An inverse spectral problem for second-order functional-differential pencils with two delays
aa r X i v : . [ m a t h . SP ] O c t AN INVERSE SPECTRAL PROBLEM FOR SECOND-ORDERFUNCTIONAL-DIFFERENTIAL PENCILS WITH TWO DELAYSS.A. Buterin , M.A. Malyugina and C.-T. Shieh Abstract.
We consider a second order functional-differential pencil with two constantdelays of the argument and study the inverse problem of recovering its coefficients from thespectra of two boundary value problems with one common boundary condition. The uniquenesstheorem is proved and a constructive procedure for solving this inverse problem along withnecessary and sufficient conditions for its solvability is obtained. Moreover, we give a surveyon the contemporary state of the inverse spectral theory for operators with delay. The pencilunder consideration generalizes Sturm–Liouville-type operators with delay, which allows us toillustrate essential results in this direction, including recently solved open questions.Key words: functional-differential equation, pencil, deviating argument, constant delay,inverse spectral problem2010 Mathematics Subject Classification: 34A55 34B07 34K29
1. Introduction and main results
Recently, there appeared considerable interest in inverse problems of spectral analysis forSturm–Liouville-type operators with constant delay: ℓy ≡ − y ′′ ( x ) + q ( x ) y ( x − a ) = λy ( x ) , < x < π, (1)under two-point boundary conditions, see [1–17], which are often adequate for modelling variousreal-world processes frequently possessing a nonlocal nature. Here q ( x ) is a complex-valuedfunction in L ( a, π ) vanishing on (0 , a ) . In particular, it is well known that specification ofthe spectra { λ n,j } , j = 0 , , of two boundary value problems for the functional-differentialequation (1) with one common boundary condition in zero, say, y (0) = y ( j ) ( π ) = 0 (2)uniquely determines the potential q ( x ) as soon as a ∈ [ π/ , π ) . Moreover, the correspondinginverse problem is overdetermined. Thus, in [5], conditions on an arbitrary increasing sequenceof natural numbers { n k } k ≥ were obtained that are necessary and sufficient for the uniquedetermination of q ( x ) by specifying the corresponding subspectra { λ n k , } and { λ n k , } . For a long time, it was an open question whether the uniqueness result for a ∈ [ π/ , π )would remain true also for a ∈ (0 , π/ . The positive answer for a ∈ [2 π/ , π/
2) was givenindependently in [8], and in [10] for the case of the Robin boundary condition in zero: y ′ (0) − hy (0) = y ( j ) ( π ) = 0 . (3)For a ∈ [ π/ , π/ , the authors of [9] have shown that the spectra of both problemsconsisting of (1) and (2) uniquely determine the potential q ( x ) on ( a, a/ ∪ ( π − a/ , π ) . Butthe strongest uniqueness result under these settings was obtained in [11], where it was provedthat q ( x ) is uniquely determined on the set ( a, a/ ∪ ( π − a, a ) ∪ ( π − a/ , π ) . Moreover, Department of Mathematics, Saratov State University, Russia, email: [email protected] Department of Mathematics, Saratov State University, Russia, email: [email protected] Department of Mathematics, Tamkang University, Taiwan, email: [email protected] he authors of [11] proved that, for the complete determination of q ( x ) , it is sufficient toadditionally specify it on (3 a/ , π/ a/
4) as well as its mean value on ( π/ a/ , π − a ) . Meanwhile, the recent paper [17] gave a negative answer to the open question formulatedabove for boundary conditions (2) as soon as a ∈ [ π/ , π/
5) by constructing a one-parametricinfinite family B of different iso-bispectral potentials q ( x ) , i.e. of those for which both bound-ary value problems possess one and the same pair of spectra. It is also interesting that potentialsin B differ on the set (3 a/ , π − a ) ∪ (2 a, π − a/
2) by an arbitrary multiplicative complexconstant. Thus, the uniqueness subdomain for q ( x ) established in [11] is unimprovable.This appeared quite unexpected taking into account that the paper [16] announced thatspecification of the spectra of both boundary value problems consisting of (1) and (3) uniquelydetermines the potential q ( x ) also for a ∈ [ π/ , π/ , see also [13]. Even though, in [16],Robin boundary conditions were imposed also in the point π, they can be easily reduced to (3).Moreover, for any fixed a ∈ (0 , π ) , in the papers [2] and [6] for the cases of boundaryconditions (2) and (3), respectively, it was established that if the spectra coincide with the onesof the corresponding problems with the zero potential, then q ( x ) is zero too.Among numerous studies devoted to inverse spectral problems for functional-differentialoperators with delay, to the best of our knowledge, [5] remains a sole work dealing with necessaryand sufficient conditions of solvability for inverse problems of this class. In particular, from theresults in [5] it follows that, unlike the classical case a = 0 , the spectra { λ n, } and { λ n, } mayhave any finite number of common eigenvalues. In Appendix A, we supplement the work [5],by applying the solvability result of the present paper to (1) for a ∈ [2 π/ , π ) . There are also works devoted to inverse problems for operators with several delays: ℓ m y ≡ − y ′′ ( x ) + m X ν =1 q ν ( x ) y ( x − a ν ) = λy ( x ) , < x < π, m > , (4)where q ν ( x ) = 0 on (0 , a ν ) for ν = 1 , n, see [18–22]. However, all potentials q ν ( x ) cannotbe completely determined simultaneously even by specifying arbitrarily many different spectra.This becomes especially obvious when all a ν are equal, but even for different a ν there remainsubintervals, where the functions q ν ( x ) cannot be distinguished (see Appendix B).An attempt to generalize the results of papers [2, 6] to the operator ℓ was made in [19],where it was claimed that if the spectra of two boundary value problems for one and the sameequation (4) for m = 2 along with the boundary conditions (2) coincide with the analogoustwo spectra corresponding to the zero potentials, then q ( x ) and q ( x ) are zeros too. There-after, the generalization of this assertion to arbitrary m > cannot be true even for different delays (see the counterexample in Appendix B).Nevertheless, one can completely recover all potentials q ν ( x ) by using the spectra of 2 m boundary value problems for m different equations specially composed from (4). For example,in [21,22] for m = 2 and a , a ∈ [ π/ , π ) , it was shown that, for recovering q ( x ) and q ( x ) , it is sufficient to specify the spectra of four boundary value problems for two equations: − y ′′ ( x ) + q ( x ) y ( x − a ) + ( − ν q ( x ) y ( x − a ) = λy ( x ) , < x < π, ν = 1 , . In the present paper, we show, in particular, that there is no need to use two differentequations if one of delayed terms depends on the spectral parameter. Specifically, we considerthe functional-differential equation with nonlinear dependence on the spectral parameter ρ : y ′′ ( x ) + ρ y ( x ) = q ( x ) y ( x − a ) + 2 ρq ( x ) y ( x − a ) , < x < π, (5)2hich generalizes equation (1). We assume that a ∈ [ π/ , π ) , a ∈ [ π/ , π ) and a + a ≥ π. The case a + a < π requires a separate investigation (see Section 3 for details). For ν = 0 , , let q ν ( x ) be a complex-valued function in W ν [ a ν , π ] , q ν ( x ) = 0 on (0 , a ν ) and π Z a q ( x ) dx = 0 . (6)For j = 0 , , we denote by { ρ n,j } the spectrum of the boundary value problem L j := L j ( q , q )that consists of equation (5) along with the boundary conditions (2). We also use one and thesame symbol { κ n } for denoting different sequences in l , and put Z := Z \{ } and Z := Z . Under our assumptions, we prove first the following theorem giving asymptotics of the spectra.
Theorem 1.
For j = 0 , and n ∈ Z j , the following asymptotics holds: ρ n,j = ρ n,j + ωπn cos ρ n,j a + α j πn sin ρ n,j a + κ n n , α , α , ω ∈ C , (7) where ρ n,j = n − j/ . Moreover, α j = α + ( − j β, α = q ( a )2 , β = q ( π )2 , ω = 12 π Z a q ( x ) dx. (8)For α = 0 , this theorem was announced in the conference papers [23, 24], which becamethe first works dealing with inverse problems for functional-differential pencils in any form.Consider the following inverse problem. Inverse Problem 1.
Given the spectra { ρ n,j } n ∈ Z j , j = 0 ,
1; find q ( x ) and q ( x ) . We note that for a = a = 0 this inverse problem was studied in [25]. The followingtheorem gives uniqueness of its solution under our present settings. Theorem 2.
Let both spectra { ρ n,j } n ∈ Z j , j = 0 , , be specified. Then the function q ( x ) isuniquely determined a.e. on the union of intervals I := ( a , a / ∪ ( π − a , a ) ∪ ( π − a / , π ) , while the function q ( x ) is uniquely determined on the entire segment [ a , π ] . In particular, both functions q ( x ) and q ( x ) are completely determined when a ≥ π/ . However, for a ∈ [ π/ , π/ , the complete uniqueness does not take place. For illustratingthis, one can use the same one-parametric family of functions B = { q ,γ ( x ) } γ ∈ C constructedin [17], for which also the problems L j ( q ,γ , q ) , j = 0 , , will possess one and the same pairof spectra for all γ ∈ C . Since, as was already mentioned above, for different values of γ thefunctions q ,γ ( x ) differ precisely on the set ( a , π ) \ cl( I ) except their common zeros, theuniqueness subdomain I cannot be refined.However, as in [11], the function q ( x ) would be determined uniquely if some a prioriinformation on it were additionally specified. Namely, the following theorem holds. Theorem 3.
Under the hypothesis of Theorem 2, the specification of the function q ( x ) on the subinterval (3 a / , π/ a / along with the value ω := π − a Z π + a q ( x ) dx. determines it also on ( π/ a / , π − a ) ∪ (2 a , π − a / . So q ( x ) is determined completely.
3s in [5], one can show that, in the case a ≥ π/ , Inverse Problem 1 is overdetermined,and also describe subspectra, whose specification would uniquely determine the functions q ( x )and q ( x ) . Some results of this type can be found in [23, 24]. But here we restrict ourself todealing with the full spectra, which surprisingly does not prevent us from obtaining necessaryand sufficient conditions for the solvability of Inverse Problem 1 for a ≥ π/ . Besides asymp-totics (7), these conditions include some restrictions on the growth of certain entire functions,which makes the inverse problem consistent even in spite of its overdetermination. Specifically,the following theorem holds.
Theorem 4.
Let a ≥ π/ . Then for any sequences of complex numbers { ρ n, } | n |∈ N and { ρ n, } n ∈ Z to be the spectra of some boundary value problems L ( q , q ) and L ( q , q ) , respectively, it is necessary and sufficient to satisfy the following two conditions:(i) For j = 0 , , the sequence { ρ n,j } n ∈ Z j has the form (7);(ii) For j, ν = 0 , , the exponential type of the function g j,ν ( ρ ) does not exceed π − a ν , where g j,ν ( ρ ) = θ j ( ρ ) + ( − j + ν θ j ( − ρ ) , (9) θ ( ρ ) = ρ ∆ ( ρ ) − ρ sin ρπ + ω cos ρ ( π − a ) − α sin ρ ( π − a ) , (10) θ ( ρ ) = ρ ∆ ( ρ ) − ρ cos ρπ − ω sin ρ ( π − a ) − α cos ρ ( π − a ) , (11) while the functions ∆ j ( ρ ) are determined by the formula ∆ j ( ρ ) = π − j Y n ∈ Z j ρ n,j − ρρ n,j exp (cid:16) ρρ n,j (cid:17) , j = 0 , . (12)For j = 0 , , the function ∆ j ( ρ ) determined by (12) is the characteristic function of theproblem L j (see the next section). The proof of Theorem 4 is constructive and gives analgorithm for solving Inverse Problem 1 (Algorithm 1 in Section 6). For proving Theorem 4, weobtain and study a transformation operator associated with equation (1), which allows reducingthe inverse problem to the so-called main vectorial integral equation. For applications of thetransformation operator approach to other classes of nonlocal operators, see the survey [26].The paper is organized as follows. In the next section, we construct a transformation op-erator for the sine-type solution of equation (5), and study the characteristic functions of theproblems L j . Therein, we also give the proof of Theorem 1. In Section 3, we derive andstudy the main equation of Inverse Problem 1. In Section 4, we prove the uniqueness theorems(Theorems 2 and 3). In Section 5, we obtain important representations for the functions deter-mined by (12) with arbitrary complex zeros of the form (7). In Section 6, we prove Theorem 4and obtain an algorithm for solving the inverse problem. In Appendix A, we provide an ana-log of Theorem 4 for the Sturm–Liouville-type operator ℓ with the delay a ∈ [2 π/ , π ) , i.e.when q ( x ) ≡ . In Appendix B, we give a counterexample, showing that specification of anyspectra does not uniquely determine all potentials in equation (4) even for m = 2 and a = a .
2. Transformation operator and characteristic functions
Let y = S ( x, ρ ) be the sine-type solution of equation (5), i.e. the solution satisfying theinitial conditions S (0 , ρ ) = 0 and S ′ (0 , ρ ) = 1 . By virtue of its uniqueness, eigenvalues of theproblem L j , j = 0 , , coincide with zeros of the entire function∆ j ( ρ ) := S ( j ) ( π, ρ ) , (13)4hich is called characteristic function of L j . We introduce the designations Q ν ( x ) := x Z a ν q ν ( t ) dt, ν = 0 , , c ( x ) := cos x, c ( x ) := sin x, where the latter two ones are aimed to be used only occasionally for convenience.The following lemma gives the transformation operator that connects the solutions ρ − sin ρx and cos ρx of the simplest equation (5), possessing zero coefficients, with the solution S ( x, ρ ) . Lemma 1.
The following representation holds: S ( x, ρ ) = sin ρxρ − Q ( x ) cos ρ ( x − a ) ρ + X ν =0 x Z a ν K ν ( x, t ) c − ν ( ρ ( x − t )) ρ dt, ≤ x ≤ π, (14) where K ν ( x, t ) = 0 in the exterior of the triangle a ν ≤ t ≤ x ≤ π as ν = 0 , , and K ( x, t ) = 12 (cid:16) q (cid:16) x − t − a (cid:17) + q (cid:16) t + a (cid:17)(cid:17) , a ≤ t ≤ x ≤ π, (15) while the kernel K ( x, t ) satisfies the following integral equation: K ( x, t ) = 12 x − t − a Z t + a q ( τ ) dτ + A ( x, t )2 , a ≤ t ≤ x ≤ π, (16) where A ( x, t ) = , a ≤ t ≤ min { x, a } , x ≤ π, x Z t q ( τ ) dτ t − a Z a K ( τ − a , η ) dη − x Z x − t + a q ( τ ) dτ τ − x )+ t − a Z a K ( τ − a , η ) dη + t Z t + a q ( τ ) dτ τ − t − a Z a K ( τ − a , η ) dη, a < t ≤ x ≤ π. (17) In particular, the following relations hold: K ( x, x ) = 0 , K ( x, a ) = Q ( x )2 , K ( x, a ) = q ( x )2 + α, A ( π, a ) = A ( π, π ) = 0 . (18) Remark 1.
Since π ≤ a , the integration variable η in (17) never exceeds 2 a . Thus,the function K ( τ − a , η ) under the integrals in (17) is determined by the formula K ( τ − a , η ) = 12 τ − η + a Z η + a q ( ζ ) dζ , a ≤ η ≤ min { τ − a , a } , τ ≤ π. (19) Proof of Lemma 1.
Clearly, the function S ( x, ρ ) obeys the following integral equation: S ( x, ρ ) = sin ρxρ + X ν =0 (2 ρ ) ν x Z a ν sin ρ ( x − t ) ρ q ν ( t ) S ( t − a ν , ρ ) dt, ≤ x ≤ π. (20)5ince a ν + a ≥ π for ν = 0 , , substituting (14) into (20), we arrive at the relation x Z a K ( x, t ) sin ρ ( x − t ) ρ dt + x Z a K ( x, t ) cos ρ ( x − t ) ρ dt − Q ( x ) cos ρ ( x − a ) ρ = X j =0 A j ( x, ρ ) , (21)where A ν ( x, ρ ) = 2 ν ρ ν − x Z a ν sin ρ ( x − t ) q ν ( t ) dt t − a ν Z cos ρτ dτ, ν = 0 , , A ( x, ρ ) = x Z a sin ρ ( x − t ) ρ q ( t ) dt t − a Z a K ( t − a , τ ) dτ t − τ − a Z cos ρξ dξ. Using the formula 2 sin ρ ( x − t ) cos ρτ = sin ρ ( x − t + τ ) + sin ρ ( x − t − τ ) and changing theintegration variables along with its order, we get A ν ( x, ρ ) = (2 ρ ) ν − x Z a ν sin ρ ( x − t ) dt x − t − aν Z t + aν q ν ( τ ) dτ, ν = 0 , , (22) A ( x, ρ ) = x Z a A ( x, t ) sin ρ ( x − t )2 ρ dt, (23)where A ( x, t ) is determined by formula (17). Moreover, integration by parts gives A ( x, ρ ) = − Q ( x ) cos ρ ( x − a ) ρ + x Z a (cid:16) q (cid:16) x − t − a (cid:17) + q (cid:16) t + a (cid:17)(cid:17) cos ρ ( x − t )2 ρ dt. (24)Substituting (22) for ν = 0 as well as (23) and (24) into (21), we arrive at (14)–(17). Relations(18) are obvious. (cid:3) The next lemma gives fundamental representations for the characteristic functions.
Lemma 2.
The following representations hold: ∆ ( ρ ) = sin ρπρ − ω cos ρ ( π − a ) ρ + α sin ρ ( π − a ) ρ + X ν =0 π − a ν Z w ,ν ( x ) c ν ( ρx ) ρ dx, (25)∆ ( ρ ) = cos ρπ + ω sin ρ ( π − a ) ρ + α cos ρ ( π − a ) ρ + X ν =0 π − a ν Z w ,ν ( x ) c − ν ( ρx ) ρ dx, (26) where w j,ν ( x ) ∈ L (0 , π − a ν ) , j, ν = 0 , , and π − a Z w , ( x ) dx = ω, π − a Z xw , ( x ) dx = α ( a − π ) , π − a Z w , ( x ) dx = − α . (27) Moreover, w ,ν ( x ) = ( − ν +1 K ν, ( π, π − x ) , w ,ν ( x ) = P ν ( π, π − x ) , ν = 0 , , (28)6 here K ν, ( x, t ) := ∂∂x K ν ( x, t ) , K ν, ( x, t ) := ∂∂t K ν ( x, t ) , P ν ( x, t ) := K ν, ( x, t ) + K ν, ( x, t ) . (29) Proof.
Integrating by parts in (14) with account of (18), and recalling (6), (8) along with(13), we get (25). Differentiating (14) with respect to x and then using integration by parts,we obtain (26). Finally, although relations (27) can be established by direct calculations, weaccept them just as a simple corollary from entireness of the functions ∆ ( λ ) and ∆ ( λ ) . (cid:3) Now we are in position to give the proof of Theorem 1.
Proof of Theorem 1.
By the standard approach involving Rouch´e’s theorem (see, e.g., [27]),using representation (25), one can show that the function ∆ ( λ ) has infinitely many zeros ofthe form ρ n, = n + ε n, , where | n | ∈ N , while ε n, → | n | → ∞ . Substituting thisrepresentation into (25), we obtainsin ρ n, π = ωn cos( n + ε n, )( π − a ) − α n sin( n + ε n, )( π − a ) + κ n n . (30)Since sin ρ n, π = sin( n + ε n, ) π = ( − n ε n, π + O ( ε n, ) as | n | → ∞ , we refine ε n, = O ( n − )for | n | → ∞ . Hence, we have the asymptotic formulaesin ρ n, π = ( − n ε n, π + O (cid:16) n (cid:17) , cos( n + ε n, )( π − a ) = ( − n cos na + O (cid:16) n (cid:17) , sin( n + ε n, )( π − a ) = ( − n +1 sin na + O (cid:16) n (cid:17) as soon as | n | → ∞ . Substituting them into (30), we arrive at ε n, = ωπn cos na + α πn sin na + κ n n , which implies (7) for j = 0 . Analogously, applying Rouch´e’s theorem to representation (26), we get ρ n, = ρ n, + ε n, , where n ∈ Z , while ε n, → | n | → ∞ . Substituting this into (26), we obtaincos ρ n, π = − ωn sin( ρ n, + ε n, )( π − a ) − α n cos( ρ n, + ε n, )( π − a ) + κ n n . (31)Since cos ρ n, π = cos( ρ n, + ε n, ) π = ( − n ε n, π + O ( ε n, ) as | n | → ∞ , we refine ε n, = O ( n − )for | n | → ∞ . Then substituting the asymptotic formulaesin( ρ n, + ε n, )( π − a ) = ( − n +1 cos ρ n, a + O (cid:16) n (cid:17) , cos( ρ n, + ε n, )( π − a ) = ( − n +1 sin ρ n, a + O (cid:16) n (cid:17) , for | n | → ∞ into (31), we arrive at ε n, = ωπn cos ρ n, a + α πn sin ρ n, a + κ n n , which implies (7) for j = 1 . (cid:3) Lemma 3.
For any a , a ∈ [0 , π ] , each function ∆ ( ρ ) and ∆ ( ρ ) of the form describedin (25)–(27) is determined by its zeros uniquely. Moreover, representation (12) holds.Proof. By virtue of Hadamard’s factorization theorem (see, e.g., [28]), we get∆ j ( ρ ) = C j ρ s j exp( b j ρ ) Y ρ n,j =0 (cid:16) − ρρ n,j (cid:17) exp (cid:16) ρρ n,j (cid:17) , j = 0 , , (32)where C j and b j are some constants, while s j is the multiplicity of the null zero ρ n,j = 0 . In particular, we have ρ j − c − j ( ρπ ) = π − j Y n ∈ Z j (cid:16) − ρρ n,j (cid:17) exp (cid:16) ρρ n,j (cid:17) , j = 0 , . (33)Dividing (32) by (33), we obtain ρ − j ∆ j ( ρ ) c − j ( ρπ ) = C j π − j exp (cid:16)(cid:16) b j + X ρ n,j =0 (cid:16) ρ n,j − ρ n,j (cid:17) − X ρ n,j =0 ρ n,j (cid:17) ρ (cid:17) × Y ρ n,j =0 (cid:16) ρ − ρ n,j (cid:17) − Y ρ n,j =0 ρ n,j ρ n,j Y ρ n,j =0 ρ n,j − ρρ n,j − ρ , j = 0 , . (34)On the other hand, (25) and (26) imply ρ − j ( c − j ( ρπ )) − ∆ j ( ρ ) → j = 0 , ρ → −∞ , which along with (34) gives C j = π − j ( − s j Y ρ n,j =0 ρ n,j Y ρ n,j =0 ρ n,j ρ n,j , b j = X ρ n,j =0 ρ n,j + X ρ n,j =0 (cid:16) ρ n,j − ρ n,j (cid:17) . Substituting this into (32), we arrive at (12). (cid:3)
3. Main equation of the inverse problem
The relations in (28) can be considered as a system of equations with respect to the functions q ( x ) and p ( x ) := q ′ ( x ) , which we refer to as main (vectorial) equation of Inverse Problems 1.By virtue of our standing assumption a + a ≥ π, for each ν ∈ { , } , the functions K ν, ( x, t )and P ν ( x, t ) depend only on q ν ( x ) , while for ν = 1 they depend even only on p ( x ) . Hence,the main equation can be splitted into two independent subsystems for ν = 0 and ν = 1 : w , ( x ) = − K , ( π, π − x ; q ) , w , ( x ) = P ( π, π − x ; q ) , (35) w , ( x ) = K , ( π, π − x ; p ) , w , ( x ) = P ( π, π − x ; p ) , (36)respectively. Here and below, in order to emphasize dependence of a certain function F ( x , x )on some function f ( x ) , sometimes we write F ( x , x ; f ) . According to (15)–(17) and (29), the subsystem (35) is nonlinear when a ∈ [ π/ , π/ , while the subsystem (36) is always linear because a ≥ π/ . Consider first the linear subsystem (36). By virtue of (15) and (29), we get K , ( x, t ) = 12 p (cid:16) x − t − a (cid:17) , K , ( x, t ) = 14 (cid:16) p (cid:16) t + a (cid:17) − p (cid:16) x − t − a (cid:17)(cid:17) , ( x, t ) = 14 (cid:16) p (cid:16) t + a (cid:17) + p (cid:16) x − t − a (cid:17)(cid:17) , a < t < x < π. Thus, the subsystem (36) is equivalent to the system w j, ( x ) = 14 (cid:16) p (cid:16) π + a − x (cid:17) − ( − j p (cid:16) π + a + x (cid:17)(cid:17) , < x < π − a , j = 0 , . Solving this linear system, we get p (cid:16) π + a − x (cid:17) = 2( w , + w , )( x ) , p (cid:16) π + a + x (cid:17) = 2( w , − w , )( x ) , x ∈ (0 , π − a ) , or, after changing the variable, we have p ( x ) = 2 ( w , + w , )( π + a − x ) , a < x < a + π , ( w , − w , )(2 x − π − a ) , a + π < x < π. (37)Thus, we arrive at the following theorem. Theorem 5.
Let a ∈ [ π/ , π ) . Then for any functions w , ( x ) , w , ( x ) ∈ L (0 , π − a ) the linear subsystem (36) has a unique solution p ( x ) ∈ L ( a , π ) , which can be constructed byformula (37) . Moreover, π Z a p ( x ) dx = 2 π − a Z w , ( x ) dx, π Z a xp ( x ) dx = π + a π Z a p ( x ) dx − π − a Z xw , ( x ) dx. (38) Proof.
It remains to prove (38). Changing the integration variable, one can easily obtainthe following relations for any integrable function f ( x ) : π + a Z a f ( π + a − x ) dx = 12 π − a Z f ( x ) dx, π Z π + a f (2 x − π − a ) dx = 12 π − a Z f ( x ) dx, (39)which along with (37) give the first relation in (38). Analogously, using (37) and the relations π + a Z a xf ( π + a − x ) dx = π − a Z π + a − x f ( x ) dx, π Z π + a xf (2 x − π − a ) dx = π − a Z π + a + x f ( x ) dx, one can obtain the second identity in (38). (cid:3) Further, differentiating (16) and (17), and taking (29) into account, we get K ,l ( x, t ) = 12 q (cid:16) x − t − a (cid:17) + ∂∂x A ( x, t ) , l = 1 , − (cid:16) q (cid:16) t + a (cid:17) + q (cid:16) x − t − a (cid:17)(cid:17) + ∂∂t A ( x, t ) , l = 2 , (40)where ∂∂x A ( x, t ) = , a ≤ t ≤ min { x, a } , x ≤ π, x Z x − t + a q ( τ ) K ( τ − a , τ − x ) + t − a ) dτ, a < t ≤ x ≤ π, (41)9nd ∂∂t A ( x, t ) = , a ≤ t ≤ min { x, a } , x ≤ π, x Z t q ( τ ) K ( τ − a , t − a ) dτ − x Z x − t + a q ( τ ) K ( τ − a , τ − x ) + t − a ) dτ − t Z t + a q ( τ ) K ( τ − a , τ − t − a ) dτ, a < t ≤ x ≤ π. (42)By virtue of (29) and (40), we get P ( x, t ) = 14 (cid:16) q (cid:16) x − t − a (cid:17) − q (cid:16) t + a (cid:17)(cid:17) + B ( x, t ) , (43)where B ( x, t ) = 12 (cid:16) ∂∂x A ( x, t ) + ∂∂t A ( x, t ) (cid:17) . Then, summing up (41) and (42) and dividing by 2 , we arrive at B ( x, t ) = 12 , a ≤ t ≤ min { x, a } , x ≤ π, x Z t q ( τ ) K ( τ − a , t − a ) dτ + x Z x − t + a q ( τ ) K ( τ − a , τ − x ) + t − a ) dτ − t Z t + a q ( τ ) K ( τ − a , τ − t − a ) dτ, a < t ≤ x ≤ π. (44)According to formula (40) for l = 2 along with (43), the subsystem (35) takes the form w j, ( x ) = 14 (cid:16) q (cid:16) π + x + a (cid:17) +( − j q (cid:16) π − x + a (cid:17)(cid:17) + u j ( x ) , ≤ x ≤ π − a , j = 0 , , (45)where u ( x ) = − ∂∂t A ( π, t ) (cid:12)(cid:12)(cid:12) t = π − x = 12 ddx A ( π, π − x ) , u ( x ) = B ( π, π − x ) . (46)Thus, by virtue of (42) and (44), we get u j ( x ) = 12 ( − j +1 π Z π − x q ( τ ) K ( τ − a , π − x − a ) dτ + π Z π + x + a q ( τ ) K ( τ − a , τ − x − π − a ) dτ +( − j π − x Z π − x + a q ( τ ) K ( τ − a , τ + x − π − a ) dτ, ≤ x < π − a , , π − a ≤ x ≤ π − a , (47)10ransforming (45) and taking (47) into account, we obtain2( w , + ( − j w , )( x ) = q (cid:16) π + a + ( − j x (cid:17) + ( u + ( − j u )( x ) , ≤ x < π − a , , π − a ≤ x ≤ π − a , for j = 0 , . After changing the variables, we get2( w , − w , )( π + a − x ) = q ( x ) , a ≤ x ≤ a ,q ( x ) + 2 v ( x ) , a < x ≤ a + π , (48)2( w , + w , )(2 x − π − a ) = q ( x ) + 2 v ( x ) , a + π ≤ x < π − a ,q ( x ) , π − a ≤ x ≤ π, (49)where (note that u (0) = 0) v ( x ) = ( u − u )( π + a − x ) , a < x ≤ a + π , ( u + u )(2 x − π − a ) , a + π < x < π − a . (50)Obviously, formulae (48) and (49) immediately give the solution q ( x ) of the subsystem (35)on I := [ a , a / ∪ [ π − a / , π ] . For a ≥ π/ , we have [ a , π ] ⊂ I and, hence, the function q ( x ) is completely obtained. For a < π/ , the subsystem (35) becomes nonlinear, and itssolvability on I := (3 a / , π − a /
2) is conditioned by the following lemma.
Lemma 4.
Let v ( x ) be determined on the interval I by formula (50) with u ( x ) and u ( x ) constructed in (47). Then v | D does not depend on q | I if and only if D ⊂ [ π − a , a ] . Here and below f | S denotes the restriction of the function f to the set S . Proof.
Substituting (47) into (50), we obtain the formulae v ( x ) = x − a Z x + a q ( τ ) K ( τ − a , τ − x )) dτ − π Z x − a q ( τ ) K ( τ − a , x − a )) dτ, a < x ≤ a + π , (51)and v ( x ) = π Z x + a q ( τ ) K ( τ − a , τ − x )) dτ, a + π < x < π − a . (52)where, according to Remark 1, the function K ( τ − a , · ) is determined by (19), i.e. K ( τ − a , τ − x )) = − K ( τ − a , x − a )) = 12 x − a Z τ − x + a q ( ζ ) dζ . v ( x ) = 12 x − a Z x + a q ( τ ) dτ x − a Z τ − x + a q ( ζ ) dζ − π Z x − a q ( τ ) dτ τ − x + a Z x − a q ( ζ ) dζ = 12 π Z x + a q ( τ ) dτ x − a Z τ − x + a q ( ζ ) dζ , a < x < π − a . (53)Thus, the function v ( x ) does not depend on q | [ π − a , a ] . Moreover, according to the firstrepresentation in (53), it depends on q | (3 a / ,π − a ) if and only if x − a > a π − x + a > a , i.e. x ∈ I \ [ π − a , a ] . Analogously, v ( x ) depends on q | (2 a ,π − a / if and only if x < π − a . Hence, v ( x ) is independent of q | I if and only if x ∈ [ π − a , a ] . (cid:3) Lemma 4 along with formulae (48) and (49) guaranties solvability of the subsystem (35) onthe set I = ( a , a / ∪ ( π − a , a ) ∪ ( π − a / , π ) . The following corollary gives a conditionof the solvability on the entire interval ( a , π ) Corollary 1. v | I does not depend on q | I if and only if a ≥ π/ . Proof.
According to Lemma 4, v | I does not depend on q | I if and only if I ⊂ [ π − a , a ] , which, in turn, is equivalent to a ≥ π/ . (cid:3) According to Corollary 1, formulae (48), (49) and (53) give the representation v ( x ) = 2 π Z x + a ( w , + w , )(2 τ − π − a ) dτ x − a Z τ − x + a ( w , − w , )( π + a − ζ ) dζ = 12 π − a Z x − π ( w , + w , )( τ ) dτ x − a − τ Z π +2 a − x ( w , − w , )( ζ ) dζ , a < x < a + π , (54)as soon as a ∈ [2 π/ , π/ . Thus, we arrive at the following theorem.
Theorem 6.
Let a ∈ [2 π/ , π ) . Then for any functions w , ( x ) , w , ( x ) ∈ L (0 , π − a ) subsystem (35) has a unique solution q ( x ) ∈ L ( a , π ) , which can be constructed by the formula q ( x ) = 2 ( w , − w , )( π + a − x ) , a < x < a , ( w , − w , )( π + a − x ) − v ( x ) , a < x < a + π , ( w , + w , )(2 x − π − a ) − v ( x ) , a + π < x < π − a , ( w , + w , )(2 x − π − a ) , π − a < x < π, (55) where the function v ( x ) is determined by formula (54). Moreover, π Z a q ( x ) dx = 2 π − a Z w , ( x ) dx. (56)12 roof. It remains to prove (56). Indeed, integrating (55) and using (39), we get π Z a q ( x ) dx = 2 π − a Z w , ( x ) dx − π − a Z a v ( x ) dx, where, according to (50), we have π − a Z a v ( x ) dx = π + a Z a ( u − u )( π + a − x ) dx + π − a Z π + a ( u + u )(2 x − π − a ) dx = π − a Z u ( x ) dx. Finally, using (46) and the last two equalities in (18), we calculate π − a Z u ( x ) dx = A ( π, a ) − A ( π, π )2 = 0 , which finishes the proof. (cid:3)
4. Proof of the uniqueness theorems
Let us begin with the following assertion.
Lemma 5.
Let a , a ∈ (0 , π ) . Then specification of any pair of sequences { ρ n,j } n ∈ Z j ,j = 0 , , of the form (7) uniquely determines the values α , α and ω. Proof.
According to (7), we arrive at the asymptotic formulae ω cos na = γ n, + γ − n, o (1) , α j sin ρ n,j a = γ n,j − γ j − n,j o (1) , | n | → ∞ , (57)where we denoted γ n,j := πn ( ρ n,j − ρ n,j ) , n ∈ Z j , j = 0 , . (58)By virtue of Lemma 3.3 in [12], the sequences { cos na } n ≥ and { sin ρ n,j a } n ≥ , j = 0 , , do not converge, i.e. each of them has at least two different partial limits. Choose increasingsequences of natural numbers { m k,l } , l = 1 , , so that r := lim k →∞ cos m k, a = 0 , r l := lim k →∞ sin ρ m k,l ,l − a = 0 , l = 2 , . (59)According to (57) and (59), we calculate the values ω, α and α by the formulae ω = lim k →∞ γ m k, , + γ − m k, , r , α j = lim k →∞ γ m k,j +2 ,j − γ j − m k,j +2 ,j r j +2 , j = 0 , , (60)which finishes the proof. (cid:3) Proof of Theorem 2.
According to Lemmas 3 and 5, under the hypothesis of the theorem,the functions ∆ j ( ρ ) , j = 0 , , as well as the numbers α , α , and ω are determine uniquely.Then, by virtue of Lemma 2, the functions w j,ν ( x ) , j, ν = 0 , , are uniquely determined too.13hus, by Theorem 5 and Lemma 4, the function p ( x ) = q ′ ( x ) is uniquely determined a.e. onthe interval ( a , π ) , while q ( x ) is so on I . Finally, taking (8) into account, we get q ( x ) = α + α + x Z a p ( t ) dt, a ≤ x ≤ π, (61)which finishes the proof. (cid:3) Proof of Theorem 3.
Following the proof of Theorem 2 in [11], we denote p := q | ( a ,π − a ) , p := q | (2 a ,π ) , R ( x, t ) := x − a Z t − x + a p ( τ ) dτ, R ( x ) := π Z x p ( t ) dt. Then, by virtue of (53) and (55) we have the relation F ( x ) = q ( x ) + π Z x + a p ( t ) dt x − a Z t − x + a p ( τ ) dτ, a < x < π − a , (62)where F ( x ) = 2 ( w , − w , )( π + a − x ) , a < x < a + π , ( w , + w , )(2 x − π − a ) , a + π < x < π − a . After changing the order of integration, relation (62) on the target intervals takes the forms: F ( x ) = p ( x ) + π − a Z x + a R ( x, t ) p ( t ) dt − π − x + a Z a R (cid:16) x + t − a (cid:17) p ( t ) dt, a < x < π − a , (63)where F ( x ) := F ( x ) + a Z π − x R (cid:16) x + t − a (cid:17) p ( t ) dt + R (cid:16) π − a (cid:17) π − x Z x − a p ( t ) dt, a < x < π − a , and F ( x ) = p ( x ) + R (cid:16) x + a (cid:17) x − a Z a p ( τ ) dτ, a < x < π − a , (64)where F ( x ) := F ( x ) − π Z x + a p ( t ) dt a Z t − x + a p ( τ ) dτ, a < x < π − a . Note that the functions F ( x ) , F ( x ) , R ( x, t ) and R ( x ) involved in (63) and (64), beingdependent only on q | I , are already known. Substituting (64) into (63) and changing the orderof integration, we obtain the integral equation F ( x ) = p ( x ) − R ( x, x ) x Z a p ( t ) dt − π − a Z x R ( x, t ) p ( t ) dt π − x + a Z a R (cid:16) x + t − a (cid:17) p ( t ) dt, a < x < π − a , (65)where F ( x ) = F ( x ) − π − a Z x + a R ( x, t ) F ( t ) dt, R ( x, t ) = π − a Z t + a R ( x, τ ) R (cid:16) τ − a (cid:17) dτ. Since, according to the hypothesis of the theorem, the value ω := ω + π + a Z a q ( x ) dx = π − a Z a p ( x ) dx is known, equation (65) takes the form F ( x ) = p ( x ) + π − a Z x R ( x, t ) p ( t ) dt − π − x + a Z a R (cid:16) x + t − a (cid:17) p ( t ) dt, a < x < π − a , (66)where both functions F ( x ) = F ( x )+ ω R ( x, x ) and R ( x, t ) = R ( x, x ) − R ( x, t ) are knowntoo. Taking into account that the function p ( x ) on the interval (3 a / , π/ a /
4) is givenby the hypothesis, we find it on ( π/ a / , π − a ) by solving there equation (66), in whichthe second integral becomes known. Finally, substituting the completely found function p ( x )into relation (64), we find p ( x ) on (2 a , π − a / , which finishes the proof. (cid:3)
5. Other representations for the infinite products
The results of this section are valid for any a , a ∈ [0 , π ] . In Section 3, we proved, in particular, that any functions ∆ ( λ ) and ∆ ( λ ) of the formsdescribed in (25)–(27) have infinitely many zeros obeying (7) for the corresponding j ∈ { , } . Moreover, these functions are determined by their zeros uniquely by formula (12). However,the functions ∆ ( λ ) and ∆ ( λ ) constructed by (12) with arbitrary sequences of complexnumbers of the form (7), generally speaking, do not have the forms as in (25)–(27). Thisfact is connected, in particular, with excessiveness of the input data of Inverse Problem 1 forrecovering the functions w j,ν ( x ) in (25) and (26). Nevertheless, the following lemma holds. Lemma 6.
Let j ∈ { , } . Then for any sequence of complex numbers { ρ n,j } n ∈ Z j obey-ing (7), the function ∆ j ( ρ ) constructed by formula (12) has the form ∆ ( ρ ) = sin ρπρ − ω cos ρ ( π − a ) ρ + α sin ρ ( π − a ) ρ + γ sin ρπρ + X ν =0 π Z w ,ν ( x ) c ν ( ρx ) ρ dx (67) for j = 0 , and ∆ ( ρ ) = cos ρπ + ω sin ρ ( π − a ) ρ + α cos ρ ( π − a ) ρ + γ cos ρπρ + X ν =0 π Z w ,ν ( x ) c − ν ( ρx ) ρ dx (68)15 or j = 1 . Here w j,ν ( x ) ∈ L (0 , π ) for j, ν = 0 , , and π Z w , ( x ) dx = ω, π Z xw , ( x ) dx = α ( a − π ) − γ π, π Z w , ( x ) dx = − α − γ . (69)Before proceeding directly to the proof of Lemma 6, we establish the following auxiliaryassertion giving some important subtle estimates that will be required for the proof. Proposition 1.
Put h n,j := β cos ρ n,j a + β sin ρ n,j a + κ n , | n | ∈ N , β j ∈ C , j = 0 , . Then a n,j := X k =0 ,n h k,j k ( n − k ) = O (cid:16) n (cid:17) , | n | → ∞ . Proof.
We have a n,j = 1 n lim N →∞ N X k =0 ,nk = − N h k,j (cid:16) k + 1 n − k (cid:17) = 1 n lim N →∞ (cid:16) N X k =0 ,nk = − N h k,j k − N − n X k = − n, k = − N − n h n + k,j k (cid:17) . One can easily calculate a n,j = 1 n ∞ X k = | n | k =1 h k,n,j k − h − n,j + h n,j n , h k,n,j = h k,j − h n + k,j + h n − k,j − h − k,j . Denote κ k,n := κ k − κ n + k + κ n − k − κ − k . It remains to note that, since h k,n,j = 4 β sin na n − j ) a ka + 4 β sin na n − j ) a ka + κ k,n and the series P ∞ k =1 sin kak converges for any a, the series P ∞ k =1 h k,n,j k , j = 0 , , are convergenttoo, and their sums are uniformly bounded with respect to n. (cid:3) Proof of Lemma 6.
Fix j ∈ { , } . Let us show first that { θ j ( ρ n,j ) } n ∈ Z j ∈ l , where θ ( ρ )and θ ( ρ ) are determined by (10) and (11), respectively. It is easy to see that θ j ( ρ n,j ) = ( ρ n,j ) − j ∆ j ( ρ n,j ) + ( − n ω n,j , ω n,j = ω cos ρ n,j a + α j sin ρ n,j a . (70)By virtue of (12) and (33), we obtain ρ − j ∆ j ( ρ ) = ρ ( ρ n,j − ρ ) c − j ( ρπ ) ρ n,j − ρ Y k ∈ Z j \{ n } ρ k,j − ρρ k,j − ρ . Thus, having put ε n,j := ρ n,j − ρ n,j = ω n,j πn + κ n n , (71)we get ( ρ n,j ) − j ∆ j ( ρ n,j ) = ( − n +1 πρ n,j ε n,j b n,j , b n,j = Y k ∈ Z j \{ n } (cid:16) ε k,j k − n (cid:17) , n ∈ Z j . πρ n,j ε n,j = ω n,j + κ n , we get θ j ( ρ n,j ) = ( − n (1 − b n,j ) ω n,j + κ n . Thus, we need toprove that { − b n,j } ∈ l . For this purpose, we choose N ∈ N so that | ε n,j | ≤ / | n | ≥ N, and represent b n,j in the form b n,j = b (1) n,j b (2) n,j , where b (1) n,j = Y k = n − j ≤| k | 6. Solution of the inverse problem In this section, besides our initial assumptions on a and a , we also assume a ≥ π/ . The preliminary work fulfilled in Sections 3 and 5 allows us to give the proof of Theorem 4. Proof of Theorem 4. By necessity, the asymptotics (7) is already established in Theorem 1.Let us prove (ii). According to Lemma 3, the functions ∆ ( ρ ) and ∆ ( ρ ) determined byformula (12) are the characteristic functions, which, by virtue of Lemma 2, have representations(25) and (26), respectively. Hence, according to (10) and (11), we have the representations θ j ( ρ ) = π − a j Z w j,j ( x ) cos ρx dx + π − a − j Z w j, − j ( x ) sin ρx dx, j = 0 , , which along with (9) give g j,j ( ρ ) = 2 π − a j Z w j,j ( x ) cos ρx dx, g j, − j ( ρ ) = 2 π − a − j Z w j, − j ( x ) sin ρx dx, j = 0 , , (74)which, in turn, implies (ii) and finishes the proof of the necessity.18or the sufficiency, we assume that some complex sequences { ρ n, } | n |∈ N and { ρ n, } n ∈ Z obeying (i) and (ii) are given. Find the values α , α and ω as in the proof of Lemma 5. Then,by formula (12), construct the functions ∆ ( ρ ) and ∆ ( ρ ) , which, according to Lemma 6, haverepresentations (67) and (68), respectively, with some numbers γ and γ and some functions w ν,j ( x ) ∈ L (0 , π ) , j, ν = 0 , , obeying (69). Using (9)–(11) and (67), (68), we calculate g , ( ρ ) = 2 π Z w , ( x ) cos ρx dx, g , ( ρ ) = 2 γ sin ρπ + 2 π Z w , ( x ) sin ρx dx,g , ( ρ ) = 2 π Z w , ( x ) sin ρx dx, g , ( ρ ) = 2 γ cos ρπ + 2 π Z w , ( x ) cos ρx dx. Thus, condition (ii) implies γ = γ = 0 and hence, by virtue of the Paley–Wiener theorem, w j,ν ( x ) = 0 a.e. on ( π − a ν , π ) for j, ν = 0 , , which along with (69) gives (27). Therefore,the functions ∆ ( λ ) and ∆ ( λ ) has the forms (25) and (26), respectively. By virtue ofTheorems 5 and 6, the subsystems (35) and (36) with these w j,ν ( x ) have unique solutions q ( x ) ∈ L ( a , π ) and p ( x ) ∈ L ( a , π ) , satisfying (56) and (38), respectively. Construct thefunction q ( x ) ∈ W [ a , π ] by the formula q ( x ) = 1 π − a π Z a dt π Z t p ( τ ) dτ − π Z x p ( t ) dt, (75)which, obviously, obeys (6). Thus, we constructed the boundary value problems L ( q , q ) and L ( q , q ) . Let ˜∆ ( λ ) and ˜∆ ( λ ) be their characteristic functions, respectively. According toLemma 2, they have the representations˜∆ ( ρ ) = sin ρπρ − ˜ ω cos ρ ( π − a ) ρ + ˜ α sin ρ ( π − a ) ρ + X ν =0 π − a ν Z ˜ w ,ν ( x ) c ν ( ρx ) ρ dx, (76)˜∆ ( ρ ) = cos ρπ + ˜ ω sin ρ ( π − a ) ρ + ˜ α cos ρ ( π − a ) ρ + X ν =0 π − a ν Z ˜ w ,ν ( x ) c − ν ( ρx ) ρ dx, (77)where ˜ ω = 12 π Z a q ( x ) dx, ˜ α j = ˜ α + ( − j ˜ β, j = 0 , , ˜ α = q ( a )2 , ˜ β = q ( π )2 , (78)˜ w , ( x ) = − K , ( π, π − x ; q ) , ˜ w , ( x ) = P ( π, π − x ; q ) , (79)˜ w , ( x ) = K , ( π, π − x ; p ) , ˜ w , ( x ) = P ( π, π − x ; p ) . (80)Comparing (79) and (80) with (35) and (36), respectively, we arrive at˜ w j,ν ( x ) = w j,ν ( x ) , j, ν = 0 , . (81)Successively using the first equality in (78), identity (56) and the first equality in (27), we get˜ ω = 12 π Z a q ( x ) dx = π − a Z w , ( x ) dx = ω. (82)19urther, using the first equality in (38) along with the third one in (27), we obtain π Z a p ( x ) dx = 2 π − a Z w , ( x ) dx = − α , (83)while the second equalities in (27) and in (38) along with (83) give π Z a xp ( x ) dx = π + a π Z a p ( x ) dx + ( π − a ) α = ( π − a ) α − ( π + a ) α . (84)On the other hand, successively using (78), (75) and (83), we get˜ α = q ( a ) − q ( π )2 = − π Z a p ( x ) dx = α , (85)while the second and the last equalities in (78) along with (75), (83) and (84) imply( π − a )( ˜ α − ˜ α ) = π Z a dt π Z t p ( τ ) dτ = π Z a ( x − a ) p ( x ) dx = ( π − a )( α − α ) . (86)By virtue of (85) and (86), we have ˜ α j = α j , j = 0 , , which along with (25), (26), (76), (77),(81) and (82) gives ˜∆ j ( ρ ) ≡ ∆ j ( ρ ) , j = 0 , . Hence, each given sequence { ρ n,j } n ∈ Z j is thespectrum of the corresponding problem L j ( q , q ) , j = 0 , . (cid:3) The Paley–Wiener theorem implies the following corollary from Theorem 4. Corollary 2. Arbitrary complex sequences { ρ n, } | n |∈ N and { ρ n, } n ∈ Z are the spectra ofsome boundary value problems L ( q , q ) and L ( q , q ) , respectively, if and only if their con-vergence exponents are equal to , and there exist some α , α , ω ∈ C such that the functions g j,ν ( ρ ) , j, ν = 0 , , determined by formulae (9)–(12) satisfy the following conditions: g j,ν ( x ) ∈ L ( −∞ , ∞ ) , | g j,ν ( ρ ) | ≤ C exp(( π − a ν ) | ρ | ) , g j,ν ( − ρ ) = ( − j + ν g j,ν ( ρ ) . (87) Proof. In addition to the proof of Theorem 4, it is sufficient to note that, by virtue of thePaley–Wiener theorem, conditions (87) are equivalent to the representations (74) with somefunctions w j,ν ( x ) ∈ L (0 , π − a ν ) , j, ν = 0 , . (cid:3) The proof of Theorem 4 gives the following algorithm for solving Inverse Problem 1. Algorithm 1. Let the spectra { ρ n,j } n ∈ Z j of some problems L j ( q , q ) , j = 0 , , be given.(i) Calculate α , α and ω by the formulae (58) and (60), in which the sequences { m k,l } ,l = 1 , , are chosen so that (59) is fulfilled;(ii) Construct the functions w j,ν ( x ) ∈ L (0 , π − a ν ) , j, ν = 0 , , in representations (25)and (26) by inverting the corresponding Fourier transforms: (cid:20) w j,j ( x ) w j, − j ( x ) (cid:21) = 1 π ∞ X n = −∞ θ j ( n ) (cid:20) cos nx sin nx (cid:21) , j = 0 , , where the functions θ ( ρ ) and θ ( ρ ) are determined by formulae (10) and (11), respectively,with ∆ ( ρ ) and ∆ ( ρ ) constructed by (12); iii) Find the functions q ( x ) ∈ L ( a , π ) and p ( x ) ∈ L ( a , π ) by formulae (54) and (55)and by formula (37), respectively, with w j,ν ( x ) , j, ν = 0 , , constructed on step (ii);(iv) Finally, construct the function q ( x ) ∈ W [ a , π ] by formula (61) or by formula (75). Remark 2. As in [5], step (ii) of Algorithm 1 can be refined by changing to recoveringthe functions w j,ν ( x ) , j, ν = 0 , , from certain subspectra depending on the values a and a . Appendix A Here we obtain an analog of Theorem 4 for the boundary value problems L j ( q ) , j = 0 , , that consist of (1) and (2): − y ′′ ( x ) + q ( x ) y ( x − a ) = λy ( x ) , < x < π, y (0) = y ( j ) ( π ) = 0 , where q ( x ) ∈ L (0 , π ) is a complex-valued function, q ( x ) = 0 on (0 , a ) , while a ∈ [2 π/ , π ) . Let { λ n,j } n ≥ be the spectrum of L j ( q ) . Consider the following inverse problem Inverse Problem A. Given { λ n, } n ≥ and { λ n, } n ≥ , find the potential q ( x ) . As was mentioned in Introduction with references to [5], Inverse Problem A is overdeter-mined. In [5] for a ∈ [ π/ , π ) , it was established, in particular, that for unique determinationof q ( x ) by the subspectra { λ n k , } k ≥ and { λ n k , } k ≥ , it is necessary and sufficient that eachof the functional systems { cos n k x } k ≥ and { sin( n k − / x } k ≥ is complete in L (0 , π − a ) . Moreover, the appropriate asymptotics along with Riesz-basisness of these two systems is suf-ficient for solvability of Inverse Problem A. In particular, solely the asymptotics is a necessaryand sufficient condition of the solvability when this Riesz-basisness is patently the case. Anal-ogous results can be obtained also for a ∈ [2 π/ , π/ . So far, these results remain sole onesdealing with the question of solvability of Inverse Problem A.Despite the overdetermination of Inverse Problem A, one can obtain necessary and sufficientconditions for it solvability given the full spectra as a particular case of Theorem 4. Theorem A. Let a ∈ [2 π/ , π ) . Then for any sequences of complex numbers { λ n, } n ≥ and { λ n, } n ≥ to be the spectra of some boundary value problems L ( q ) and L ( q ) , respectively,it is necessary and sufficient to satisfy the following two conditions:(i) For j = 0 , , the following asymptorics holds: λ n,j = (cid:16) n − j ω cos( n − j/ aπn + κ n n (cid:17) , ω ∈ C , where, as before, one and the same symbol { κ n } denotes different sequences in l ; (ii) The exponential types of the functions θ ( ρ ) and θ ( ρ ) do not exceed π − a, where θ ( ρ ) = ρ ∆ ( ρ ) − ρ sin ρπ + ω cos ρ ( π − a ) , θ ( ρ ) = ρ ∆ ( ρ ) − ρ cos ρπ − ω sin ρ ( π − a ) , (88)∆ j ( ρ ) = π − j ∞ Y n =1 λ n,j − ρ ( n − j/ , j = 0 , . (89)As Corollary 2 from Theorem 4, one can obtain the following corollary from Theorem A. Corollary A. Let a ∈ [2 π/ , π ) . Then arbitrary sequences of complex numbers { λ n, } n ≥ and { λ n, } n ≥ are the spectra of some boundary value problems L ( q ) and L ( q ) , respectively,if and only if their convergence exponents are equal to / , and the functions θ j ( ρ ) , j = 0 , , determined by (88) and (89) satisfy the following conditions: θ j ( x ) ∈ L ( −∞ , ∞ ) , | θ j ( ρ ) | ≤ C exp(( π − a ) | ρ | ) , θ j ( − ρ ) = ( − j θ j ( ρ ) . ppendix B Let π/ ≤ a ≤ a < π and consider the boundary value problem B for the equation − y ′′ ( x ) + q ( x ) y ( x − a ) + q ( x ) y ( x − a ) = λy ( x ) , < x < π, (90)where q ν ( x ) = 0 on (0 , a ν ) and q ν ( x ) ∈ L ( a ν , π ) , along with the two-point boundaryconditions of the general form: U ν ( y ) := h ,ν y (0) + h ,ν y ′ (0) + H ,ν y ( π ) + H ,ν y ′ ( π ) = 0 , ν = 1 , , (91)with arbitrary complex coefficients h j,ν and H j,ν . Let S ( x, λ ) and C ( x, λ ) be solutions ofequation (90) under the initial conditions S (0 , λ ) = C ′ (0 , λ ) = 0 and S ′ (0 , λ ) = C (0 , λ ) = 1 . The spectrum sp( B ) of the problem B coincides with zeros of its characteristic function∆( λ ) := (cid:12)(cid:12)(cid:12)(cid:12) U ( S ) U ( C ) U ( S ) U ( C ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) h , + H , ∆ ( λ ) + H , ∆ ( λ ) h , + H , Θ ( λ ) + H , Θ ( λ ) h , + H , ∆ ( λ ) + H , ∆ ( λ ) h , + H , Θ ( λ ) + H , Θ ( λ ) (cid:12)(cid:12)(cid:12)(cid:12) , where ∆ j ( λ ) = S ( j ) ( π, λ ) is the characteristic function of the boundary value problem B j := B j ( q , q ) consisting of equation (90) and the boundary conditions (2), while Θ j ( λ ) = C ( j ) ( π, λ )is the characteristic function of the problem for equation (90) under the boundary conditions y ′ (0) = y ( j ) ( π ) = 0 . The following proposition means, actually, that specification of sp( B ) gives no additionalinformation on the potentials q ( x ) and q ( x ) after specifying the spectra sp( B ) and sp( B ) . Proposition B. Specification of the functions ∆ ( λ ) and ∆ ( λ ) along with the coefficients h j,ν and H j,ν for j = 0 , and ν = 1 , uniquely determines the function ∆( λ ) . Proof. It is sufficient to show that specification of the functions ∆ ( λ ) and ∆ ( λ ) uniquelydetermines the functions Θ ( λ ) and Θ ( λ ) . Indeed, analogously to Lemma 2 one can obtainthe following representations∆ ( λ ) = ∆ ( λ ) + X ν =1 π − a ν Z w ,ν ( x ) cos ρxρ dx, ∆ ( λ ) := sin ρπρ − X ν =1 ω ν cos ρ ( π − a ν ) ρ , (92)∆ ( λ ) = ∆ ( λ ) + X ν =1 π − a ν Z w ,ν ( x ) sin ρxρ dx, ∆ ( λ ) := cos ρπ + X ν =1 ω ν sin ρ ( π − a ν ) ρ , (93)where ρ = λ and ω ν = 12 π Z a ν q ν ( x ) dx, w j,ν ( x ) = 14 (cid:16) q ν (cid:16) π + x + a ν (cid:17) + ( − j q ν (cid:16) π − x + a ν (cid:17)(cid:17) . Moreover, in a similar way, one can get also the representationsΘ ( λ ) = ∆ ( λ ) − X ν =1 π − a ν Z w ,ν ( x ) sin ρxρ dx, Θ ( λ ) = − λ ∆ ( λ ) + X ν =1 π − a ν Z w ,ν ( x ) cos ρx dx. ( λ ) = 2∆ ( λ ) − ∆ ( λ ) , Θ ( λ ) = λ (∆ ( λ ) − ( λ )) , which finish the proof because ∆ j ( λ ) is determined by specifying ∆ j ( λ ) . (cid:3) Obviously, no set of spectra determines the functions q ( x ) and q ( x ) separately if a = a . The following example shows that they cannot be completely distinguished also when a = a . Example B. Let q ( x ) = , a < x < a + a , , a + a < x < a + π , − , a + π < x < π − a − a , , π − a − a < x < π, q ( x ) = − , a < x < a + π , , a + π < x < π, (94)and, hence, q ( x ) = − q (cid:16) x − a − a (cid:17) , a < x < π. (95)Then the spectra of the problems B ( q , q ) and B ( q , q ) coincide with the ones of B (0 , B (0 , , respectively. Indeed, according to the relations L ( ρ ) := ∆ ( λ ) + iρ ∆ ( λ ) , ∆ ( λ ) = L ( ρ ) − L ( − ρ )2 iρ , ∆ ( λ ) = L ( ρ ) + L ( − ρ )2 , specification of both spectra is equivalent to specification of the function L ( ρ ) , which, inturn, is the characteristic function of the Regge-type problem for equation (90) along with theboundary conditions y (0) = y ′ ( π ) + iρy ( π ) = 0 . (96)On the other hand, the following representation holds (see, e.g., [19]): L ( ρ ) exp( − iρπ ) − X ν =1 ω ν iρ exp( − iρa ν ) − X ν =1 exp( iρa ν )2 iρ π Z a ν q ν ( x ) exp( − iρx ) dx, (97)which also can be easily obtained by using (92) and (93). According to (94), we have2 ω = a π Z a a dx − π − a − a Z a π dx = 0 , ω = − a π Z a dx + π Z a π dx = 0 . Moreover, by virtue of (94) and (95), we get X ν =1 exp( iρa ν ) π Z a ν q ν ( x ) exp( − iρx ) dx = exp( iρa ) π − a − a Z a a q ( x ) exp( − iρx ) dx − exp( iρa ) π Z a q (cid:16) x − a − a (cid:17) exp( − iρx ) dx = 0 . L ( ρ ) = exp( iρπ ) , which is the characteristic function of theproblem (90), (96) with the zero potentials. Thus, specification of the spectra of the problems B ( q , q ) and B ( q , q ) does not uniquely determine the functions q ( x ) and q ( x ) . Finally, we note that Example B refutes both Theorem 3.1 in [19] and Theorem 3.1 in [20].Moreover, according to this counterexample along with Proposition B, the functions q ( x ) and q ( x ) cannot be uniquely determined by specifying any set of the spectra of boundary valueproblems having the form (90) and (91). Acknowledgement. This research was partially supported by the Ministry of Science andTechnology of Taiwan under Grant no. 108-2115-M-032-005-. The first author was supportedby Grant 20-31-70005 of the Russian Foundation for Basic Research. The second authors wassupported by Grant 1.1660.2017/4.6 of the Ministry of Science and Higher Education of theRussian Federation. References [1] Pikula M. 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