An inverse Sturm–Liouville-type problem with constant delay and non-zero initial function

. We suggest a new statement of the inverse spectral problem for Sturm–Liouville-type operators with constant delay. This inverse problem consists in recovering the coeﬃcient (often referred to as potential) of the delayed term in the corresponding equation from the spectra of two boundary value problems with one common boundary condition. However, all studies in this direction focus on the case of the zero initial function, i.e. they exploit the assumption that the potential vanishes on the corresponding subinterval. In the present paper, we waive that assumption in favor of a continuously matching initial function, which leads to appearing an additional term with frozen argument in the equation. For the resulting new inverse problem, we pay a special attention to the situation when one of the spectra is given only partially. Suﬃcient conditions and necessary conditions on the corresponding subspectrum for the unique determination of the potential are obtained, and a constructive procedure for solving the inverse problem is given. In parallel, we obtain the characterization of the spectra for the zero initial function and the Neumann common boundary condition, which is found to include an additional restriction as compared with the case of the Dirichlet common condition.

Various equations with delay have been actively studied from the last century in connection with numerous applications (see, e.g., [20][21][22][23][24][25][26]).One of specific features of equation (1) for a > 0 is its underdetermination since the argument of the unknown function y(x) may go beyond the segment [0, π].In order to overcome this issue, one can specify an initial function, i.e. to assume that y(x) = f (x) for x ∈ (−a, 0] with some known f (x).Alternatively, one can assume that q(x) = 0 on (0, a), which actually corresponds to specifying f = 0.However, we intentionally distinguish these two ways.Indeed, rewriting equation (1) in the form − y ′′ (x) + q + (x)y(x − a) = λy(x) − r(x), 0 < x < π, where r(x) = q − (x)f (x − a) and q − (x) = q(x), x ∈ (0, a), 0, x ∈ (a, π), q + (x) = 0, x ∈ (0, a), q(x), x ∈ (a, π), shows that f = 0 leads to a non-homogenous equation, while f = 0 deals with the corresponding homogenous one.Thus, for posing an eigenvalue problem, it is natural to choose the latter, i.e. to assume that q(x) = 0 on (0, a).In particular, the previous studies of inverse problems for (1) were focused namely on this case, i.e. the reconstruction of q(x) was actually carried out only on (a, π) since on (0, a) it was a priori assumed to be equal zero.Meanwhile, admitting a non-zero f also may be appropriate but one should deal with a "linear" initial function, i.e. when f is linearly dependent on y as, e.g., This example is quite natural from the point of view of the general theory [22] because it ensures a continuous continuation of the solution y(x) to [−a, 0) whenever g(x) ∈ C[−a, 0] and g(0) = 1.Such continuation, however, is not always required (see, e.g., [25]).So one can consider more general forms of the initial function such as, e.g., f (x) = Ly(x) with a linear operator L acting from L 2 (0, π) to L ∞ (−a, 0).Then for keeping L in frames of a perturbation, a natural requirement would be its relative compactness [27] with respect to the minimal operator of double differentiation.In particular, one can take Ly(x) = F (y)g(x), where F (y) is a linear functional relatively bounded to that operator.For example, We will focus, however, on the special case (4).
Since the functions q − (x) and g(x − a) enter only in their product p(x), they cannot be recovered simultaneously from any spectral information.Moreover, the reconstruction of q − (x) on any subinterval (α, β) ⊂ (0, a) can be possible only if g(x) = 0 a.e. on (α − a, β − a).For those reasons, we consider without loss of generality the canonical situation when g(x) ≡ 1.
The main results of the present paper (Theorems 1-3) are restricted to the case a ≥ π/2.In accordance with [13,14], the solution of Inverse Problem 1 may be non-unique for a ∈ (0, 2π/5), while the case a ∈ [2π/5, π/2) requires an additional investigation.For the future reference, however, we will mark those auxiliary assertions below whose proofs automatically extend to any wider ranges of a than just [π/2, π).
Everywhere below, one and the same symbol {κ n } will denote different sequences in l 2 .
The following theorem gives basic necessary conditions for the solvability of Inverse Problem 1.
Theorem 1.For j = 0, 1, the following asymptotics holds Here, the constant ω is determined by the formula Moreover, if the spectra {λ n,0 } n≥0 and {λ n,1 } n≥0 correspond to one and the same q − (x), then where while the functions ∆ 0 (λ) and ∆ 1 (λ) are determined by the formulae Condition (7) actually means that Inverse Problem 1 remains overdetermined as in the case q − = 0 (see [6,15]).As will be seen below, it is sufficient to specify only one full spectrum and an appropriate part of the other one.For example, we consider also the following problem.
Here, {n k } k∈N is an increasing sequence of non-negative integers.The next theorem gives sufficient conditions as well as necessary conditions on {n k } k∈N for the uniqueness of q(x).
Since the system {sin(n + 1/2)x} n≥0 is complete in L 2 (0, π), this theorem, obviously, implies the unique determination of q(x) by both complete spectra as in Inverse Problem 1.
We note that the gap between the sufficient and the necessary conditions in Theorem 2 is actually caused by imposing the common Neumann boundary condition (at 0).By the same reason, the conditions in Theorem 1 do not suffice for the solvability of Inverse Problem 1.
In the case of the Dirichlet common condition, necessary and sufficient conditions for the solvability of the corresponding inverse problem were obtained in [15] when q − = 0. Here, we provide such conditions in the same case q − = 0 but for the Neumann common condition, which brings to them an additional item.Specifically, the following theorem holds.Theorem 3. Arbitrary complex sequences {λ n,0 } n≥0 and {λ n,1 } n≥0 of the form (5) sharing one and the same ω ∈ C are the spectra of the problems B 0 (q) and B 1 (q), respectively, with q(x) = 0 a.e. on (0, a) if and only if the exponential types of the functions θ 0 (ρ) and θ 1 (ρ) determined by (8) and (9) do not exceed π − a and the following relation is fulfilled: The latter relation is an additional characterizing condition, which is unnecessary in the Dirichlet case [15].We note that the relevant difference between both cases was pointed out in [12] (see Remark 2 therein).
The paper is organized as follows.In the next section, we construct transformation operators for a fundamental system of solutions of the homogeneous equation in (2), i.e. when r(x) = 0.In Section 3, Green's function of the Cauchy problem for the non-homogeneous equation ( 2) under the zero initial conditions is constructed.In Section 4, we study the characteristics functions of the problems B j (q) and prove Theorem 1. Proofs of Theorems 2 and 3 are given in Section 5 along with a constructive procedure for solving the inverse problems.

Transformation operators
Let C(x, λ) and S(x, λ) be solutions of the homogeneous equation in (2), i.e. the equation under the initial conditions As in the local case a = 0, they form a fundamental system of solutions of equation (11).Throughout the paper, f ′ and f (j) denote the derivatives with respect to the first argument: In this section, we obtain representations for the functions C(x, λ) and S(x, λ) involving the so-called transformation operators, which connect them with the corresponding solutions of the simplest equation with the zero potential.Specifically, the following lemma holds.
Lemma 1.Let a ≥ π/2.The functions S(x, λ) and C(x, λ) admit the representations where ρ 2 = λ and Proof.The assertion for S(x, λ) is a particular case of Lemma 1 in [15].So we will prove only ( 13) and (15).It is easy to see that the Cauchy problem for C(x, λ) is equivalent to the integral equation Taking into account that a ≥ π/2, we calculate which finishes the proof.
The following corollary can be easily checked by direct calculations.
Corollary 1.The following representations hold: where

Green's function of the Cauchy operator
Here, we obtain the solution z(x, λ) = z(x, λ; r) of the Cauchy problem for the nonhomogeneous equation ( 2) with an arbitrary free term r(x) under the zero initial conditions In the next section, we will need representations for z(π, λ; q − ) and z ′ (π, λ; q − ).
As in the local case a = 0, the function z(x, λ) is expected to have the form where G(x, t, λ) is the corresponding Green function.Let us find an explicit formula for it.
The following lemma holds for any a ∈ [0, π].
Lemma 2. For each fixed t ∈ [0, π), the function is a solution of the Cauchy problem where Proof.Since the function G(x, t, λ) is uniquely determined by the representation (20), one has the right to impose any restrictions on it that will finally lead to (20).In particular, it is natural to assume that G(x, t, λ) is sufficiently smooth and obeys the conditions Then substituting ( 20) into (2) and taking the arbitrariness of r(x) into account, we obtain the relations which along with (24), in turn, guaranty that ( 20) is a solution of the problem (2) and (19).Substituting x + t into the above three relations instead of x, we get Combining ( 25) and ( 27) and taking ( 21) into account we rewrite: while (26) takes the form Using the designation (23) along with initial conditions (24), we arrive at (22).Finally, note that, after solving the Cauchy problem (22) in the standard way (see, e.g., [14]), it is easy to see that G(x, t, λ) is a continuous function with respect to all arguments.Hence, the integral in (20) exists and gives a solution to the Cauchy problem (2) and (19).
By substituting ( 28) and ( 29) into (20) and changing the order of integration, we obtain where r(x) = 0 for x < 0. Further, differentiating ( 28) and ( 29) with respect to x, we arrive at the formulae we analogously obtain the representation

Characteristic functions
Consider the entire functions The next lemma holds for any a ∈ [0, π].
As usual, we call ∆ j (λ) characteristic function of the problem B j (q).The following lemma based on the two preceding sections gives representations for both characteristic functions.
In the rest part of this section, we provide auxiliary facts about arbitrary functions of the form ( 34) and (35), and give the proof of Theorem 1.
The next assertion for a = 0 can be found in [18] but the proof does not depend on the value of a as soon as it ranges within [0, 2π].Lemma 7. Any functions of the forms ( 34) and ( 35) are determined by their zeros uniquely.Moreover, the representations in ( 9) hold.Now, we are in position to give the proof of Theorem 1.
Statements analogous to the next lemma are often used for finding necessary and sufficient conditions for the solvability of inverse problems, i.e. a characterization of the spectral data (see Remark 2 in [37]).For its proof, we will follow a new simple idea suggested in [37].Lemma 8.For j = 0, 1, let {λ n,j } n≥0 be arbitrary complex sequences of the form (5). Then the function ∆ j (λ) constructed by the corresponding formula in (9) has the form (34) or (35), respectively.
For the general case, it is sufficient to note that multiplying ∆ 1 (λ) with any function preserves the form (35) and changes only w 1 (x).Indeed, we have where The function H(λ) is entire as soon as λ n,1 are zeros of ∆ 1 (λ).Moreover, in the ρ -plane, we, obviously, have Thus, by virtue of the Paley-Wiener theorem (see, e.g., [38]), it has the form which finishes the proof completely.
Finally, let us give one more auxiliary assertion, which will be used in the proof of Theorem 2. Let {n k } k∈N be an increasing sequence of non-negative integers.Without loss of generality, assume that multiple elements in the subspectrum {λ n k ,0 } k∈N are neighboring, i.e.
where m k is the multiplicity of the value λ n k ,0 in this subspectrum.Put and consider the functional system σ := {s n (x)} n∈N , where Lemma 9.The system σ is complete (is a Riesz basis) in H b := L 2 (0, b) if and only if so is the system σ 0 .Moreover, they have equal defects, i.e. dim( Proof.The first assertion of the lemma coincides with the second assertion of Lemma 1 in [6].For proving the second one, let there exist d linearly independent entire functions h ν (λ), ν = 1, d, of the form whose zeros have the common part {(n k + 1/2) 2 } k∈N .Consider the meromorphic function Then the function hν (λ) := F (λ)h ν (λ) also has the form Indeed, as in the proof of Lemma 2 in [37], one can show that for each fixed δ > 0. Obviously, the function hν (λ), after removing the singularities, is entire and, by the latter estimate, we have in the ρ -plane.Moreover, the maximum modulus principle for analytic functions implies ρ hν (ρ 2 ) = o(exp(|Im ρ|b)) as ρ → ∞.Hence, by the Paley-Wiener theorem [38], we have (47).
Proof.Adding up equations (37) and (38) and then subtracting one from the other, we get Changing the variable, we arrive at the relations Then changing the order of integration in the last two formulae we obtain the system Using the designations one can rewrite the latter system as the Volterra integral equation which possesses a unique solution q + (x) ∈ L 2 (a, π).
Proof of Theorem 2. First of all, note that, due to (5), the value ω is always determined by specifying {λ n,1 } n≥0 via the formula where the natural sequence {ñ k } is chosen so that | cos ñk a| ≥ c > 0. Alternatively, in accordance with (35), one can use the formula where ∆ 1 (λ) is constructed by the second representation in (9).(i) Let the system σ 0 be complete in H. Since, according to Lemma 7, the characteristic function ∆ 1 (λ) is uniquely determined by its zeros, so is also w 1 (x) in (35).By virtue of (36), the function w 0 (x) coincides with w 1 (x) a.e. on (π − a, π), i.e. it becomes known too.
By differentiating (34) ν = 0, m k − 1 times and substituting λ = λ n k ,0 for k ∈ S, we arrive at the relations where m k , S and s n (x) were defined before Lemma 9 and Hence, by virtue of Lemma 9, the function w 0 (x) is determined uniquely also on (0, π − a).Thus, it remains to recall representations (3) and (36) as well as to apply Lemma 10.