [gol1], [mch1], [was1] and the referencestherein). For example, turning points connected with physical situations in which zeroscorrespond to the limit of motion of a wave mechanical particle bound by a potential field.Turning points appear also in elasticity, optics, geophysics and other branches of naturalsciences. Moreover, a wide class of differential equations with Bessel-type singularities and

their perturbations can be reduced to differential equations having turning points. Inverseproblems for equations with turning points also help to study blow-up behavior of solutionsfor nonlinear evolution equations in mathematical physics [Con1].

In order to study the inverse problem in this section we use the method of spectral map-pings described in Section 1.6 for the classical Sturm-Liouville equation without turningpoints. The presence of turning points in the differential equation produces essential qual-itative modifications in the method. An important role in this investigation is played bythe special fundamental system of solutions for equation (4.8.1) constructed in [ebe4]. This

fundamental system of solutions gives us an opportunity to obtain the asymptotic behavior



274

of the so-called Weyl solutions and the Weyl function for the boundary value problem Land to solve the corresponding inverse problems. In Subsection 4.8.2 we prove uniquenesstheorems, and in Subsection 4.8.3 we provide a constructive procedure for the solution of the inverse problem.

Let ε > 0 be fixed, sufficiently small and let D0ε = [0, x1 − ε], Dνε = [xν + ε, xν +1 − ε]

for 1 ≤ ν ≤ m−1, Dmε = [xm+ε, 1], Dε =m

ν =0

Dνε , and I νε = Dν −1,ε∪[xν −ε, xν +ε]∪Dνε .

We distinguish four different types of turning points: For 1 ≤ ν ≤ m

T ν =

I, if ν is even and R2(x)(x − xν )−ν < 0 in I νε ,

II, if ν is even and R2(x)(x − xν )−ν > 0 in I νε ,

III, if ν is odd and R2(x)(x − xν )−ν < 0 in I νε ,

IV, if ν is odd and R2(x)(x−

xν )−ν > 0 in I νε ,

is called type of xν . Further we set for 1 ≤ ν ≤ m

µν =1

2 + ν ,

θν =

1 if µν >1

4,

1 − δ 0 (with δ 0 > 0 arbitrary small) if µν =

1

4 ,4µν if µν <

1

4,

and 0 < θ0 = minθν |1 ≤ ν ≤ m . We also denote

I + = x : R2(x) > 0, I − = x : R2(x) < 0,

ξ (x) =

0, for x ∈ I +1, for x

∈I −

,

γ ν =

2sin

πµν

2, for T ν = III,IV,

sin πµν , for T ν = I, II,

K ±(x) = (

xν∈(0,x)

γ −1ν )exp(±i

π

4(ξ (x) − ξ (0))),

K ∗±(x) = (

xν∈(0,x)

γ ν ) exp(±iπ

4(ξ (x) + ξ (0))),

R2

+(x) = max(0, R2

(x)), R2

−(x) = max(0, −R2

(x)).Clearly,

K ±(x)K ∗±(x) = exp(±iπ

2ξ (x)) =

1, if x ∈ I +± i, if x ∈ I −.

Let

S k = ρ : arg ρ ∈ [kπ

4,

(k + 1)π

4],

σδ

s=

ρ : arg ρ

∈[sπ

2 −δ,

sπ

2+ δ ]

, δ > 0,



275

σδ =

s

σδs , S δk = S k \ σδ, S δ =

1k=−2

S δk.

Below it is sufficient to consider the sectors S k and S δk for k = −2, −1, 0, 1 only.It is shown in [ebe4] that for each fixed sector S k (k = −2, −1, 0, 1) there exists a

fundamental system of solutions of (1) z 1(x, ρ), z 2(x, ρ), x ∈ I, ρ ∈ S k such that thefunctions

(x, ρ) → z (s)

j (x, ρ) ( j = 1, 2; s = 0, 1)

are continuous for x ∈ I, ρ ∈ S k and holomorphic for each fixed x ∈ I with respect toρ ∈ S k ; moreover for |ρ| → ∞, ρ ∈ S k, x ∈ Dε, j = 1, 2,

z ( j)1 (x, ρ) = (±iρ) j|R(x)| j−1/2(exp(∓i

π

2ξ (x)) j exp(ρ

x

0|R−(t)|dt)

× exp(±iρ x

0|R+(t)|dt)K ±(x)κ(x, p), (4.8.4)

z ( j)2 (x, ρ) = (∓iρ) j|R(x)| j−1/2(exp(∓i

π

2ξ (x)) j exp(−ρ

x

0|R−(t)|dt)

× exp(∓iρ x

0|R+(t)|dt)K ∗±(x)κ(x, ρ), (4.8.5)

z 1(x, ρ) z 2(x, ρ)z 1(x, ρ) z 2(x, ρ)

= ∓(2iρ)[1]. (4.8.6)

Here and in the sequel:(i) the upper or lower signs in formulas correspond to the sectors S −2, S −1 or S 0, S 1 , re-spectively;

(ii) [1]def = 1 + O(

1

ρθ0) uniformly in x ∈ Dε ,

(iii) one and the same symbol κ(x, ρ) denotes various functions such that:(1) uniformly in x ∈ Dε, κ(x, ρ) = O(1) as |ρ| → ∞, ρ ∈ S k,(2) for each fixed δ > 0, κ(x, ρ) = [1] as

|ρ

| → ∞, ρ

∈S δk.

4.8.2. Let ϕ(x, λ) be the solution of (4.8.1) under the initial conditions ϕ(0, λ) =1, ϕ(0, λ) = h and denote

∆(λ) = ϕ(1, λ) + h1ϕ(1, λ). (4.8.7)

The function λ → ∆(λ) is entire in λ , and its zeros coincide with the eigenvalues λnn≥0

of the boundary value problem L . The functions x → ϕ(x, λn) are eigenfunctions of L .Let Φ(x, λ) be the solution of (4.8.1) under the boundary conditions U (Φ) = 1, V (Φ) =

0 . We set M (λ) = Φ(0, λ) . The functions x → Φ(x, λ) and λ → M (λ) are called the Weyl solution and the Weyl function for the boundary value problem (4.8.1)-(4.8.3), respectively.Clearly

ϕ(x, λ), Φ(x, λ) ≡ 1 (4.8.8)

for all x and λ, where y, z := yz − yz.

The inverse problem studied in this section is formulated as follows. Suppose that thefunction R2 is known a priori. Our goal is to find q (x), h and h1 from the given Weyl

function M .



276

In order to formulate and prove the uniqueness theorem for the solution of this inverseproblem we agree that together with L = L(R2(x), q (x), h , h1) we consider a boundaryvalue problem L = L(R2(x), q (x), h, h1) of the same form (4.8.1)-(4.8.3) but with differentcoefficients. If a certain symbol denotes an object related to L , then the correspondingsymbol with tilde will denote the analogous object related to L .

Theorem 4.8.1. If M = M then q (x) = q (x) for x ∈ I, h = h and h1 = h1 .

Thus, the specification of the Weyl function M uniquely determines L .

Proof. Let us define the matrix P (x, λ) = [P jk (x, λ)] j,k=1,2 by the formula

P (x, λ)

ϕ(x, λ) Φ(x, λ)

ϕ(x, λ) Φ(x, λ)

=


. (4.8.9)


P 11(x, λ) = ϕ(x, λ)Φ(x, λ) − Φ(x, λ)ϕ(x, λ)

P 12(x, λ) = Φ(x, λ)ϕ(x, λ) − ϕ(x, λ)Φ(x, λ)

(4.8.10)

Let us now study the asymptotic behavior of ϕ(x, λ), Φ(x, λ) and P 1 j(x, λ) as |λ| → ∞ .For this purpose we use the above-mentioned fundamental system of solutions z 1(x, ρ), z 2(x, ρin each fixed sector S k, k = −2, −1, 0, 1 .

From the initial conditions on ϕ(x, λ) we calculate

ϕ(x, λ) =1

w(λ)(U (z 2(x, ρ))z 1(x, ρ) − U (z 1(x, ρ))z 2(x, ρ)), (4.8.11)

where

w(λ) =

z 1(0, ρ) z 2(0, ρ)z 1(0, ρ) z 2(0, ρ)

.

Using (4.8.4)-(4.8.6) we get for |ρ| → ∞, ρ ∈ S k

U (z 1(x, ρ)) = (± i ρ)|R(0)|1/2 exp(∓ i π2

ξ (0))κ(ρ),

U (z 2(x, ρ)) = (∓ iρ)|R(0)|1/2κ(ρ),w(λ) = ∓2iρ[1];

(4.8.12)

here and in the sequel κ(ρ) = O(1) for ρ ∈ S k and κ(ρ) = [1] for ρ ∈ S δk as |ρ| → ∞ .Substituting (4.8.4), (4.8.5) and (4.8.12) into (4.8.11) we conclude that for |ρ| → ∞, ρ ∈

S k, x ∈ Dε, j = 0, 1,

ϕ( j)(x, λ) = 12

(± iρ) j|R(0)|1/2|R(x)| j−1/2(exp(∓ i π2

ξ (x))) j

× exp(ρ x

0|R−(t)|dt) exp(± iρ

x

0|R+(t)|dt)K ±(x)κ(x, ρ). (4.8.13)

Consequently, taking (4.8.7) into account, we have

∆(λ) =1

2(±iρ)|R(0)R(1)|1/2 exp(ρ

1

0|R−(t)|dt)



277

× exp(±iρ 1

0|R+(t)|dt)K ±(1)κ(ρ), ρ ∈ S k, |ρ| → ∞. (4.8.14)

From (4.8.13) and (4.8.14) we obtain the estimates:

|ϕ( j)(x, λ)

| ≤C

|ρ

| j

|exp(ρ

x

0 |R−

(t)

|dt)

× exp(±iρ x

0|R+(t)|dt)|, ρ ∈ S k, x ∈ Dε, (4.8.15)

and

|∆(λ)| ≤ C |ρ exp(ρ 1

0|R−(t)|dt)exp(±iρ

1

0|R+(t)|dt)|, ρ ∈ S k. (4.8.16)

Here and below one and the same symbol C denotes various positive constants in estimates.Moreover, it can be shown that

|∆(λ)| ≥ C δ|ρ exp(ρ 1

0|R−(t)|dt) exp(±iρ

1

0|R+(t)|dt)|, λ ∈ Gδ, (4.8.17)

whereGδ = λ : |λ − λn| ≥ δ, n ≥ 0, δ > 0.

It follows from (4.8.15) and (4.8.16) that the functions ϕ( j)(x, ·) and ∆ are entire functionsof order 1/2.

Let us go on to the Weyl solution Φ(x, λ) . Applying the boundary conditions to Φ(x, λ)

we calculateΦ(x, λ) =

1

∆0(ρ)(V (z 2(x, ρ))z 1(x, ρ) − V (z 1(x, ρ))z 2(x, ρ)) (4.8.18)

where

∆0(ρ) =

U (z 1(x, ρ)) U (z 2(x, ρ))V (z 1(x, ρ)) V (z 2(x, ρ))

.

Taking (4.8.4)-(4.8.6) into account we obtain from (4.8.18) that for |ρ| → ∞, ρ ∈ S δ, x ∈Dε, j = 0, 1,

Φ( j)(x, λ) = (∓ iρ) j−1|R(0)|−1/2|R(x)| j−1/2(exp(∓ iπ

2ξ (x))) j

× exp(−ρ x

0|R−(t)|dt)exp(∓ iρ

x

0|R+(t)|dt)K ∗±(x)[1] (4.8.19)

It follows from the boundary conditions on ϕ(x, λ) and Φ(x, λ) that

Φ(x, λ) = S (x, λ) + M (λ)ϕ(x, λ), (4.8.20)

where S (x, λ) is the solution of (4.8.1) satisfying the conditions S (0, λ) = 0, S (0, λ) = 1 .Using the fundamental system of solutions z 2(x, ρ), z 2(x, ρ) we get

S (x, λ) =1

w(λ)(z 1(0, ρ)z 2(x, ρ) − z 2(0, ρ)z 1(x, ρ)).

By virtue of (4.8.4)-(4.8.6), we infer that for ρ ∈ S k, x ∈ Dε, j = 0, 1,

S

( j)

(x, λ) =

1

2(± iρ)

j

−1

|R(0)|−1/2

|R(x)| j

−1/2

(exp(∓ i

π

2 ξ (x))

j

exp(± i

π

2 ξ (0))



278

× exp(± iπ

2ξ (0)) exp(ρ

x


x

0|R+(t)|dt)K ±(x)κ(x, ρ) (4.8.21)

and|S ( j)(x, λ)| ≤ C |ρ| j−1| exp(ρ

x


x

0|R+(t)|dt)|. (4.8.22)

It follows from (4.8.22) that the functions S ( j)

(x, ·) are entire of order 1/2 . Using (4.8.10),(4.8.13) and (4.8.19) we get

P 11(x, λ) = [1], P 12(x, λ) = O(1

ρθ0), |ρ| → ∞, ρ ∈ S δ, x ∈ Dε. (4.8.23)

Furthermore, substituting (4.8.20) into (4.8.10), we calculate

P 11(x, λ) = ϕ(x, λ)S (x, λ) − S (x, λ)ϕ(x, λ) + (M (λ) − M (λ)ϕ(x, λ)ϕ(x, λ),

P 12(x, λ) = S (x, λ)ϕ(x, λ) − ϕ(x, λ)S (x, λ) + (M (λ) − M (λ))ϕ(x, λ)ϕ(x, λ).

By the hypothesis of Theorem 4.8.1, M (λ) ≡ M (λ) , and consequently the functionsP 11(x, λ) and P 12(x, λ) are entire in λ of order 1/2 . It follows from this and (4.8.23) thatP 11(x, λ) ≡ 1, P 12(x, λ) ≡ 0 . Substituting into (4.8.9) we obtain ϕ(x, λ) ≡ ϕ(x, λ), Φ(x, λ) ≡Φ(x, λ) for all x and λ . Consequently, q (x) = q (x) for x ∈ I, h = h and h1 = h1 .Theorem 4.8.1 is proved. 2

Let λ

1

nn≥0 be the sequence of eigenvalues of the boundary value problem L1 forequation (4.8.1) with the boundary conditions y(0) = V (y) = 0 . Now we consider theinverse problem of recovering q (x), h and h1 from the given two spectra λnn≥0 of Land λ1

nn≥0 of L1 .

Theorem 4.8.2. If λn = λn, λ1n = λ1

n for all n ≥ 0 , then q (x) = q (x) for x ∈ I, h =h and h1 = h1 .

Proof. The function ∆ is entire of order 1/2 , and ∆(λ) is uniquely determined byits zeros

λn

n

≥0 via the formula

∆(λ) = A∞

n=0

(1 − λ

λn) ( if λn = 0 ∀n),

where the constant A is uniquely determined with the help of the asymptotic formula(4.8.14), (the case when ∆(0) = 0 requires minor modifications).

The eigenvalues λ1nn≥0 of the boundary value problem L1 coincide with zeros of the

entire function ∆1(λ) = S (1, λ) + h1S (1, λ) of order 1/2 . It follows from (4.8.21) that

∆1(λ) = 12|R(0)|−1/2|R(1)|1/2 exp(± i π

2ξ (0)) exp(ρ

1

0|R−(t)|dt)

× exp(± iρ 1

0|R+(t)|dt)K ±(1)κ(ρ). (4.8.24)

The function ∆1 is uniquely determined by its zeros λ1nn≥0 via the formula

∆1(λ) = A1

∞

n=0

(1

−

λ

λ1

n

),



279

where the constant A1 is uniquely determined with the help of (4.8.24).Since V (Φ) = 0 , it follows from (4.8.20) that

M (λ) = −∆1(λ)

∆(λ). (4.8.25)

By hypothesis of Theorem 4.8.2, λn = λn, λ1n = λ1

n for all n ≥ 0 , and consequently∆(λ) ≡ ∆(λ), ∆1(λ) ≡ ∆1(λ) . From this, in view of (4.8.25), we get M (λ) ≡ M (λ) .Using Theorem 4.8.1 we conclude that q (x) = q (x) for x ∈ I, h = h and h1 = h1 ; henceTheorem 4.8.2 is proved. 2

We consider now the inverse problem of recovering L from the so-called spectral data.For simplicity, we confine ourselves to the case when all zeros of ∆(λ) are simple.

Denote

αn = 1

0R2(x)ϕ2(x, λn)dx.

The data λn, αnn≥0 are called the spectral data of the boundary value problem L .

Theorem 4.8.3. If λn = λn, αn = αn for all n ≥ 0 , then q (x) = q (x) for x ∈I, h = h and h1 = h1 .

Proof. Since−Φ(x, λ) + q (x)Φ(x, λ) = λR2(x)Φ(x, λ),

−ϕ(x, λn) + q (x)ϕ(x, λn) = λnR2(x)ϕ(x, λn),

we get(Φ(x, λ)ϕ(x, λn) − Φ(x, λ)ϕ(x, λn)) = (λ − λn)R2(x)Φ(x, λ)ϕ(x, λn)

and hence

(λ − λn) 1

0R2(x)Φ(x, λ)ϕ(x, λn)dx =

1

0(Φ(x, λ)ϕ(x, λn) − Φ(x, λ)ϕ(x, λn)) = 1.

In view of (4.8.20) and (4.8.25), we have

(λ − λn) 1

0R2(x)S (x, λ)ϕ(x, λn)dx − (λ − λn)

∆1(λ)

∆(λ)

1

0R2(x)ϕ(x, λ)ϕ(x, λn)dx = 1.

Let λ → λn , then∆1(λn)

∆(λn)

1

0R2(x)ϕ2(x, λn)dx = −1,

where ∆(λ) = ddλ∆(λ) . Hence

αn = − ∆(λn)

∆1(λn). (4.8.26)

It follows from (4.8.25) that the Weyl function M is meromorphic with simple poles in thepoints λn, n ≥ 0 . Using (4.8.26) we calculate

Resλ=λnM (λ) =1

αn. (4.8.27)



280

Furthermore, by virtue of (4.8.19), we derive

M (λ) =1

(∓ iρ)|R(0)| · exp(±iπ

2ξ (0)), |ρ| → ∞, ρ ∈ S δ,

and consequently |M (λ)| ≤ C

|ρ| , |ρ| → ∞, ρ ∈ S δ. (4.8.28)

Let us consider the function λ → N (λ)def = M (λ) − M (λ) . By hypothesis of Theorem 4.8.3

and (4.8.27), we have

Res λ=λnN (λ) =1

αn

− 1

αn= 0.

Thus, the function N is an entire function of order 1/2 . On the other hand, in view of

(4.8.28),

|N (λ)| ≤ C

|ρ| , ρ ∈ S δ, |ρ| → ∞.

Consequently, N (λ) ≡ 0 , i.e. M (λ) ≡ M (λ) . Using Theorem 4.8.1 we obtain q (x) = q (x)for x ∈ I, h = h and h1 = h1 . Theorem 4.8.3 is proved. 2

Remark 4.8.1. Theorem 4.8.2 and 4.8.3 are generalizations of results by Borg andMarchenko (see Section 1.4) for the Sturm-Liouville equation without turning points where

R2(x) ≡ 1 .

Remark 4.8.2. We can also apply this method to investigate the inverse problems forequation (4.8.1) with turning points in 0 or (and) 1 .

4.8.3. Let us now go on to constructing the solution of the inverse problem. The centralrole for solving the inverse problem is played by the so-called main equation of the inverseproblem which connects the spectral characteristics with the corresponding solutions of the

differential equation. We give a derivation of the main equation, which is a linear equationin the Banach space m of bounded sequences. Moreover we prove the unique solvabilityof the main equation. For simplicity, we confine ourselves to the most important particularcase when m = 1 and T 1 = IV , i.e. the weight-function changes sign exactly once. Thegeneral case can be treated analogously.

For deriving the main equation of the inverse problem we need more precise asymptoticsfor the solutions of equation (4.8.1). For definiteness, everywhere below ρ ∈ S 0 ∪ S 1 (thecase ρ ∈ S −1 ∪ S −2 is considered in the same way). It has been shown in [ebe4] that for

|ρ| → ∞, j = 0, 1 the following asymptotic formulas are valid

z ( j)1 (x, ρ) = ρ j|R(x)| j−1/2 exp(ρ

x

0|R(t)|dt)[1], x < x1,

z ( j)1 (x, ρ) = (−iρ) j|R(x)| j−1/2 exp(ρ

x1

0|R(t)|dt)

1

2csc

πµ1

2exp(i

π

4)

×(exp(−iρ x

x1|R(t)|dt)[1] + (−1) j+1i exp(iρ

x

x1|R(t)| dt)[1], x > x1,

(4.8.29)



281

z ( j)2 (x, ρ) = (−ρ) j|R(x)| j−1/2(−i exp(−ρ

x

0|R(t)|dt)[1]

+(−1) j exp(ρ x

0|R(t)|dt)exp(−2ρ

x1

0|R(t)|dt)[1]), x < x1,

z ( j)2 (x, ρ) = (iρ) j|R(x)| j−1/22sin

πµ1

2exp(−i

π

4) exp(−ρ

x1

0|R(t)|dt)

× exp(iρ x

x1|R(t)|dt)[1], x > x1.

(4.8.30)

Here, as above, [1] = 1 + O(1

ρθ0) uniformly in x ∈ Dε . It follows from (4.8.11), in view of

(4.8.6), (4.8.29)-(4.8.30), that

ϕ( j)(x, λ) =1

2ρ j|R(0)|1/2|R(x)| j−1/2(exp(ρ

x

0|R(t)|dt)[1]

+(

−1) j exp(

−ρ

x

0 |R(t)

|dt)[1]), x < x1,

ϕ( j)(x, λ) =1

2(iρ) j|R(0)|1/2|R(x)| j−1/2(A1(ρ) exp(iρ

x

x1|R(t)|dt)[1]

+(−1) jA2(ρ)exp(−iρ x

x1|R(t)|dt)[1]), x > x1,

(4.8.31)

where

A2(ρ) =1

2csc

πµ1

2exp(i

π

4)(exp(ρ

x1

0|R(t)|dt)[1]

−i exp(−ρ x1

0 |R(t)|dt)[1]),A1(ρ) = −iA2(ρ) + 2 sin

πµ1

2exp(i

π

4)exp(−ρ

x1

0|R(t)| dt)[1].

(4.8.32)

Remark 4.8.3. Let ξ = x

x1|R(t)|dt . It follows from results of [ebe4] that (4.8.29)-

(4.8.31) are also valid uniformly for |ρξ | ≥ 1 with [1] = 1 + O(1

(ρξ )θ0) ; moreover for

|ρξ | ≤ 1 we have the estimates

|ϕ(x, λ)| ≤ C |R(x)|−1/2| exp(ρ x

0|R(t)|dt)|, x ≤ x1

|ϕ(x, λ)| ≤ C |R(x)|−1/2| exp(ρ x1

0|R(t)|dt)|, x ≥ x1.

(4.8.33)

Lemma 4.8.1. (i) The spectrum λk of the boundary value problem L consists of two sequences of eigenvalues: λk = λ+

k ∪ λ−k , k ∈ N , such that

ρ±k =

λ±k = ρ±k,0 + O(

1

kθ0), k → +∞, (4.8.34)

where

ρ+k,0 = (

1

x1|R(t)|dt)−1(k +

1

4)π, ρ−k,0 = (

x1

0|R(t)|dt)−1(k +

1

4)πi.

(ii) Denote α±k = 1

0R2(x)ϕ2(x, λ±k )dx , i.e. αk = α+

k ∪ α−k . Then

α+k =

1

2|R(0)|

1

x1|R(t)|dt(

1

2(csc

πµ1

2)2ω2

k(1 + O(1

kσ)), k → +∞,

α−k = −1

2 |R(0)| x1

0 |R(t)|dt(1 + O(

1

kσ )), k → +∞,

(4.8.35)



282

where ωk = exp(ρ+

k,0

x1

0|R(t)|dt), σ = min(1 − δ 0, θ0).

Proof. Substituting (4.8.31) into (4.8.7) we calculate

∆(λ) = 12

(iρ)|R(0)R(1)|1/2(A1(ρ) exp(iρ x

x1|R(t)|dt)[1]

−A2(ρ) exp(−iρ x

x1|R(t)|dt)[1]). (4.8.36)

Let ρ ∈ S 0 . It follows from (4.8.32) that

A1(ρ) =1

2csc

πµ1

2exp(−i

π

4)exp(ρ

x1

0|R(t)|dt),

A2(ρ) =1

2csc

πµ1

2exp(i

π

4) exp(ρ

x1

0|R(t)|dt)[1].

Hence, by virtue of (4.8.36), the equation ∆(λ) = 0 can be rewritten in the form

exp(2iρ 1

x1|R(t)|dt) = i[1].

This equation has a countable set of roots ρ

+

k such that ρ

+

k = ρ

+

k,0 + O(

1

kθ0 ) . Analogously,if ρ ∈ S 1 then the equation ∆(λ) = 0 can be transformed to

exp(2ρ x1

0|R(t)|dt) = i[1];

therefore this equation has a countable set of roots ρ−k such that ρ−k = ρ−k,0 + O(1

kθ0) .

Let us now consider α−k . Put α−k = α−k,1 + α−k,2 , where

α−k,1 = x1

0R2(x)ϕ2(x, λ−k )dx, α−k,2 =

1

x1R2(x)ϕ2(x, λ−k )dx.

Denote I k,1 = x ∈ [0, x1] : |ξρ−k | ≥ 1, I k,2 = x ∈ [0, x1] : |ξρ−k | ≤ 1 , where ξ = x

x1|R(t)|dt . According to (4.8.31),

I k,1

R2(x)ϕ2(x, λ−k )dx = −1

4|R(0)|

I k,1

|R(x)|(exp(ρ

x

0|R(t)|dt)[1]

+exp(−ρ x

0|R(t)|dt)[1])2

/ρ=ρ−k

dx.

The change of variables gives

I k,1

R2(x)ϕ2(x, λ−k )dx = −1

4|R(0)|

ξ1

1/|ρ|(exp(ρ(ξ 1 − ξ ))(1 + O((ρξ )−θ0))

+exp(−

ρ(ξ 1−

ξ ))(1 + O((ρξ )−θ0))2dξ |ρ=ρ

−

k

,



283

where ξ 1 = x1

0|R(t)|dt . Hence

I k,1

R2(x)ϕ2(x, λ−k )dx = −1

2|R(0)|

x1

0|R(x)|dx(1 + O(

1

kσ)).


|

I k,2

R2(x)ϕ2(x, λ−k )dx| ≤

I k,2

|R(x) exp(2ρ x

0|R(t)|dt)|

ρ=ρ−k

|dx.

In view of (4.8.34), the exponential is bounded, and consequently,

|

I k,2R2(x)ϕ2(x, λ−k )dx| ≤

I k,2

|R(x)|dx = 1/|ρ|

0dξ |ρ=ρ−

k= O(

1

k).

Thus, we arrive at

α−k,1 = −1

2|R(0)|

x1

0|R(x)|dx(1 + O(

1

kσ)). (4.8.37)

Let us now estimate α−k,2 . Denote J k,1 = x ∈ [x1, 1] : |ξρ−k | ≥ 1, J k,2 = x ∈ [x1, 1] :|ξρ−k | ≤ 1 . In the same way as above one can show that

|

J k,2

R2(x)ϕ2(x, λ−k )dx| = O(1

k).

Thenα−k,2 =

J k,1

R2(x)ϕ2(x, λ−k )dx + O(1

k).

Taking (4.8.31) into account, we get

α−k,2 = −|R(0)|4

J k,1

(A1(ρ) exp(iρ x

x1|R(t)|dt)[1]

+A2(ρ)exp(−iρ x

x1 |R(t)|dt)[1])2

ρ=ρ−k dx + O(

1

k ).

The integral seems to be unbounded as k → +∞ . But it follows from (4.8.36) and (4.8.32)that

A2(ρ)

A1(ρ) |ρ=ρ−k

= exp(2iρ−k 1

x1|R(t)|dt)[1], (4.8.38)

A1(ρ−k ) = 2exp(iπ

4)sin

πµ1

2exp(−ρ−k

x1

0|R(t)|dt)[1]. (4.8.39)

Thenα−k2 = −|R(0)|

4A1(ρ−k )

J k,1

|R(x)|(exp(iρ x

x1|R(t)|dt)[1]

+ exp(2iρ 1

x1|R(t)|dt) exp(−iρ

x

x1|R(t)|dt)[1])2

/ρ=ρ−k

dx + O(1

k).

After changing variables x → ξ = x

x1|R(t)|dt and integrating we arrive at the estimate

α−k,2 = O(

1

kσ ).



284

Combining this with (4.8.37), we obtain (4.8.35) for α−k . For α+k the proof is analogous;

hence Lemma 4.8.1 is proved. 2

Now we go on to the derivation of the main equation of the inverse problem. We assumethat the spectral data

λk, αk

k≥0 of L are given. Let L = L(R2(x), q (x), h, h1) be

a certain known model boundary value problem with the same weight-function R2(x) andwith the spectral data λk, αkk≥0 .

Lemma 4.8.2. For each fixed x = x1 , and k → +∞ the following estimates hold:(i) If x < x1 then

ϕ(x, λ−k ) = O(1), ϕ(x, λ−k ) = O(1), (4.8.40)

ϕ(x, λ−k ) − ϕ(x, λ−k ) = O(ρ−k − ρ−k ), (4.8.41)

ϕ(x, λ+k ) = O(exp(ρ+

k,0 x

0 |R(t)

|dt)),

ϕ(x, λ+k ) = O(exp(ρ+

k,0

x

0|R(t)|dt)),

(4.8.42)

ϕ(x, λ+k ) − ϕ(x, λ+

k ) = O(|ρ+k − ρ+

k | exp(ρ+k,0

x

0|R(t)|dt)). (4.8.43)

(ii) If x > x1 then

ϕ(x, λ−k ) = O(exp(iρ−k,0

x

x1|R(t)|dt)), (4.8.44)

ϕ(x, λ−k ) = O(k−θ0 exp(−iρ−k,0 x

x1|R(t)|dt)), (4.8.45)

ϕ(x, λ+k ) = O(ωk), ϕ(x, λ+

k ) = O(ωk), (4.8.46)

ϕ(x, λ+k ) − ϕ(x, λ+

k ) = O(|ρ+k − ρ+

k |ωk). (4.8.47)

We mark the difference in the estimates (4.8.44) and (4.8.46) for the functions ϕ(x, λ−k )and ϕ(x, λ−k ) ; the real part of the exponent in (4.8.44) is negative but the real part of theexponent in (4.8.46) is positive.

Proof. Let x < x1 . The estimates (4.8.40) and (4.8.42) follow from (4.8.31) andthe asymptotics of λ+

k and λ−k . Using (4.8.40), (4.8.42) and Schwarz’s lemma we obtain(4.8.41) and (4.8.43).

Let now x > x1 . For λ = λ−k and λ = λ−k we rewrite (4.8.31) as follows

ϕ(x, λ) =1

2|R(0)|1/2|R(x)|−1/2A1(ρ)(exp(iρ

x

x1|R(t)|dt)[1]

+A2(ρ)A1(ρ) exp(−iρ x

x1|R(t)|dt)[1]). (4.8.48)

From this, using (4.8.38) and (4.8.39), we obtain (4.8.44). Furthermore, it is easy to showthat

A2(ρ)

A1(ρ) |ρ=ρ−k

= O(1

kθ0), A1(ρ−k ) = O(1).

Substituting into (4.8.48) we arrive at (4.8.45). The estimates (4.8.46) and (4.8.47) areproved in the same way as (4.8.42) and (4.8.43). Lemma 4.8.2 is proved. 2



285

Lemma 4.8.3. For each fixed x ∈ I, the following relations hold

ϕ(x, λ) = ϕ(x, λ) +

k

ϕ(x, λ), ϕ(x, λk)αk(λ − λk)

ϕ(x, λk) − ϕ(x, λ), ϕ(x, λk)αk(λ − λk)

ϕ(x, λk)

, (4.8.49)

ϕ(x, λ), ϕ(x, µ)λ − µ

− ϕ(x, λ), ϕ(x, µ)λ − µ

+

k

ϕ(x, λ), ϕ(x, λk)ϕ(x, λk), ϕ(x, µ)αk(λ − λk)(λk − µ)

− ϕ(x, λ), ϕ(x, λk)ϕ(x, λk), ϕ(x, µ)αk(λ − λk)(λk − µ)

= 0.

(4.8.50)

We omit the proof of Lemma 4.8.3 since the arguments here are the same as in Lemma1.6.3.

Denote λk0 = λk, λk1 = λk, αk0 = αk, αk1 = αk, ϕkj (x) = ϕ(x, λkj ), ϕkj (x) =ϕ(x, λkj ) ,

P ni,kj(x) =ϕni(x), ϕkj (x)αkj (λnj − λkj )

=1

αkj

x

0R2(t)ϕni(t)ϕkj (t)dt,


=1

αkj

x

0R2(t)ϕni(t)ϕkj (t)dt.


ϕni(x) = ϕni(x) + k

(P ni,k0(x)ϕk0(x) − P ni,k1(x)ϕk1(x)), (4.8.51)

P ni,j(x) − P ni,j(x) +

k

(P ni,ko(x)P ko,j(x) − P ni,k1(x)P k1,j(x)) = 0. (4.8.52)

Let x < x1 . Denote

γ k(x) =

exp(ρ+k,0

x

0|R(t)|dt), for λk = λ+

k

1, for λk = λ−k ,

γ k = γ k(x1), ξ k = |ρk − ρk| +|αk − αk|

γ 2k.

Clearly, ξ k = O(k−σ) . It follows from (4.8.40)-(4.8.43) that

|ϕkj (x)| ≤ Cγ k(x), |ϕk0(x) − ϕk1(x)| ≤ Cξ kγ k(x),

|P ni,kj(x)| ≤ Cγ n(x)γ k(x)

(|n

−k|

+ 1)γ 2k

. (4.8.53)

Let V be a set of indices v = (k, j), k ∈ N, j = 0, 1 . Define the vector ψ(x) =[ψv(x)]v∈V := [ψ00, ψ01, ψ10, ψ11, . . .]T and similarly the block matrix

H (x) = [H u,v(x)]u,v∈V =

H n0,k0(x) H no,k1(x)H n1,k0(x) H n1,k1(x)

n,k∈N

,

u = (n, i), v = (k, j) , where

ψk0(x) = (ϕk0(x) − ϕk1(x))(ξ kγ k(x))−1

, ψk1(x) = ϕk1(x)(γ k(x))−1

,



286

H n0,k0(x) = (P n0,k0(x) − P n1,k0(x))ξ kγ k(x)(ξ nγ n(x))−1,

H n1,k0(x) = P n1,k0(x)ξ kγ k(x)(γ n(x))−1, H n1,k1(x) = (P n1,k0(x) − P n1,k1(x))γ k(x)(γ n(x))−1,

H n0,k1(x) = (P n0,k0(x) − P n1,k0(x) − P n0,k1(x) + P n1,k1(x))γ k(x)(ξ nγ n(x))−1.

Analogously we define ψ(x), H (x) . It follows from (4.8.53) and Schwarz’s lemma that foreach fixed x ∈ (0, x1),

|ψni(x)| ≤ C, |H ni,kj(x)| ≤ Cξ k(|n − k| + 1)

. (4.8.54)

Similarly,

|ψni(x)| ≤ C, |H ni,kj(x)| ≤ Cξ k(|n − k| + 1)

. (4.8.55)

Let us consider the Banach space m of bounded sequences α = [αv]v∈V with the normαm = supv∈V

|αv| . It follows from (4.8.54) and (4.8.55) that for each fixed x ∈ (0, x1) the

operators E + H (x) and E − H (x) (here E is the identity operator), acting from m tom , are linear bounded operators, and

H (x), H (x) ≤ C supn

k

ξ k(|n − k| + 1

) < ∞.

Theorem 4.8.4. For each fixed x ∈ (0, x1) , the vector ψ(x) ∈ m is the solution of the equation

ψ(x) = (E + H (x))ψ(x) (4.8.56)

in the Banach space m . The operator E + H (x) has a bounded inverse operator, i.e.equation (4.8.56) is uniquely solvable.

Proof. Indeed, taking into account our notations, we can rewrite (4.8.51) and (4.8.52) inthe form

ψ(x) = (E + H (x))ψ(x), (E + H (x))(E − H (x)) = E.Interchanging places for L and L we obtain analogously

ψ(x) = (E − H (x))ψ(x), (E − H (x))(E + H (x)) = E.

Hence the operator (E + H (x))−1 exists, and it is a linear bounded operator. 2

Equation (4.8.56) is called the main equation of the inverse problem. Solving (4.8.56)we find the vector ψ(x) , and consequently, the functions ϕni(x) . Since ϕni(x) = ϕ(x, λni)

are the solutions of (4.8.1), we can construct the function q (x) for x ∈ (0, x1) and thecoefficient h . Thus, the inverse problem has been solved for the interval x ∈ (0, x1) . Forx > x1 we can act analogously, starting from (4.8.51)-(4.8.52) and using (4.8.44)-(4.8.47)instead of (4.8.40)-(4.8.43), and construct q (x) for x ∈ (x1, 1) and the coefficient h1 .

Remark 4.8.4. To construct q (x) for x ∈ (x1, 1) we can act also in another way.Suppose that, using the main equation of the inverse problem, we have constructed q (x)for x ∈ (0, x1) and h . Consequently, the solutions ϕ(x, λ) and S (x, λ) are known forx

∈[0, x1] . By virtue of (4.8.20), the solution Φ(x, λ) is known for x

∈[0, x1] too. Let



287

ψ(x, λ) be the solution of (4.8.1) under the conditions ψ(1, λ) = 1, ψ(1, λ) = −h1 . Clearly,

Φ(x, λ) = −ψ(x, λ)

∆(λ). Thus, using q (x) for x ∈ [0, x1] , we can constant the functions

δ j (λ) = ψ( j−1)(x1, λ), j = 1, 2.

The functions δ j(λ) are characteristic functions of the boundary value problems Q j forequation (4.8.1) on x ∈ (x1, 1) with the conditions y( j−1)(x1) = y(1) + h1y(1) = 0 . Thus,we can reduce our problem to the inverse problem of recovering q (x), x ∈ (x1, 1) fromtwo spectra of boundary value problems Q j on the interval (x1, 1) . In this problem theweight-function R2(x) does not change sign. We can treat this inverse problem by the samemethod as above. In this case the main equation will be simpler. We note that the casewhen the weight-function does not change sign were studied in more general cases in [yur22]

and other papers.



288

REFERENCES

[abl1] Ablowitz M.J. and Segur H., Solitons and the Inverse Scattering Transform, SIAM,

Philadelphia, 1981.

[abl2] Ablowitz M. and Segur H., The inverse scattering transform: semi-infinite interval, J.Math. Phys. 16 (1975), 1054-1056.

[abl3] Ablowitz M.J. and Clarkson P.A., Solitons, Nonlinear Evolution Equations and InverseScattering, London Math. Soc. Lecture Note Series, 149, Cambridge University Press,Cambridge, 1991.

[agr1] Agranovich Z.S. and Marchenko V.A., The Inverse Problem of Scattering Theory,Gordon and Breach, New York, 1963.

hm1] Ahmad F. and Razzaghi M., A numerical solution to the Gelfand-Levitan-Marchenkoequation, Differential equations and computational simulations, II (Mississippi State,MS, 1995). Appl. Math. Comput. 89 (1998), no. 1-3, 31-39.

[ain1] Ainola L., An inverse problem on eigenoscillations of elastic covers, Appl. Math. Mech.2 (1971), 358-365.

[akt1] Aktosun T., Inverse Schrodinger scattering on the line with partial knowledge of thepotential, SIAM J. Appl. Math. 56, no. 1 (1996), 219-231.

[akt2] Aktosun T., Klaus M. and van der Mee C., Recovery of discontinuities in a non-homogeneous medium, Inverse Problems, 12 (1996), 1-25.

[akt3] Aktosun T., Klaus M. and van der Mee C., Inverse scattering in one-dimensionalnonconservative media, Integral Equations Operator Theory 30 (1998), no. 3, 279-316.

[akt4] Aktosun, Tuncay; Sacks, Paul E. Inverse problem on the line without phase informa-tion. Inverse Problems 14 (1998), no. 2, 211-224.

[ale1] Alekseev A.A., Stability of the inverse Sturm-Liouville problem for a finite interval,Dokl. Akad. Nauk SSSR 287 (1986), 11-13; English transl. in Soviet Math.Dokl. 33(1986), no.2, 299-301.

[ale2] Alekseev A.A. and Lorenzi A., Sturm-Liouville inverse problems with functionally sym-metric coefficients, Boll. Unione Mat. Ital. A (7) 8 (1994), no. 1, 11-23.

[alp1] Alpay D. and Gohberg I., Inverse spectral problem for differential operators with ra-tional scattering matrix functions, J. Diff. Equat. 118 (1995), no. 1, 1-19.

mb1] Ambarzumian V.A., Uber eine Frage der Eigenwerttheorie, Zs. f. Phys. 53 (1929),690-695.

mo1] Amour L., Inverse spectral theory for the AKNS system with separated boundaryconditions, Inverse Problems 9 (1993), no. 5, 507-523.



289

and1] Andersson L.-E., Inverse eigenvalue problems with discontinuous coefficients, InverseProblems 4 (1988), no. 2, 353-397.

and2] Andersson L.E., Inverse eigenvalue problems for a Sturm-Liouville equation in impedanceform, Inverse Problems 4 (1988), 929-971.

nd1] Anderssen R.S., The effect of discontinuities in density and shear velocity on the asymp-totic overtone structure of tortional eigenfrequencies of the Earth, Geophys. J.R. astr.Soc. 50 (1997), 303-309.

ang1] Anger G., Inverse Problems in Differential Equations, Plenum Press, New York, 1990.

[ani1] Anikonov Y.E., Multidimensional Inverse and Ill-Posed Problems for Differential Equa-tions, Utrecht, VSP, 1995.

[aru1] Arutyunyan T.N., Isospectral Dirac operators, Izv. Nats. Akad. Nauk Armenii Mat.29 (1994), no. 2, 3-14; English transl. in J. Contemp. Math. Anal., Armen. Acad.Sci. 29 (1994), no. 2, 1-10;

[atk1] Atkinson F., Discrete and Continuous Boundary Problems, Academic Press, New York,1964.

[bae1] Baev A.V., Solution of inverse problems of dissipative scattering theory, Dokl. Akad.Nauk SSSR, 315 (1990), no. 5, 1103-1104; English transl. in Soviet Phys. Dokl. 35

(1990), no.12, 1035-1036.

[bar1] Baranova E.A., On recovery of higher-order ordinary differential operators from theirspectra, Dokl. Akad. Nauk SSSR, 205 (1972), 1271-1273; English transl. in SovietMath. Dokl. 13 (1972).

aR1] Barcilon V., Explicit solution of the inverse problem for a vibrating string, J. Math.Anal. Appl. 93 (1983), no. 1, 222-234.

Bar1] Bari N.K., Biorthogonal systems and bases in a Hilbert space, Uchenye Zapiski Moskov.Gos. Univ., Matematika, 148, 4 (1951), 69-107.

bAr1] Barnes D.C., The inverse eigenvalue problem with finite data, SIAM J. Math. Anal.22 (1991), no. 3, 732-753.

[bea1] Beals R., Deift P. and Tomei C., Direct and Inverse Scattering on the Line, Math.Surveys and Monographs, v.28. Amer. Math. Soc. Providence: RI, 1988.

[bea2] Beals R., The inverse problem for ordinary differential operators on the line, Amer. J.

Math. 107 (1985), 281-366.

[bea3] Beals R. and Coifman R.R., Scattering and inverse scattering for first order systems,Comm. Pure Appl. Math., 37 (1984), 39-90.

[bea4] Beals R. and Coifman R.R., Scattering and inverse scattering for first-order systemsII, Inverse Problems 3 (1987), no. 4, 577-593.

[bea5] Beals R., Henkin G.M. and Novikova N.N., The inverse boundary problem for the

Rayleigh system, J. Math. Phys. 36 (1995), no. 12, 6688-6708.



290

bek1] Bekhiri S.E., Kazaryan A.R. and Khachatryan I.G., On the reconstruction of a regulartwo-term differential operator of arbitrary even order from the spectrum, Erevan. Gos.Univ. Uchen. Zap. Estestv. Nauk, 181 (1994), no. 2, 8-22.

[Bel1] Belishev M.I., An inverse spectral indefinite problem for the equation y + zp(x)y = 0

in the interval. Funkt. Anal. i Prilozhen. 21 (1987), no. 2, 68-69; English transl. inFunct. Anal. Appl. 21 (1987), no. 2, 146-148.

[bel1] Bellmann R. and Cooke K., Differential-difference Equations, Academic Press, NewYork, 1963.

[ber1] Berezanskii Y.M., The uniqueness theorem in the inverse spectral problem for theSchrodinger equation, Trudy Moskov. Mat. Obshch. 7 (1958), 3-51.

[ber2] Berezanskii Y.M., Expansions in eigenfunctions of Selfadjoint Operators, Translationsof Math. Monographs, vol.17, AMS, Providence, RI, 1968; (transl. from Russian:Naukova Dumka, Kiev, 1965.

[ber3] Berezanskii Y.M., Integration of nonlinear difference equations by the inverse problemmethod, Dokl. Akad. Nauk SSSR 281 (1985), 16-19.

[blo1] Blokh M.S., On the determination of a differential equation from its spectral matrix-function, Dokl. Akad. Nauk SSSR 92 (1953), 209-212.

Bor1] Borg G., Eine Umkehrung der Sturm-Liouvilleschen Eigenwertaufgabe, Acta Math. 78(1946), 1-96.

[bor1] Borovikov V.A., On the Sturm-Liouville problem with non-regular asymptotics of eigenvalues, Matem. Zametki, 50, no.6 (1991), 43-51; English transl. in Math. Notes50 (1991), no. 5-6, 1243-1249.

ou1] Boumenir A., The inverse spectral problem for the generalized second-order operator,Inverse Problems 10 (1994), no. 5, 1079-1097.

ou2] Boumenir A., Inverse spectral problem for the Laguerre differential operator, J. Math.Anal. Appl. 224 (1998), no. 2, 218-240.

bou1] Boutet de Monvel, A. and Shepelsky D., Inverse scattering problem for anisotropicmedia, J. Math. Phys. 36 (1995), no. 7, 3443-3453.

[bro1] Browne P.J. and Sleeman B.D., Inverse nodal problems for Sturm-Liouville equationswith eigenparameter dependent boundary conditions, Inverse Problems 12 (1996), no.

4, 377-381.

[bro2] Browne P.J. and Sleeman B.D., A uniqueness theorem for inverse eigenparameter de-pendent Sturm-Liouville problems, Inverse Problems 13 (1997), no. 6, 1453-1462.

buk1] Bukhgeim A.L., Introduction to the Inverse Problem Theory, Nauka, Novosibirsk, 1988.

[car1] Carlson R., Inverse spectral theory for some singular Sturm-Liouville problems, J. Diff.Equations 106 (1993), 121-140.



291

[car2] Carlson R., A Borg-Levinson theorem for Bessel operators, Pacific J. Math. 177 (1997),no. 1, 1-26.

[cau1] Caudill L.F., Perry P.A. and Schueller A.W. Isospectral sets for fourth-order ordinarydifferential operators, SIAM J. Math. Anal. 29 (1998), no. 4, 935-966.

au1] Caudrey P., The inverse problem for the third-order equation, Phys. Lett. 79A (1980),264-268.

[cha1] Chadan K. and Sabatier P.C., Inverse Problems in Quantum Scattering Theory, 2nded., Texts and Monographs in Physics. Springer-Verlag, New York-Berlin, 1989.

[cha2] Chadan K. and Musette M., Inverse problems in the coupling constant for the Schrodin-ger equation, Inverse Problems 5 (1989), no. 3, 257-268.

[cha3] Chadan K., Colton D., Paivarinta L. and Rundell W., An Introduction to Inverse Scat-tering and Inverse Spectral Problems. SIAM Monographs on Mathematical Modelingand Computation. Society for Industrial and Applied Mathematics, Philadelphia, PA,1997.

ha1] Chakravarty N.K., A necessary and sufficient condition for the existence of the spectralmatrix of a differential system, Indian J. Pure Appl. Math. 25 (1994), no. 4, 365-380.

Che1] Chern H. and Shen C.-L., On the n -dimensional Ambarzumyan‘s theorem, Inverse

Problems 13 (1997), no. 1, 15-18.

hu1] Chugunova M.V., An inverse boundary value problem on a finite interval, Funct. Anal.,35, 113-122, Ulyanovsk. Gos. Ped. Univ., Ulyanovsk, 1994.

[cla1] Clancy K. and Gohberg I., Factorization of Matrix Functions and Singular IntegralOperators, Birkhauser, Basel, 1981.

[cod1] Coddington E. and Levinson N., Theory of Ordinary Differential Equations, McGraw-

Hill Book Company, New York, 1955.[col1] Coleman C.F. and McLaughlin J.R., Solution of the inverse spectral problem for an

impedance with integrable derivative I, II, Comm. Pure Appl. Math. 46 (1993), no.2, 145-184, 185-212.

on1] Constantin A., On the inverse spectral problem for the Camassa-Holm equation, J.Funct. Anal. 155 (1998), no. 2, 352-363.

[con1] Conway J.B., Functions of One Complex Variable, 2nd ed., vol.I, Springer-Verlag, New

York, 1995.

[cox1] Cox S. and Knobel R., An inverse spectral problem for a nonnormal first order differ-ential operator, Integral Equations Operator Theory 25 (1996), no. 2, 147-162.

dah1] Dahlberg B. and Trubowitz E., The inverse Sturm-Liouville problem III, Comm. PureAppl. Math. 37 (1984), 255-267.

[dar1] Darwish A.A., The inverse scattering problem for a singular boundary value problem,

New Zeland J. Math. 22 (1993), 37-56.



292

[deg1] Degasperis A. and Shabat A., Construction of reflectionless potentials with infinitediscrete spectrum, Teoret. i Matem. Fizika, 100, no. 2 (1994), 230-247; Englishtransl.: Theoretical and Mathem. Physics, 100, no. 2 (1994), 970-984.

[dei1] Deift P. and Trubowitz E., Inverse scattering on the line, Comm. Pure Appl. Math.

32 (1979), 121-251.

[dei2] Deift P., Tomei C. and Trubowitz E., Inverse scattering and the Boussinesq equation,Comm. Pure Appl. Math. 35 (1982), 567-628.

[dei3] Deift P. and Zhou X., Direct and inverse scattering on the line with arbitrary singu-larities, Comm. Pure Appl. Math. 44 (1991), no.5, 485-533.

den1] Denisov A.M., The uniqueness of the solution of some inverse problems (Russian), Zh.

Vych. Mat. Mat. Fiz., 22, no.4 (1982), 858-864.[dip1] DiPrima R.C. and Habetler G.J., A completeness theorem for non-selfadjoint eigen-

value problems in hydrodynamic stability, Arch. Rational Mech. Anal. 34 (1969),218-227.

[dor1] Dorren H.J.S., Muyzert E.J. and Snieder R.K., The stability of one-dimensional inversescattering, Inverse Problems 10, no. 4, (1994), 865-880.

dub1] Dubrovskii V.V. and Sadovnichii V.A., Some properties of operators with a discretespectrum, Diff. Uravneniya 15 (1979), no.7, 1206-1211; English transl.: Diff. Equations15 (1979), no. 7, 854-858.

[ebe1] Eberhard W., Freiling G. and Schneider A., On the distribution of the eigenvalues of aclass of indefinite eigenvalue problem, J. Diff. Integral Equations 3 (1990), 1167-1179.

[ebe2] Eberhard W. and Freiling G., The distribution of the eigenvalues for second-ordereigenvalue problems in the presence of an arbitrary number of turning points, Resultsin Mathematics, 21 (1992), 24-41.

[ebe3] Eberhard W. and Freiling G., An expansion theorem for eigenvalue problems withseveral turning points, Analysis, 13 (1993), 301-308.

[ebe4] Eberhard W., Freiling G. and Schneider A., Connection formulas for second-orderdifferential equations with a complex parameter and having an arbitrary number of turning point, Math. Nachr., 165 (1994), 205-229.

[ebe5] Eberhard W., Freiling G., Stone-regulare Eigenwertprobleme. Math. Z. 160 (1978),

139-161.

[elc1] Elcrat A. and Papanicolaou V.G., On the inverse problem of a forth-order selfadjointbinomial operator, SIAM J. Math. Anal. 28, no. 4 (1997), 886-896.

[elr1] El-Reheem Z.F.A., On the scattering problem for the Sturm-Liouville equation on thehalf-line with sign valued weight coefficient, Applicable Analysis, 57 (1995), 333-339.

[ere1] Eremin M.S., An inverse problem for a second-order integro-differential equation with

a singularity, Differen. Uravneniya, 24 (1988), 350-351.



293

[fab1] Fabiano R.H., Knobel R. and Lowe B.D., A finite-difference algorithm for an inverseSturm-Liouville problem, IMA J. Numer. Anal. 15, no. 1 (1995), 75-88.

[fad1] Faddeev L.D., On a connection the S -matrix with the potential for the one-dimensionalSchrodinger operator, Dokl. Akad. Nauk SSSR 121 (1958), 63-66.

[fad2] Faddeev L.D., Properties of the S -matrix of the one-dimensional Schrodinger equa-tion, Trudy Mat.Inst.Steklov, 73 (1964) 314-336; English transl. in Amer. Math. Soc.Transl. (2) 65 (1967), 139-166.

[fag1] Fage M.K., Solution of certain Cauchy problem via enlargement of number of indepen-dent variables, Mat. Sb. 46(88) (1958), 261-290.

[fag2] Fage M.K., Integral representations of operator-valued analytic functions of one inde-

pendent variable, Trudy Moskov. Mat. Obshch. 8 (1959), 3-48.[FY1] Freiling G. and Yurko V.A., Inverse problems for differential equations with turning

points, Inverse Problems, 13 (1997), 1247-1263.

[FY2] Freiling G. and Yurko V.A., On constructing differential equations with singularitiesfrom incomplete spectral information, Inverse Problems, 14 (1998), 1131-1150.

[FY3] Freiling G. and Yurko V.A., Inverse spectral problems for differential equations on thehalf-line with turning points, J. Diff. Equations, 153 (1999).

[gak1] Gakhov F.D., Boundary Value Problems, 3rd ed., Nauka, Moscow, 1977; English transl.of 2nd ed., Pergamon Press, Oxford and Addison-Wesley, Reading, Mass., 1966.

[gar1] Gardner G., Green J., Kruskal M. and Miura R., A method for solving the Korteweg-deVries equation, Phys. Rev. Letters, 19 (1967), 1095-1098.

[gas1] Gasymov M.G. and Levitan B.M., Determination of o differential equation by two of its spectra, Usp. Mat. Nauk 19, no. 2 (1964), 3-63; English transl. in Russian Math.

Surveys 19 (1964), 1-64.

[gas2] Gasymov M.G., Determination of Sturm-Liouville equation with a singular point fromtwo spectra, Dokl. Akad. Nauk SSSR, 161 (1965), 274-276; English transl. in Sov.Math., Dokl. 6 (1965), 396-399.

[gas3] Gasymov M.G. and Levitan B.M., The inverse problem for a Dirac system, Dokl. Akad.Nauk SSSR, 167 (1966), 967-970.

[gas4] Gasymov M.G., The inverse scattering problem for a system of Dirac equations of order2n, Trudy Moskov. Mat. Obst. 19 (1968), 41-119; English transl. in Trans. Mosc.Math. Soc. 19 (1968).

[gas5] Gasymov M.G. and Gusejnov G.S., Determination of diffusion operators according tospectral data, Dokl. Akad. Nauk Az. SSR 37, no. 2, (1981) 19-23.

[gel1] Gelfand I.M. and Levitan B.M., On the determination of a differential equation fromits spectral function, Izv. Akad. Nauk SSSR, Ser. Mat. 15 (1951), 309-360; English

transl. in Amer. Math. Soc. Transl. (2) 1 (1955).



294

[ges1] Gesztesy F. and Simon B., Uniqueness theorems in inverse spectral theory for one-dimensional Schrodinger operators, Trans. Amer. Math. Soc. 348 (1996), no. 1,349-373.

[ges2] Gesztesy F. and Simon B., Inverse spectral analysis with partial information on the

potential. The case of an a.c. component in the spectrum. Helv. Phys. Acta 70(1997), no.1-2, 66-71.

[gla1] Gladwell G.M.L., Inverse Problems in Scattering, An introduction. Solid Mechanicsand its Applications 23, Kluwer Acad. Publ., Dordrecht, 1993.

goh1] Gohberg I.C. and Krein M.G., Introduction to the Theory of Linear NonselfadjointOperators, Nauka, Moscow, 1965; English transl.: Transl. Math. Monographs, vol.18,Amer. Math. Soc., Providence, RI, 1969.

goh2] Gohberg I.C., Kaashoek, M.A. and Sakhnovich A.L., Pseudo-canonical systems withrational Weyl functions: explicit formulas and applications, J. Diff. Equat. 146 (1998),no. 2, 375-398.

[gol1] Goldenveizer A.L., Lidsky V.B. and Tovstik P.E., Free Vibration of Thin Elastic Shells,Nauka, Moscow, 1979.

gon1] Gonchar A.A., Novikova N.N. and Khenkin G.M., Multipoint Pade approximants inan inverse Sturm-Liouville problem, Mat. Sb. 182 (1991), no. 8, 1118-1128; Englishtransl. in Math. USSR, Sb. 73 (1992), no. 2, 479-489.

[gre1] Grebert B. and Weder R., Reconstruction of a potential on the line that is a prioriknown on the half-line, SIAM J. Appl. Math. 55, no. 1 (1995), 242-254.

[gri1] Grinberg N.I., The one-dimensional inverse scattering problem for the wave equation,Mat. Sb. 181 (1990), no. 8, 1114-1129; English transl. in Math. USSR, Sb. 70 (1991),no. 2, 557-572.

[gus1] Guseinov G.S., The determination of the infinite nonselfadjoint Jacobi matrix from itsgeneralized spectral function, Mat. Zametki, 23 (1978), 237-248.

[gus2] Guseinov G.S. and Tuncay H., On the inverse scattering problem for a discrete one-dimensional Schrodinger equation, Comm. Fac. Sci. Univ. Ankara Ser. A1 Math.Statist. 44 (1995), no. 1-2, 95-102 (1996).

us1] Guseinov I.M. and Nabiev I.M., Solution of a class of inverse Sturm-Liouville boundaryvalue problems, Mat. Sbornik, 186 (1995), no. 5, 35-48; English transl. in Sb. Math.

186 (1995), no. 5, 661-674.

[hal1] Hald O.H., Discontinuous inverse eigenvalue problems, Comm. Pure Appl. Math. 37(1984), 539-577.

[hal2] Hald O.H. and McLaughlin J.R., Solutions of inverse nodal problems, Inverse Problems,5 (1989), 307-347.

[hal3] Hald O.H. and McLaughlin J.R., Inverse problems: recovery of BV coefficients from

nodes, Inverse Problems, 14 (1998), no. 2, 245-273.



295

[hel1] Hellwig G., Partial Differential Equations, Teuber, Stuttgart, 1977.

en1] Henrici P., Applied and Computational Complex Analysis, vol.2, John Wiley and Sons,New York, 1977.

[hin1] Hinton D.B., Jordan A.K., Klaus M. and Shaw J.K., Inverse scattering on the line fora Dirac system, J. Math. Phys. 32 (1991), no. 11, 3015-3030.

[hoc1] Hochstadt H., The inverse Sturm-Liouville problem, Comm. Pure Appl. Math. 26(1973), 715-729.

[hoc2] Hochstadt H., On the well-posedness of the inverse Sturm-Liouville problems, J. Diff.Eq. 23 (1977), no. 3, 402-413.

[hoe1] Hoenders B.J. On the solution of the phase retrieval problem, J. Math. Phys. 16, no.9

(1975), 1719-1725.

[hur1] Hurt N.E. Phase Retrieval and Zero Crossing, Kluwer, Dordrecht, 1989.

[isa1] Isaacson E.L. and Trubowitz E., The inverse Sturm-Liouville problem I, Comm. PureAppl. Math. 36 (1983), 767-783.

[isa2] Isaacson E.L., McKean H.P. and Trubowitz E., The inverse Sturm-Liouville problemII, Comm. Pure Appl. Math. 37 (1984), 1-11.

[Isa1] Isakov V., Inverse Problems for Partial Differential equations, Springer-Verlag, New-York, 1998.

[jer1] Jerri A.J., Introduction to Integral Equations with Applications, Dekker, New York,1985.

[kab1] Kabanikhin S.I. and Bakanov G.B., A discrete analogue of the Gelfand-Levitan method.(Russian) Doklady Akad. Nauk 356 (1997), no.2, 157-160.

[kau1] Kaup D., On the inverse scattering problem for cubic eigenvalue problems, Stud. Appl.Math. 62 (1980), 189-216.

[kay1] Kay J. and Moses H., The determination of the scattering potential from the spectralmeasure function I, II, III., Nuovo Cimento 2 (1955), 917-961 and 3 (1956), 56-84,276-304.

[kaz1] Kazarian A.R. and Khachatryan I.G., The inverse scattering problem for higher-orderoperators with summable coefficients, I, Izvestija Akad. Nauk Armenii, Mat., 29. no.

5 (1994), 50-75; English transl. in J. Contemporary Math. Anal. Armenian Acad.Sci., 29, no. 5 (1994), 42-64.

[kaz2] Kazarian A.R. and Khachatryan I.G., The inverse scattering problem for higher-orderoperators with summable coefficients, II, Izvestija Akad. Nauk Armenii, Mat., 30. no.1 (1995), 39-65; English transl. in J. Contemporary Math. Anal. Armenian Acad.Sci., 30, no. 1 (1995), 33-55.

kha1] Khachatryan I.G., The recovery of a differential equation from its spectrum, Funkt.

Anal. i Prilozhen. 10 (1976), no. 1, 93-94.



296

kha2] Khachatryan I.G., On transformation operators for higher-order differential equations,Izv. Akad. Nauk Armyan. SSR, Ser. Mat. 13 (1978), no. 3, 215-237.

kha3] Khachatryan I.G., On some inverse problems for higher-order differential operators onthe half-line, Funkt. Anal. i Prilozhen. 17 (1983), no. 1, 40-52; English transl. in

Func. Anal. Appl. 17 (1983).

kha4] Khachatryan I.G., Necessary and sufficient conditions of solvability of the inverse scat-tering problem for higher-order differential equations on the half-line, Dokl. Akad.Nauk Armyan. SSR 77 (1983), no. 2, 55-58.

Ha1] Khanh B.D., A numerical resolution of the Gelfand-Levitan equation, J. Comput.Appl. Math. 72, no. 2 (1996), 235-244.

ha1] Khasanov A.B., The inverse problem of scattering theory on the half-axis for a sys-tem of difference equations, Boundary value problems for nonclassical equations of mathematical physics, 266-295, 423, ”Fan”, Tashkent, 1986.

Khr1] Khruslov E.Y. and Shepelsky D.G., Inverse scattering method in electromagneticsounding theory. Inverse Problems, 10 (1994), no.1, 1-37.

[kir1] Kirsch A., An Introduction to the Mathematical Theory of Inverse Problems, AppliedMathematical Sciences, 120, Springer-Verlag, Berlin, 1996.

[kli1] Klibanov M.V. and Sacks P.E., Use of partial knowledge of the potential in the phaseproblem of inverse scattering, J. Cmput. Phys. 112 (1994), 273-281.

kno1] Knobel R. and Lowe B.D., An inverse Sturm-Liouville problem for an impedance, Z.Angew. Math. Phys. 44 (1993), no. 3, 433-450.

kob1] Kobayashi M., A uniqueness proof for discontinuous inverse Sturm-Liouville problemswith symmetric potentials, Inverse Problems, 5 (1989), no. 5, 767-781.

kob2] Kobayashi M., An algorithm for discontinuous inverse Sturm-Liouville problems withsymmetric potentials, Comput. Math. Appl. 18 (1989), no. 4, 349-356.

[kra1] Kravchenko K.V., Inverse spectral problems for differential operators with nonlocalboundary conditions, Ph.D. Thesis, Saratov Univ., Saratov, 1998, 101pp.

[kre1] Krein M.G., Solution of the inverse Sturm-Liouville problem, Dokl. Akad. Nauk SSSR,76 (1951), 21-24.

[kre2] Krein M.G., On a method of effective solution of the inverse problem, Dokl. Akad.

Nauk SSSR, 94 (1954), 987-990.

Kre1] Kreyszig E., Introductory Functional Analysis with Applications, John Wiley and Sons,New York, 1978.

[kru1] Krueger R.J., Inverse problems for nonabsorbing media with discontinuous materialproperties, J.Math.Phys. 23 (1982), no. 3, 396-404.

kud1] Kudishin P.M., Recovery of differential operators with a singular point, Proc. Conf.

dedicated to 90th Anniversary of L.S.Pontryagin, MSU, Moscow, 1998, p.66.



297

[kur1] Kurasov P., Scattering matrices with finite phase shift and the inverse scattering prob-lem, Inverse Problems, 12 (1996), no. 3, 295-307.

[lap1] Lapwood F.R. and Usami T., Free Oscillations of the Earth, Cambridge UniversityPress, Cambridge, 1981.

[lav1] Lavrentjev M.M., Reznitskaya K. and Yakhno V., One-Dimensional Inverse Problemsof Mathematical Physics, Nauka: Sibirsk. Otdel., Novosibirsk, 1982; English transl.:AMS Translations, Series 2, 130. American Mathematical Society, Providence, R.I.,1986.

[lax1] Lax P., Integrals of nonlinear equations of evolution and solitary waves, Comm. PureAppl. Math. 21 (1968), 467-490.

[lee1] Lee J.-H., On the dissipative evolution equations associated with the Zakharov-Shabatsystem with a quadratic spectral parameter, Trans. Amer. Math. Soc. 316 (1989),no. 1, 327-336.

[lei1] Leibenson Z.L., The inverse problem of spectral analysis for higher-order ordinarydifferential operators, Trudy Moskov. Mat. Obshch. 15 (1966), 70-144; English transl.in Trans. Moscow Math. Soc. 15 (1966).

[lei2] Leibenson Z.L., Spectral expansions of transformations of systems of boundary value

problems, Trudy Moskov. Mat. Obshch. 25 (1971), 15-58; English transl. in Trans.Moscow Math. Soc. 25 (1971).

[leo1] Leontjev A.F., A growth estimate for the solution of a differential equation for largeparameter values and its application to some problems of function theory, Sibirsk. Mat.Zh. 1 (1960), 456-487.

Lev1] Levinson N., The inverse Sturm-Liouville problem, Math. Tidsskr. 13 (1949), 25-30.

[lev1] Levitan B.M., Theory of Generalized Translation Operators, 2nd ed., Nauka, Moscow,1973; English transl. of 1st ed.: Israel Program for Scientific Translations, Jerusalem,1964.

[lev2] Levitan B.M., Inverse Sturm-Liouville Problems, Nauka, Moscow, 1984; English transl.,VNU Sci.Press, Utrecht, 1987.

[lev3] Levitan B.M. and Sargsjan I.S., Introduction to Spectral Theory, Nauka, Moscow,1970; English transl.: AMS Transl. of Math. Monographs. 39, Providence, RI, 1975.

[lev4] Levitan B.M. and Sargsjan I.S., Sturm-Liouville and Dirac Operators, Nauka, Moscow,1988; English transl.: Kluwer Academic Publishers, Dordrecht, 1991.

[li1] Li S.Y., The eigenvalue problem and its inverse spectrum problem for a class of differ-ential operators, Acta Math. Sci. (Chinese) 16 (1996), no.4, 391-403.

[lit1] Litvinenko O.N. and Soshnikov V.I., The Theory of Heterogenious Lines and theirApplications in Radio Engineering, Moscow: Radio, 1964.



298

[low1] Lowe B.D., Pilant M. and Rundell W., The recovery of potentials from finite spectraldata, SIAM J. Math. Anal. 23 (1992), no. 2, 482-504.

mal1] Malamud M.M., Similarity of Volterra operators and related problems in the theoryof differential equations of fractional orders, Trudy Moskov. Mat. Obshch. 55 (1994),

73-148; English transl. in Trans. Moscow Math. Soc. 1994, 57-122 (1995).

mal2] Malamud M.M., A connection between the potential matrix of the Dirac system andits Wronskian, Dokl. Akad. Nauk, 344, no. 5 (1995), 601-604; English transl. in Dokl.Math. 52, no. 2 (1995), 296-299.

mar1] Marchenko V.A., Sturm-Liouville Operators and their Applications, Naukova Dumka,Kiev, 1977; English transl., Birkhauser, 1986.

mar2] Marchenko V.A., Spectral Theory of Sturm-Liouville Operators, Naukova Dumka,Kiev, 1972.

mar3] Marchenko V.A., Some problems in the theory of a second-order differential operator,Dokl. Akad. Nauk SSSR, 72 (1950), 457-460.

mar4] Marchenko V.A., Some problems in the theory of linear differential operators, TrudyMoskov. Mat. Obshch. 1 (1952), 327-420.

mar5] Marchenko V.A., Expansion in eigenfunctions of nonselfadjoint singular second-orderdifferential operators, Mat. Sb. 52(94) (1960), 739-788.

mar6] Marchenko V.A. and Maslov K.V., Stability of recovering the Sturm-Liouville operatorfrom the spectral function, Mat. Sb. 81 (123) (1970), 525-551.

mar7] Marchenko V.A. and Ostrovskii I.V., A characterization of the spectrum of the Hilloperator, Mat. Sb. 97 (1975), 540-606; English transl. in Math. USSR-Sb. 26 (1975),4, 493-554.

mat1] Matsaev V.I., The existence of operator transformations for higher-order differentialoperators, Dokl. Akad. Nauk SSSR 130 (1960), 499-502; English transl. in SovietMath. Dokl. 1 (1960), 68-71.

ch1] McHugh J., An historical survey of ordinary linear differential equations with a largeparameter and turning points, Arch. Hist. Exact. Sci., 7 (1970), 277-324.

mcl1] McLaughlin J.R., An inverse eigenvalue problem of order four, SIAM J. Math. Anal.7 (1976), no. 5, 646-661.

mcl2] McLaughlin J.R., Analytical methods for recovering coefficients in differential equa-tions from spectral data, SIAM Rev. 28 (1986), 53-72.

mcl3] McLaughlin J.R., On uniqueness theorems for second order inverse eigenvalue prob-lems, J. Math. Anal. Appl. 118 (1986), no. 1, 38-41.

mcl4] McLaughlin J.R., Inverse spectral theory using nodal points as data - a unique result,J. Diff. Eq. 73 (1988), 354-362.



300

pod1] Podolskii V.E., Reconstruction of the Sturm-Liouville operator from its weighted zeta-function, Dokl. Akad. Nauk SSSR, 313 (1990), no. 3, 559-562; English tran. in SovietMath. Dokl. 42 (1991), no. 1, 53-56.

[pos1] Poschel J. and Trubowitz E., Inverse Spectral Theory, Academic Press, New York,

1987.

pov1] Povzner A.Y., On differential equations of Sturm-Liouville type on a half-axis, Mat.Sb. 23 (65) (1948), 3-52; English transl. in Amer. Math. Soc. Transl. (1) 4 (1962),24-101.

[Pri1] Prilepko A.I. and Kostin A.B., Some inverse problems for parabolic equations withfinal and integral observation. (Russian) Matemat. Sbornik 183 (1992), no. 4, 49-68;translation in Russian Acad. Sci. Sb. Math. 75 (1993), no. 2, 473-490.

[Pri2] Prilepko A.I. and Kostin, A.B., On some problems of the reconstruction of a boundarycondition for a parabolic equation. I,II. (Russian) Differ. Uravnenija. 32 (1996), no.1, 107-116; no. 11, 1519-1528; translation in Differential Equations 32 (1996), no. 1,113-122; no. 11, 1515-1525.

[pri1] Privalov I.I., Introduction to the Theory of Functions of a Complex Variable, 11th ed.,Nauka, Moscow, 1967; German trans. of 9th ed., Vols I,II,III, Teubner, Leipzig, 1958,1959.

[pro1] Provotorov V.V., On the solution of an inverse problem for the Sturm-Liouville op-erator with boundary conditions inside the interval, Functional-differential equations,132-137, Perm. Politekh. Inst., Perm, 1989.

ram1] Ramm A.G., Multidimensional Inverse Scattering Problems. Pitman Monographs andSurveys in Pure and Applied Mathematics, 51. Longman Scientific and Technical,Harlow; co-published in the United States with John Wiley and Sons, Inc., New York,1992.

[rei1] Reid, W.T., Riccati differential equations. Academic Press, New York, London, 1972.

[rja1] Rjabushko T.I., Stability of recovering the Sturm-Liouville operator from two spectra,Theory of Functions, Funct. Analysis and their Applications, 16, 186-198, Kharkov,1972.

[rof1] Rofe-Beketov F.S., The spectral matrix and the inverse Sturm-Liouville problem, The-ory of Functions, Funct. Analysis and their Applications, 4, 189-197, Kharkov, 1967.

rom1] Romanov V.G., Inverse Problems in Mathematical Physics, Nauka, Moscow, 1984;English transl.: VNU Scince Press, Utrecht, 1987.

rom2] Romanov, V.G.; Kabanikhin, S.I., Inverse problems for Maxwell’s equations. Inverseand Ill-posed Problems Series. Utrecht: VSP, iii, 249 p.

[run1] Rundell W. and Sacks P.E., On the determination of potentials without bound statedata, J. Comput. Appl. Math. 55 (1994), no. 3, 325-347.



301

[sac1] Sacks P.E., Recovery of singularities from amplitude information, J. Math. Phys. 38(1997), no. 7, 3497-3507.

[sad1] Sadovnichii V.A., On some formulations of inverse spectral problems, Uspekhi Mat.Nauk 38 (1983), no. 5, p.132.

[sak1] Sakhnovich L.A., The inverse problem for differential operators of order n > 2 withanalytic coefficients, Mat. Sb. 46 (88) (1958), 61-76.

[sak2] Sakhnovich L.A., The transformation operator method for higher-order equations, Mat.Sb. 55 (97) (1961), 347-360.

[sak3] Sakhnovich L.A., On an inverse problem for fourth-order equations, Mat. Sb. 56 (98)(1962), 137-146.

[sak4] Sakhnovich L.A., Evolution of spectral data and nonlinear equations, Ukrain. Math.Zh. 40 (1988), 533-535.

[sak5] Sakhnovich L.A., Inverse problems for equations systems, Matrix and operator valuedfunctions, 202-211, Oper. Theory Adv. Appl., 72, Birkhauser, Basel, 1994.

sAk1] Sakhnovich A., Canonical systems and transfer matrix-functions, Proc. Amer. Math.Soc. 125 (1997), no. 5, 1451-1455.

[sha1] Shabat A.B., Inverse scattering problem for a system of differential equations, Funkt.Anal. Prilozh. 9 (1975), 75-78; English transl. in Func.Anal.Appl 9 (1975), 244-247.

[sha2] Shabat A.B., An inverse scattering problem, Differ. Uravneniya 15 (1979), no. 10,1824-1834; English transl. in Differ. Equations 15 (1979), no. 10, 1299-1307.

She1] Shen C.L. and Tsai T.M., On a uniform approximation of the density function of a string equation using eigenvalues and nodal points and some related inverse nodalproblems, Inverse Problems, 11 (1995), no. 5, 1113-1123.

[she1] Shepelsky D.G., The inverse problem of reconstruction of the medium‘s conductivity ina class of discontinuous and increasing functions, Spectral operator theory and relatedtopics, 209-232, Advances in Soviet Math., 19, Amer. Math. Soc., Providence, RI,1994.

[shk1] Shkalikov A.A. and Tretter C., Spectral analysis for linear pencils of ordinary differ-ential operators, Math. Nachr. 179 (1996), 275-305.

[shu1] Shubin C., An inverse problem for the Schrodinger equation with a radial potential, J.

Diff. Equations 103 (1993), no. 2, 247-259.

[sjo1] Sjoeberg A., On the Korteweg-de Vries equation: existence and uniqueness, J. Math.Anal. Appl. 29 (1970), 569-579.

[sta1] Stashevskaya V.V., On inverse problems of spectral analysis for a certain class of differential equations, Dokl. Akad. Nauk SSSR, 93 (1953), 409-412.

[suk1] Sukhanov V.V., The inverse problem for a selfadjoint differential operator on the line,

Mat. Sb. 137 (179) (1988), 242-259; English transl. in Math. USSR Sb. 65 (1990).



302

[tak1] Takhtadjan L.A. and Faddeev L.D., Hamiltonian Methods in the Theory of Soli-tons, Nauka, Moscow, 1986; English transl.: Springer Series in Soviet Mathematics.Springer-Verlag, Berlin-New York, 1987.

[Tik1] Tikhonov A.N., On uniqueness of the solution of a electroreconnaissance problem,

Dokl. Akad. Nauk SSSR, 69 (1949), 797-800.

[tik1] Tikhonravov A.V. The accuracy obtainable in principle when solving synthesis prob-lems, Zh. Vychisl. Mat. i Mat. Fiz., 22, no. 6 (1982), 1421-1433; English transl.:USSR Comput. Maths. Math. Phys., 22, no. 6 (1982), 143-157.

[tik2] Tikhonravov A.V. The synthesis of laminar media with specified amplitude and phaseproperties, Zh. Vychisl. Mat. i Mat. Fiz. 25, no.11 (1985), 1674-1688 (Russian);English transl.: USSR Comput. Maths. Math. Phys., 25, no.11 (1985), 55-64.

[tik3] Tikhonravov A.V., Amplitude-phase properties of the spectral coefficients of laminarmedia, Zh. Vychisl. Mat. i Mat. Fiz., 25, no.2 (1985), 442-450 (Russian); Englishtransl. in USSR Comput. Maths. Math. Phys., 25, no.2 (1985), 77-83.

[tre1] Tretter C., On λ -Nonlinear Boundary Eigenvalue Problems, Mathematical Research,vol.71, Berlin: Akad. Verlag, 1993.

was1] Wasow W. Linear Turning Point Theory, Springer, Berlin, 1985.

win1] Winkler H., The inverse spectral problem for canonical systems, Integral EquationsOperator Theory, 22 (1995), no. 3, 360-374.

[yak1] Yakubov V.Y., Reconstruction of the Sturm-Liouville equation with a summable weight,Uspekhi Mat. Nauk, 51 (1996), no. 4 (310), 175-176; English transl. in Russ. Math.Surv. 51 (1996), no. 4, 758-759.

am1] Yamamoto M., Inverse spectral problem for systems of ordinary differential equationsof first order I, J. Fac. Sci. Univ. Tokyo Sect. IA Math. 35 (1988), no. 3, 519-546.

am2] Yamamoto M., Inverse eigenvalue problem for a vibration of a string with viscous drag.J. Math. Anal. Appl., 152 (1990), no.1, 20-34.

yan1] Yang X.-F., A solution of the inverse nodal problem, Inverse Problems, 13 (1997),203-213.

you1] Young R.M., An introduction to Nonharmonic Fourier Series, Academic Press, NewYork, 1980.

[yur1] Yurko V.A., Inverse Spectral Problems for Linear Differential Operators and theirApplications, Gordon and Breach, New York, 2000.

[yur2] Yurko V.A., An inverse problem for second-order differential operators with regularboundary conditions, Matematicheskie Zametki, 18 (1975), no. 4, 569-576; Englishtransl. in Math.Notes, 18 (1975), no. 3-4, 928-932.

[yur3] Yurko V.A., On stability of recovering the Sturm-Liouville operators, Differ. Equations

and Theory of Functions, Saratov Univ., Saratov 3 (1980), 113-124.



303

[yur4] Yurko V.A., Reconstruction of fourth-order differential operators, Differ. Uravneniya,19 (1983), no. 11, 2016-2017.

[yur5] Yurko V.A., Boundary value problems with a parameter in the boundary conditions,Izv. Akad. Nauk Armyan. SSR, Ser. Mat., 19 (1984), no. 5, 398-409; English transl.

in Soviet J. Contemporary Math. Anal., 19 (1984), no. 5, 62-73.

[yur6] Yurko V.A., An inverse problem for integrodifferential operators of the first order,Funct. Anal., 22, 144-151, Ulyanovsk. Gos. Ped. Univ., Ulyanovsk, 1984.

[yur7] Yurko V.A., An inverse problem for integral operators, Matematicheskie Zametki, 37(1985), no. 5, 690-701; English transl. in Math. Notes, 37 (1985), no. 5-6, 378-385.

[yur8] Yurko V.A., Uniqueness of the reconstruction of binomial differential operators from

two spectra, Matematicheskie Zametki, 43 (1988), no. 3, 356-364; English transl. inMath. Notes, 43 (1988), no. 3-4, 205-210.

[yur9] Yurko V.A., Reconstruction of higher-order differential operators, Differ. Uravneniya,25 (1989), no. 9, 1540-1550; English transl. in Differential Equations, 25 (1989), no.9, 1082-1091.

ur10] Yurko V.A., Inverse Problem for Differential Operators, Saratov University, Saratov,1989.

ur11] Yurko V.A., Recovery of differential operators from the Weyl matrix, Doklady AkademiiNauk SSSR, 313 (1990), no. 6, 1368-1372; English transl. in Soviet Math. Dokl., 42(1991), no. 1, 229-233.

ur12] Yurko V.A., A problem in elasticity theory, Prikladnaya matematika i mekhanika, 54(1990), no. 6, 998-1002; English transl. in J. Appl. Math. Mech., 54 (1990), no. 6,820-824.

ur13] Yurko V.A., An inverse problem for integro-differential operators, Matematicheskiezametki, 50 (1991), no. 5, 134-146; English transl. in Math. Notes, 50 (1991), no. 5-6,1188-1197.

ur14] Yurko V.A., An inverse problem for differential operators on the half-line, IzvestijaVusshikh Uchebnykh Zavedenii, seriya matematika, no. 12 (1991), 67-76; Englishtransl. in Soviet Math. (Iz.VVZ), 35 (1991), no. 12, 67-74.

ur15] Yurko V.A., Solution of the Boussinesq equation on the half-line by the inverse problemmethod, Inverse Problems, 7 (1991), 727-738.

ur16] Yurko V.A., Recovery of nonselfadjoint differential operators on the half-line from theWeyl matrix, Mat. Sbornik, 182 (1991), no. 3, 431-456; English transl. in Math.USSR Sbornik, 72 (1992), no. 2, 413-438.

ur17] ) Yurko V.A., Inverse problem for differential equations with a singularity, Differ.Uravneniya, 28 (1992), no. 8, 1355-1362; English transl. in Differential Equations, 28(1992), 1100-1107.



304

ur18] Yurko V.A., On higher-order differential operators with a singular point, Inverse Prob-lems, 9 (1993), 495-502.

ur19] Yurko V.A., Inverse problem for selfadjoint differential operators on the half-line, Dok-lady RAN, 333 (1993), no. 4, 449-451; English transl. in Russian Acad. Sci Dokl.

Math. 48 (1994), no. 3, 583-587.

ur20] Yurko V.A., On differential operators with nonseparated boundary conditions, Funkt.Analiz i Prilozheniya, 28 (1994), no. 4, 90-92; English transl. in Funct. Anal. Appl.,28 (1994), no. 4, 295-297.

ur21] Yurko V.A., On determination of selfadjoint differential operators on the half-line,Matematicheskie Zametki, 57 (1995), no. 3, 451-462; English transl. in Math. Notes,57 (1995), no. 3-4, 310-318.

ur22] Yurko V.A., On higher-order differential operators with a singularity, Mat. Sbornik,186 (1995), no. 6, 133-160; English transl. in Sbornik; Mathematics 186 (1995), no. 6,901-928.

ur23] Yurko V.A., On integration of nonlinear dynamical systems by the inverse problemmethod, Matematicheskie Zametki, 57 (1995), no. 6, 945-949; English transl. in Math.Notes, 57 (1995), no. 6, 672-675.

ur24] Yurko V.A., On higher-order difference operators, Journal of Difference Equations andApplications, 1 (1995), 347-352.

ur25] Yurko V.A., On inverse problems for nonlinear differential equations, Differ. Urav-neniya, 31 (1995), no. 10, 1768-1769; English transl. in Differential Equations, 31(1995), no. 10, 1741-1743.

ur26] Yurko V.A., An inverse problems for operators of a triangular structure, Results inMathematics, 30 (1996), no. 3/4, 346-373.

ur27] Yurko V.A., An inverse problem for systems of differential equations with nonlineardependence on the spectral parameter, Differ. Uravneniya, 33 (1997), no. 3, 390-395;English transl. in Differential Equations, 33 (1997), no. 3, 388-394.

ur28] Yurko V.A., Integral transforms connected with differential operators having singular-ities inside the interval, Integral Transforms and Special Functions, 5 (1997), no. 3-4,309-322.

ur29] Yurko V.A., On recovering Sturm-Liouville differential operators with singularitiesinside the interval, Matematicheskie Zametki, 64 (1998), no. 1, 143-156; English transl.in Math. Notes, 64 (1998), no. 1, 121-132.

ur30] Yurko V.A., Inverse spectral problems for differential operators and their applications,J. Math. Sciences (to appear).

ur31] Yurko V.A., An inverse problem for differential equations of the Orr-Sommerfeld type,Mathematische Nachrichten 211 (2000), 177-183.



305

inverse sturm-liouville problems and their applications

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