Inverse spectral problems for Hill-type operators with frozen argument
aa r X i v : . [ m a t h . SP ] A ug INVERSE SPECTRAL PROBLEMS FOR HILL-TYPEOPERATORS WITH FROZEN ARGUMENTSergey Buterin and Yi-Teng Hu Abstract.
We study two inverse problems of recovering the complex-valued square-integrable potential q ( x ) in the operator − y ′′ ( x ) + q ( x ) y ( a ) , y ( ν ) (0) = γy ( ν ) (1) for ν = 0 , , where γ ∈ C \ { } . The first problem involves only single spectrum as the input data. Weobtain complete characterization of the spectrum and prove that its specification determines q ( x ) uniquely if and only if γ = ± . For the rest cases: periodic and antiperiodic ones, wedescribe classes of iso-spectral potentials and provide restrictions under which the uniquenessholds. The second inverse problem deals with recovering q ( x ) from the two spectra relatedto γ = ± . We obtain necessary and sufficient conditions for its solvability and establish thatuniqueness holds if and only if a = 0 , . For a ∈ (0 , , we describe classes of iso-bispectralpotentials and provide restrictions under which the uniqueness resumes. Algorithms for solvingboth inverse problems are provided. Key words : Sturm–Liouville-type operator, functional-differential operator, frozen argu-ment, inverse spectral problem. : 34A55 34K29
1. Introduction
Let { λ n } n ≥ be the spectrum of the boundary value problem L ( q ( x ) , a, γ ) of the form ℓy := − y ′′ ( x ) + q ( x ) y ( a ) = λy ( x ) , < x < , (1) y ( ν ) (0) = γy ( ν ) (1) , ν = 0 , , (2)where q ( x ) ∈ L (0 ,
1) is a complex-valued function, λ is the spectral parameter and γ ∈ C \ { } , while a ∈ [0 , . The operator generated by the functional-differential expression ℓ equipped withboundary conditions (2) is called Hill-type operator with frozen argument .In the present paper we study inverse spectral problems of recovering the potential q ( x ) . The most complete results in the inverse spectral theory are known for differential operators(see, e.g., monographs [1–4]). In particular, for classical Hill’s operator the inverse problem wasstudied in [1, 5–9] and other works. For operators with frozen argument as well as for otherclasses of non-local operators the classical methods of the inverse spectral theory do not work.Equation (1) also belongs to the so-called loaded differential equations. Loaded equationsappear in mathematics, physics, mathematical biology, etc. (see, e.g., [10–12] and referencestherein) and are characterized by the presence of a trace of the unknown function. Consider,for example, the loaded parabolic partial differential equation ∂∂t u ( x, t ) = ∂ ∂x u ( x, t ) + f ( x, t ) , f ( x, t ) = − q ( x ) u ( a, t ) , < x < , t > , (3)involving the trace u ( a, t ) of the unknown function u ( x, t ) and subject to the initial condition u ( x,
0) = ϕ ( x ) , < x < , (4) Department of Mathematics, Saratov State University, Astrakhanskaya 83, Saratov 410012, Russia, email:[email protected] Department of Applied Mathematics, School of Science, Nanjing University of Science and Technology,Nanjing, 210094, Jiangsu, China, email: [email protected]
1s well as the homogeneous boundary conditions ∂ α ∂x α u ( x, t ) (cid:12)(cid:12)(cid:12) x =0 = ∂ β ∂x β u ( x, t ) (cid:12)(cid:12)(cid:12) x =1 = 0 , t > , (5)where α, β ∈ { , } are fixed. It is well-known that the initial-boundary value problem (3)–(5)models heat conduction in a rod of unit length possessing an external distributed heat source f ( x, t ) . In our case, this heat source is described by the function − q ( x ) u ( a, t ) , i.e. its power isproportional to the temperature u ( a, t ) at the fixed point a of the rod. Such a model can beimplemented by an electric conductive rod of a constant thermal conductivity, but possessinga variable electrical resistance R ( x ) = − q ( x ) independent on the temperature of the rod. Inorder to create the desired heat source, the voltage U ( t ) applied to the ends of the rod shouldbe proportional to p u ( a, t ) : U ( t ) = R p u ( a, t ) , R = − Z q ( ξ ) dξ, where R is the full resistance of the rod. Indeed, the current I ( t ) in the rod does not dependon x and, by Ohm’s law, we have I ( t ) = U ( t ) /R = p u ( a, t ) , while the power of the created heatsource in each point x ∈ [0 ,
1] of the rod equals to P ( x, t ) = I ( t ) R ( x ) = − q ( x ) u ( a, t ) = f ( x, t ) . Similarly, the loaded hyperbolic equation ∂ ∂t u ( x, t ) = ∂ ∂x u ( x, t ) − q ( x ) u ( a, t ) , < x < , t > , (6)under the initial conditions u ( x,
0) = ϕ ( x ) , ∂∂t u ( x, t ) (cid:12)(cid:12)(cid:12) t =0 = ψ ( x ) , < x < , (7)as well as the boundary conditions (5), models an oscillatory process under a damping externalforce in each point x proportional both to the function q ( x ) and to the displacement u ( a, t )of the oscillating system in the fixed point a, where the displacement sensor may be installed.Here belong, for example, vibrations of a string possessing a constant density, but the variablemagnetization characterized by the function q ( x ) , in a magnetic field of a variable strengthdepending on the displacement u ( a, t ) . By the Fourier method of separation of variables, the mentioned above initial-boundaryvalue problems can be reduced to one and the same eigenvalue problem B := B ( q ( x ) , a, α, β )consisting of the ordinary functional-differential equation with frozen argument (1) and theseparated boundary conditions y ( α ) (0) = y ( β ) (1) = 0 . (8)We note that some analogous as well as different models leading to ordinary functional-differential equations with one or several frozen arguments were given, e.g., in [10, 11].Various aspects of inverse problems for operators with frozen argument were studied in[13–23]. In particular, in [21] the problem B was considered for a ∈ [0 , ∩ Q , i.e. a = j/k with some mutually prime integers j and k. It was established that unique recoverability of thepotential q ( x ) from the spectrum of B depends on the values α, β and sometimes on the parityof k. This implies two cases: non-degenerate and degenerate ones (see also [18, 19]), dependingon whether the inverse problem is uniquely solvable or not, respectively. In the degenerate case,asymptotically k -th part of the spectrum degenerates in the sense that each k -th eigenvaluecarries no information on the potential. In [23], it was shown that each irrational a ∈ (0 , α, β ∈ { , } .
2n the present paper, we consider the situation of non-separated boundary conditions (2)and begin our study with the following inverse problem.
Inverse Problem 1.
Given { λ n } n ≥ , a and γ ; find q ( x ) . For example, the case γ = 1 corresponds to the initial-boundary value problem for the heatequation (3) under the initial condition (4) as well as the periodic boundary conditions u (0 , t ) = u (1 , t ) , ∂∂x u ( x, t ) (cid:12)(cid:12)(cid:12) x =0 = ∂∂x u ( x, t ) (cid:12)(cid:12)(cid:12) x =1 , t > , which models heat conduction in a thin closed ring of unit length parameterized by the variable x ∈ [0 , , whose values 0 and 1 correspond to one and the same physical point of the ring.Analogously, by using equation (6) along with the initial conditions (7), one can model thecorresponding oscillatory process. Each of these two models hints that it is sufficient to studythe case a = 0 . Moreover, this sufficiency remains also for any γ = 0 (see Lemma 4 in Section 3).This property allows making no distinguishing between rational and irrational cases, whichdiffers from the case of separated boundary conditions (8).For any γ = 0 , we obtain complete characterization of the spectrum and prove that itsspecification determines q ( x ) uniquely if and only if γ = ± . For γ = ± , we establish thatprecisely half of the spectrum degenerates, and describe classes of iso-spectral potentials. More-over, we provide restrictions on the potential under which the uniqueness resumes. The proofis constructive and gives algorithms for solving Inverse Problem 1.Further, we study recovering the potential q ( x ) from the periodic and the antiperiodicspectra. For α = 0 , , we denote by { λ n,α } n ≥ the spectrum of the boundary value problem L α ( q ( x ) , a ) := L ( q ( x ) , a, ( − α ) and consider the following inverse problem. Inverse Problem 2.
Given { λ n,α } n ≥ , α = 0 , , and a ; find q ( x ) . We prove a uniqueness theorem and obtain a constructive procedure for solving this inverseproblem along with necessary and sufficient conditions for its solvability. In particular, it isestablished that the solution is unique if and only if a ∈ { , } . For a ∈ (0 , , we describeclasses of iso-bispectral potentials and provide restrictions under which the uniqueness holds.In the latter case, characterization of the spectra, besides asymptotics, include a restriction onthe type of the sum of the characteristic functions being an entire function of order 1 / . The paper is organized as follows. In the next section, we study the characteristic functionof the problem L ( q ( x ) , a, γ ) and derive the so-called main equation of the inverse problem.Asymptotics of the spectrum is established also in Section 2. In Section 3, we study InverseProblem 1, while Inverse Problem 2 is investigated in Section 4. In Appendix A, we proveRiesz basisness of one auxiliary functional system.
2. Characteristic function. Main equation
Consider the solutions C ( x, λ ) and S ( x, λ ) of equation (1) under the initial conditions C ( a, λ ) = S ′ ( a, λ ) = 1 , S ( a, λ ) = C ′ ( a, λ ) = 0 . Having put ρ = λ, we arrive at the representations C ( x, λ ) = cos ρ ( x − a ) + Z xa sin ρ ( x − t ) ρ q ( t ) d t, S ( x, λ ) = sin ρ ( x − a ) ρ . (9)Clearly, eigenvalues of the problem (1), (2) coincide with zeros of the entire function∆( λ ) = (cid:12)(cid:12)(cid:12)(cid:12) C (0 , λ ) − γC (1 , λ ) S (0 , λ ) − γS (1 , λ ) C ′ (0 , λ ) − γC ′ (1 , λ ) S ′ (0 , λ ) − γS ′ (1 , λ ) (cid:12)(cid:12)(cid:12)(cid:12) , (10)3hich is called characteristics function . Consider the Wronski-type determinant W ( x, λ ) := h C ( x, λ ) , S ( x, λ ) i , where h y ( x ) , z ( x ) i = y ( x ) z ′ ( x ) − y ′ ( x ) z ( x ) . Using (9), it is easy to calculate W ( x, λ ) = 1 − Z a − x q ( a − t ) sin ρtρ dt. (11)Unlike the Wronskian for the classical Sturm–Liouville equation, the function W ( x, λ ) dependson x and, moreover, may vanish for some values of x ∈ [0 , . However, we need the designation W ( x, λ ) only for brevity. The next lemma gives a fundamental representation for ∆( λ ) . Remark 1.
Since the spectra of the problems L ( q ( x ) , a, γ ) and L ( q (1 − x ) , − a, γ − ) , obviously, coincide, without loss of generality, one can assume that a ∈ [0 , / . Lemma 1.
The characteristic function ∆( λ ) of the problem (1), (2) has the form ∆( λ ) = 1 + γ − γ cos ρ − Z w ( x ) sin ρxρ dx, w ( x ) ∈ L (0 , . (12) Moreover, the following representation holds: w ( x ) = γ q ( a + x ) + q ( a − x ) , x ∈ (0 , a ) ,γq ( a + 1 − x ) + γ q ( a + x ) , x ∈ ( a, − a ) ,γ ( q ( a + 1 − x ) + q ( a − x )) , x ∈ (1 − a, , (13) where, without loss of generality, we assumed a ≤ . Proof.
According to (10) and the definition of W ( x, λ ) , we have∆( λ ) = W (0 , λ ) + γ ∆ , ( λ ) − γ ∆ , ( λ ) + γ W (1 , λ ) , (14)where ∆ α,β ( λ ) = (cid:12)(cid:12)(cid:12)(cid:12) C ( α ) (0 , λ ) S ( α ) (0 , λ ) C ( β ) (1 , λ ) S ( β ) (1 , λ ) (cid:12)(cid:12)(cid:12)(cid:12) . By virtue of Lemma 1 in [21], for α = β we have∆ α,β ( λ ) = ( − α cos ρ + Z w α,β ( x ) sin ρxρ dx, w α,β ( x ) ∈ L (0 , , (15)where (for 2 a ≤ w α,β ( x ) = 12 q (1 − a + x ) − q (1 − a − x ) , x ∈ (0 , a ) , ( − β q (1 + a − x ) − q (1 − a − x ) , x ∈ ( a, − a ) , ( − β (cid:16) q (1 + a − x ) + q ( x − a ) (cid:17) , x ∈ (1 − a, . (16)Using (11) and (14)–(16), we arrive at (12) and (13). (cid:3) After assuming w ( x ) to be known, relation (13) can be considered as a linear functionalequation with respect to q ( x ) , which we refer to as main equation of the inverse problem.Denote α := 1 π arccos 1 + γ γ ∈ { z : Re z ∈ [0 , , Im z ≥ } ∪ { z : Re z ∈ (0 , , Im z < } . (17)4he following theorem describes behavior of the spectrum. Theorem 1.
The spectrum { λ n } n ≥ of the problem L ( q ( x ) , a, γ ) has the form λ k = (2 k + α ) π + κ k , k ≥ , λ k − = (2 k − α ) π + κ k − , k ≥ , { κ n } n ≥ ∈ l . (18) Moreover, if γ = ± (i.e. if α = 0 , ), then κ k − = 0 , k ≥ , (19) i.e. in both periodic and antiperiodic cases, the spectrum degenerates in the sense of (19).Proof. Rewrite (12) in the form∆( λ ) = 4 γ sin ρ + πα ρ − πα − Z w ( x ) sin ρxρ dx, w ( x ) ∈ L (0 , . (20)Thus, by the known method involving Rouch´e’s theorem (see, e.g., [3]) one can prove thatany function of the form (12) has infinitely many zeros λ n , n ≥ , which with account ofmultiplicities have the form λ n = ρ n , ρ k = (2 k + α ) π + ε k , k ≥ , ρ k − = (2 k − α ) π + ε k − , k ≥ , ε n = o (1) . (21)Further we consider two cases.(i) Let γ = ± , i.e. sin πα = 0 . Substituting (21) into (20), we arrive at the relation4 γ (cid:16) ( − n sin πα cos ε n γ γ sin ε n (cid:17) sin ε n Z w ( x ) sin ρ n xρ n dx. Since sin ρ = ρ + O ( ρ ) and cos ρ = 1 + O ( ρ ) as ρ → , we refine that { nε n } n ≥ ∈ l , whichalong with (21) gives (18) for γ = ± . (ii) Let γ ∈ {− , } . Then, by virtue of (13), we have w ( x ) = ( − − γ w (1 − x ) . (22)For γ = 1 , this implies Z w ( x ) sin ρx dx = Z w ( x ) (cid:16) sin ρx + sin ρ (1 − x ) (cid:17) dx = 2 sin ρ Z w (cid:16) − x (cid:17) cos ρx dx. Since α = 0 , we rewrite (20) in the form∆( λ ) = 2 ρ sin ρ (cid:16) ρ sin ρ − Z w (cid:16) − x (cid:17) cos ρx dx (cid:17) , (23)which yields (18) and (19) for γ = 1 . Similarly, for γ = − , formula (22) gives Z w ( x ) sin ρx dx = Z w ( x ) (cid:16) sin ρx − sin ρ (1 − x ) (cid:17) dx = − ρ Z w (cid:16) − x (cid:17) sin ρx dx, which along with (20) and α = 1 implies∆( λ ) = 2 cos ρ (cid:16) ρ Z w (cid:16) − x (cid:17) sin ρxρ dx (cid:17) . γ = − . (cid:3) The characteristic function is uniquely determined by the spectrum. In more detail, thefollowing lemma takes place.
Lemma 2.
Any function ∆( λ ) of the form (12) is uniquely determined by its zeros { λ n } n ≥ .Moreover, the following representation holds: ∆( λ ) = (1 − γ ) ∞ Y n =0 λ n − λλ n , γ = 1 , ( λ − λ ) ∞ Y n =1 λ n − λλ n , γ = 1 , (24) where λ n = ( ρ n ) , n ≥ , ρ k = (2 k + α ) π, k ≥ , ρ k − = (2 k − α ) π, k ≥ . (25) Proof.
By virtue of Hadamard’s factorization theorem (see, e.g., [24]), formula (12) implies∆( λ ) = Cλ s Y λ n =0 (cid:16) − λλ n (cid:17) , (26)where C is some constant, while s is the algebraic multiplicity of the null zero λ n = 0 . Inparticular, we have∆ ( λ ) := 1 + γ − γ cos ρ = (1 − γ ) ∞ Y n =0 (cid:16) − λλ n (cid:17) , γ = 1 ,λ ∞ Y n =1 (cid:16) − λλ n (cid:17) , γ = 1 . (27)Let γ = 1 . Then, dividing (27) by (26), we obtain∆ ( λ )∆( λ ) = (1 − γ ) C Y λ n =0 (cid:16) λ − λ n (cid:17) Y λ n =0 λ n λ n Y λ n =0 λ n − λλ n − λ , (28)while (12) and (27) imply ∆ ( λ ) / ∆( λ ) → λ → −∞ , which along with (28) gives C = (1 − γ ) ( − s Y λ n =0 λ n Y λ n =0 λ n λ n . Substituting this into (26), we arrive at (24) for γ = 1 . The case γ = 1 is treated similarly. (cid:3) For proving solvability of the inverse problems we will need also the following lemma.
Lemma 3.
Let γ = ± . Then for any complex sequence { λ n } n ≥ of the form (18), the func-tion ∆( λ ) constructed by formula (24) has the form (12) with some function w ( x ) ∈ L (0 , . Let γ ∈ {− , } (i.e. α ∈ { , } ). Then for any complex sequence { λ n } n ≥ of the form (18)and satisfying the degeneration condition (19), the function ∆( λ ) constructed by formula (24)has the form (12) with some function w ( x ) ∈ L (0 , obeying (22).Proof. Let us first consider the case γ = 1 , in which one should actually prove representa-tion (23), where we have 2 ρ sin ρ ∞ Y k =1 (cid:16) − λ (2 πk ) (cid:17) . α = 0 and (19), it is sufficient to prove the relation( λ − λ ) ∞ Y k =1 λ k − λ (2 πk ) = 2 ρ sin ρ − Z w (cid:16) − x (cid:17) cos ρx dx, which, in turn, can be established similarly to Lemma 3.3 in [25] or obtained as its corollary.The case γ = − γ = ± { ρ n ∆( λ n ) } n ≥ ∈ l . Indeed, according to (24), (25) and (27), we have∆( λ ) = ∆ ( λ ) ∞ Y k =0 λ k − λλ k − λ = λ n − λρ n + ρ · ∆ ( λ ) ρ n − ρ ∞ Y k = nk =0 λ k − λλ k − λ . Substituting λ = λ n therein and using (18), we get ρ n ∆( λ n ) = a n b n κ n , where a n = 12 lim ρ → ρ n ∆ ( λ ) ρ n − ρ = γ ( − n +1 sin απ, b n = ∞ Y k = nk =0 λ k − λ n λ k − λ n = ∞ Y k = nk =0 (cid:16) κ k λ k − λ n (cid:17) . Using (17), it is easy to show that | λ k − λ n | ≥ C γ ( k + 1) for k = n, where C γ > γ. Hence, | b n | ≤ C. Further, note that the functional system { sin ρ n x } n ≥ coincides upto signs with the system { sin(2 n + α ) πx } n ∈ Z , which, in turn, forms a Riesz basis in L (0 ,
1) assoon as α / ∈ Z . Indeed, for real non-integer α ’s this fact was proved in [27], while for the generalcase the corresponding proof is provided in Appendix A (see Lemma A1). Thus, there exists aunique function w ( x ) ∈ L (0 ,
1) such that∆( λ n ) = − Z w ( x ) sin ρ n xρ n dx, n ≥ . Consider the function θ ( λ ) = − Z w ( x ) sin ρxρ dx. It remains to show that ∆( λ ) = ∆ ( λ ) + θ ( λ ) . For this purpose, we introduce the function σ ( λ ) = ∆( λ ) − ∆ ( λ ) − θ ( λ )∆ ( λ ) = F ( λ ) − − θ ( λ )∆ ( λ ) , F ( λ ) = ∞ Y k =0 λ k − λλ k − λ . After removing singularities, the function σ ( λ ) is entire in λ. As in the proof of Lemma 3.3in [25], one can show that | F ( λ ) | < C δ for λ ∈ G δ := { λ = ρ : | ρ − (2 n + α ) π | ≥ δ, n ∈ Z } ,δ > , and F ( λ ) → λ → −∞ . Moreover, we have ρθ ( λ ) = o (∆ ( λ )) in G δ as soon as | λ | → ∞ . Thus, by virtue of the maximum modulus principle, the function σ ( λ ) is boundedand consequently, according to Liouville’s theorem, it is constant. Since σ ( λ ) → λ → −∞ , we get σ ( λ ) ≡ , which finishes the proof. (cid:3)
3. Recovering from one spectrum
As mentioned in Introduction, when γ = 1 , studying Inverse Problem 1 for any a can,obviously, be reduced to the case a = 0 . The following lemma reveals this possibility also forany other γ = 0 . Lemma 4.
For any a ∈ (0 , , the spectrum of the problem L ( q ( x ) , a, γ ) coincides with thespectrum of L ( q a , , γ ) , where the function q a ( x ) is determined by the formula q a ( x ) = q ( x + a ) , x ∈ (0 , − a ) , γ q ( x + a − , x ∈ (1 − a, . (29)7 roof. Denoting y a ( x ) = y ( x + a ) , x ∈ (0 , − a ) , γ y ( x + a − , x ∈ (1 − a, , (30)we note that, by virtue of the boundary conditions (2), y a ( x ) ∈ W [0 , . Moreover, since y ( x ) ∈ W [0 , , we have y ( ν ) a (0) = γy ( ν ) a (1) , ν = 0 , . (31)Further, using (29) and (30), we rewrite equation (1) in the form − y ′′ a ( x ) + q a ( x ) y a (0) = λy a ( x ) , < x < . (32)It remains to note that one and the same λ can be eigenvalue of the boundary value problem(1), (2) and the problem (31), (32) only simultaneously and only of the same multiplicity. (cid:3) Now we proceed directly to studying Inverse Problem 1. Here and in the subsequent section,along with the boundary value problem L := L ( q ( x ) , a, γ ) we consider a problem ˜ L := L (˜ q, a, γ )of the same form but with a different potential ˜ q ( x ) . We agree that if a certain symbol β denotesan object related to the problem L , then this symbol with tilde ˜ β will denote the correspondingobject related to ˜ L . The following uniqueness theorem holds.
Theorem 2.
Let γ = ± . Then coincidence of the spectra: { λ n } n ≥ = { ˜ λ n } n ≥ impliescoincidence of the potentials: q ( x ) = ˜ q ( x ) a.e. on (0 , . Let γ ∈ {− , } . Assume that there exists an operator K : L (0 , / → L (0 , / withinjective I + γK, such that q a (cid:16) − x (cid:17) = K (cid:16) q a (cid:16)
12 + x (cid:17)(cid:17) , ˜ q a (cid:16) − x (cid:17) = K (cid:16) ˜ q a (cid:16)
12 + x (cid:17)(cid:17) a . e . on (cid:16) , (cid:17) , (33) where I is the identity operator and the function q a ( x ) is determined by formula (29). Thencoincidence of the spectra: { λ n } n ≥ = { ˜ λ n } n ≥ implies q ( x ) = ˜ q ( x ) a.e. on (0 , . Proof.
Let { λ n } n ≥ = { ˜ λ n } n ≥ . Then, by virtue of (12) and (24), we always have w ( x ) =˜ w ( x ) a.e. on (0 , . According to Lemmas 1 and 4, the main equation (13) takes the form w ( x ) = γq a (1 − x ) + γ q a ( x ) , x ∈ (0 , , (34)where q a ( x ) is determined by (29). This can also be checked directly using (13) and (29).Clearly, equation (34) is equivalent to the linear system w ( x ) = γq a (1 − x ) + γ q a ( x ) ,w (1 − x ) = γ q a (1 − x ) + γq a ( x ) , ) x ∈ (cid:16) , (cid:17) , (35)whose determinant equals to γ (1 − γ ) . Thus, the system (35) is non-degenerate if and only if γ = ± . Hence, assuming γ = ± q ( x ) = ˜ q ( x ) a.e. on (0 , . Let γ ∈ {− , } . Then, with accordance to (22), the equations in (35) are equivalent.Making in the first equation of (35) the change of variable x → / − x and using (33), we get w (cid:16) − x (cid:17) = γq a (cid:16)
12 + x (cid:17) + γ K (cid:16) q a (cid:16)
12 + x (cid:17)(cid:17) , w (cid:16) − x (cid:17) = γ ˜ q a (cid:16)
12 + x (cid:17) + γ K (cid:16) ˜ q a (cid:16)
12 + x (cid:17)(cid:17) (36)a.e. on (0 , / . Since I + γK is injective, we get q a ( x ) = ˜ q a ( x ) a.e. on (1 / , . Finally, using(33) again, we arrive at q a ( x ) = ˜ q a ( x ) a.e. on (0 , / , which finishes the proof. (cid:3) emark 2. It is easy to see that specification of the spectrum { λ n } n ≥ uniquely determinesalso the constant γ. Remark 3.
Clearly, the second part of Theorem 2 remains true also if one applies theoperator K to the left-hand sides of the equalities in (33) instead of the right-hand ones.For γ ∈ { , − } , condition (33) may mean, in particular, the oddness (for γ = −
1) orthe evenness (for γ = 1) of the function q a ( x ) with respect to the midpoint of the interval(0 , , i.e. q a (1 / − x ) = γq a (1 / x ), 0 < x < / . However, the case K = − γI is notcovered by condition (33) and not eligible. Indeed, according to (36), in this case the spectrumof L ( q ( x ) , a, γ ) coincides with the one of L (0 , a, γ ) and, hence, carries no information on thepotential q ( x ) . The case of constant K (i.e. when K ( f ) is independent of f ) corresponds to apriori specification of q a ( x ) on the subinterval (0 , / . The next theorem means that Theorem 1 gives complete characterization of the spectrum.
Theorem 3.
Let γ = ± . Then for any fixed a ∈ [0 , and any sequence of complexnumbers { λ n } n ≥ of the form (18), there exists a unique function q ( x ) ∈ L (0 , such that { λ n } n ≥ is the spectrum of the corresponding boundary value problem L ( q ( x ) , a, γ ) . Let γ ∈ {− , } . Then for any fixed a ∈ [0 , and any complex sequence { λ n } n ≥ of theform (18), obeying the degeneration condition (19), there exists q ( x ) ∈ L (0 , (not unique)such that { λ n } n ≥ is the spectrum of the corresponding problem L ( q ( x ) , a, γ ) . Proof.
Using { λ n } n ≥ , we construct the function ∆( λ ) by formula (24). By virtue ofLemma 3, it has the representation (12) with a certain function w ( x ) ∈ L (0 , . If γ ∈ {− , } , then this w ( x ) additionally obeys condition (22). Thus, in any case, the linear system (35)is consistent. Consider some its solution q a ( x ) ∈ L (0 , , which is not unique if and only if γ ∈ {− , } . Find the function q ( x ) ∈ L (0 ,
1) determined by formula (29) and consider thecorresponding problem L := L ( q ( x ) , a, γ ) . Let us show that { λ n } n ≥ is its spectrum.Indeed, by virtue of Lemma 4, the spectrum of L coincides with the one of the problem L a := L ( q a , , γ ) , where the function q a ( x ) is determined by formula (29). Let ∆ a ( λ ) be thecharacteristic function of L a . According to Lemma 1, it has the representation∆ a ( λ ) = 1 + γ − γ cos ρ − Z w a ( x ) sin ρxρ dx, where w a ( x ) = γq a (1 − x ) + γ q a ( x ) , x ∈ (0 , . Comparing this with (34), we get w a ( x ) = w ( x )a.e. on (0 , . Thus, we have ∆ a ( λ ) ≡ ∆( λ ) , i.e. the spectrum of L a as well as the one of L coincide with { λ n } n ≥ . (cid:3) The proof of Theorem 3 is constructive and gives algorithms for solving the inverse problem.First, we provide an algorithm for the case γ = ± . Algorithm 1.
Let the spectrum { λ n } n ≥ of some problem L ( q ( x ) , a, γ ) , γ = ± λ ) by formula (24).2. Calculate the function w ( x ) ∈ L (0 , , inverting the Fourier transform in (12): w ( x ) = 2 ∞ X k =1 f ( πk ) sin πkx, where f ( ρ ) = ρ (1 + γ − γ cos ρ − ∆( ρ )) .
3. Find the function q a ( x ) ∈ L (0 ,
1) by solving the linear non-degenerate system (35): q a ( x ) = γw ( x ) − w (1 − x ) γ − γ .
9. Construct the potential q ( x ) by inverting (29): q ( x ) = ( γq a ( x − a + 1) , x ∈ (0 , a ) ,q a ( x − a ) , x ∈ ( a, . (37)The following algorithm deals with the case γ ∈ {− , } . Unlike the previous one, fordefiniteness it requires specifying also an operator K appeared in Theorem 2. Algorithm 2.
Assume that γ ∈ {− , } and let the spectrum { λ n } n ≥ of a boundary valueproblem L ( q ( x ) , a, γ ) along with the operator K in (33) with bijective I + γK be given.1. Construct the function w ( x ) by implementing the first two steps of Algorithm 1.2. Find the function q a ( x ) on the interval (0 , /
2) by the formula q a (cid:16) − x (cid:17) = K (cid:16) ( I + γK ) − (cid:16) γ w (cid:16) − x (cid:17)(cid:17)(cid:17) , < x < .
3. Find the function q a ( x ) on the interval (1 / ,
1) by the formula q a (cid:16)
12 + x (cid:17) = 1 γ w (cid:16) − x (cid:17) − γq (cid:16) − x (cid:17) , < x < , and construct the potential q ( x ) by formula (37).The latter algorithm also allows one to describe the set of all iso-spectral potentials q ( x ) , i.e. of those for which the corresponding problems L ( q ( x ) , a, γ ) (with fixed a ∈ [0 ,
1] and γ ∈ {− , } ) have one and the same spectrum { λ n } n ≥ . For this purpose, on the second stepof Algorithm 2 one should use a constant operator K, i.e. for which there exists a function p ( x ) ∈ L (0 , /
2) such that K ( f ( x )) = p ( x ) (38)for all f ( x ) ∈ L (0 , / . Indeed, the following theorem holds.
Theorem 4.
If the function p ( x ) in (38) ranges over L (0 , / , then the correspondingfunctions q ( x ) constructed by Algorithm 2 form the set of all iso-spectral potentials for the givenspectrum { λ n } n ≥ . Proof.
It is clear that, for the operator K of the form (38) and for any p ( x ) ∈ L (0 , / , Algorithm 2 gives iso-spectral potentials q ( x ) . On the other hand, by virtue of Theorem 2, noother iso-spectral potentials exist. (cid:3)
4. Recovering from two spectra
As can be seen in the preceding section, the spectrum of the problem L ( q ( x ) , a, γ ) possessessufficient (and necessary) information for unique reconstruction of the potential q ( x ) if andonly if γ = ± . Here, we study Inverse Problem 2, dealing with recovering q ( x ) from bothspectra { λ n,α } n ≥ , α = 0 , , of the periodic and the antiperiodic boundary value problems L α ( q ( x ) , a ) = L ( q ( x ) , a, ( − α ) , α = 0 , . Denote by ∆ α ( λ ) the characteristic function of L α ( q ( x ) , a ) . By virtue of Lemma 1, we have∆ α ( λ ) = 2 − − α cos ρ − Z w α ( x ) sin ρxρ dx, w α ( x ) ∈ L (0 , , (39)where the functions w α ( x ) are determined by the formulae w α ( x ) = q ( a + x ) + q ( a − x ) , x ∈ (0 , a ) , ( − α q (1 + a − x ) + q ( a + x ) , x ∈ ( a, − a ) , ( − α ( q (1 + a − x ) + q ( x − a )) , x ∈ (1 − a, , α = 0 , , (40)10hich were referred to as the main equations for Inverse Problem 1. Moreover, Lemma 2 gives∆ ( λ ) = ( λ − λ , ) ∞ Y n =1 λ n, − λλ n , ∆ ( λ ) = 4 ∞ Y n =0 λ n, − λλ n , (41)where λ n , n ≥ , α = 0 , , depend on α and are determined by (25).In particular, we will show that Inverse Problem 2 is uniquely solvable without any addi-tional assumptions on q ( x ) if and only if a ∈ { , } . Moreover, the following theorem alongwith Theorem 1 gives necessary and sufficient conditions for its solvability in this special case.
Theorem 5.
Let a ∈ { , } . Then for any complex sequences { λ n,α } n ≥ , α = 0 , , of theform λ k,α = (2 k + α ) π + κ k,α , { κ k,α } ∈ l , k ≥ , λ k − ,α = (2 k − α ) π , k ≥ , α = 0 , , (42) there exists a unique complex-valued function q ( x ) ∈ L (0 , such that the given sequences arethe spectra of the corresponding boundary value problems L α ( q ( x ) , a ) , α = 0 , . Proof.
According to Remark 1, it is sufficient to consider the case a = 0 . For α = 0 , , construct the entire functions ∆ α ( λ ) by formulae (41). According to Lemma 3, these functionshave the form (39) with some functions w α ( x ) ∈ L (0 ,
1) obeying condition (22), which takesthe form w α ( x ) = ( − α w α (1 − x ) , < x < , α = 0 , . (43)Solving the system of main equations (40), which in the present case a = 0 takes the form w ( x ) = q (1 − x ) + q ( x ) , w ( x ) = − q (1 − x ) + q ( x ) , < x < , (44)we get q ( x ) = w ( x ) + w ( x )2 . (45)Denote by ˜∆ α ( λ ) the characteristic functions of the constructed problems L α ( q ( x ) , , α = 0 , . According to Lemma 1, they have the form˜∆ α ( λ ) = 2 − − α cos ρ − Z ˜ w α ( x ) sin ρxρ dx, α = 0 , , (46)where the functions ˜ w α ( x ) are determined by the formulae˜ w ( x ) = q (1 − x ) + q ( x ) , ˜ w ( x ) = − q (1 − x ) + q ( x ) , < x < . Comparing them with (39) and (44), we arrive at ˜∆ α ( λ ) ≡ ∆ α ( λ ) , α = 0 , , i.e. the spectra ofthe constructed problems coincide with the given sequences, respectively.The uniqueness follows from uniqueness of solution of the system (44). (cid:3) The following algorithm gives solution of Inverse Problem 2 for a = 0 . According to Re-mark 1, in the case a = 1 this algorithm gives q (1 − x ) instead of q ( x ) . Algorithm 3.
For α = 0 , , let the spectra { λ n,α } n ≥ of the problems L α ( q ( x ) ,
0) withsome common potential q ( x ) ∈ L (0 ,
1) be given. Then it can be found by the following steps.1. Construct the functions ∆ ( λ ) and ∆ ( λ ) by formulae (41).2. Calculate the functions w ( x ) and w ( x ) , inverting the Fourier transform in (39): w α ( x ) = 2 ∞ X k =1 f α ( πk ) sin πkx, α = 0 , , f α ( ρ ) = ρ (2 − − α cos ρ − ∆ α ( ρ )) .
3. Find the function q ( x ) by formula (45).As mentioned above, we claim that 0 and 1 are unique values of a in the segment [0 , a ∈ (0 , , one should put someadditional restrictions on the potential q ( x ) . Specifically, the following uniqueness theoremholds. According to Remark 1, for definiteness, we consider the case a ∈ (0 , / . Theorem 6.
Let a ∈ (0 , / and let there exist an operator P : L (0 , a ) → L (0 , a ) withinjective I + P, such that q ( a − x ) = P ( q ( a + x )) , ˜ q ( a − x ) = P (˜ q ( a + x )) a . e . on (0 , a ) . (47) Then coincidence of the spectra: { λ n,α } n ≥ = { ˜ λ n,α } n ≥ , α = 0 , , implies coincidence of thepotentials: q ( x ) = ˜ q ( x ) a.e. on (0 , . Proof.
The coincidence of the spectra implies w α ( x ) = ˜ w α ( x ) a.e. on (0 ,
1) for α = 0 , . Thus, by virtue of the first line in (40) and (47), we get P ( q ( a + x )) + q ( a + x ) = P (˜ q ( a + x )) + ˜ q ( a + x ) a . e . on (0 , a ) , which along with the injectivity of I + P and with (47) gives q ( x ) = ˜ q ( x ) a.e. on (0 , a ) . Further, summing up (40) for x ∈ ( a, − a ) and for different α ’s, we get q ( a + x ) = w ( x ) + w ( x )2 = ˜ w ( x ) + ˜ w ( x )2 = ˜ q ( a + x ) , a . e . on ( a, − a ) , i.e. q ( x ) = ˜ q ( x ) a.e. on (2 a, , which finishes the proof. (cid:3) Remark 4.
According to Remark 1, for a > / , condition (47) takes the form q ( a + x ) = P ( q ( a − x )) , ˜ q ( a + x ) = P (˜ q ( a − x )) a . e . on (0 , − a ) . (48)Moreover, as in Remark 3, the operator P can alternatively be applied to the left-hand sidesof the equalities in (47) and (48).The following theorem gives necessary and sufficient conditions for solvability of InverseProblem 2 in the case a ∈ (0 , . Theorem 7.
Fix a ∈ (0 , . Then for arbitrary sequences of complex numbers { λ n, } n ≥ and { λ n, } n ≥ to be the spectra of the problems L ( q ( x ) , a ) and L ( q ( x ) , a ) , respectively, witha common potential q ( x ) ∈ L (0 , (not unique) it is necessary and sufficient to have theform (42) and to satisfy the condition (∆ ( λ ) + ∆ ( λ )) exp( − max { − a, a }| ρ | ) = O (1) , λ → −∞ , (49) where the functions ∆ ( λ ) and ∆ ( λ ) are determined by formulae (41).Proof. According to Remark 1, it is sufficient to consider the case a ∈ (0 , / . By necessity,(42) is already established in Theorem 1. Further, by virtue of (39), we get∆ ( λ ) + ∆ ( λ ) = 4 − Z ( w ( x ) + w ( x )) sin ρxρ dx, (50)which along with (40) gives∆ ( λ ) + ∆ ( λ ) = 4 − Z − a ( w ( x ) + w ( x )) sin ρxρ dx, { λ n, } n ≥ and { λ n, } n ≥ , obeying the conditions of the theorem be given.For α = 0 , , construct the entire functions ∆ α ( λ ) by formulae (41). According to Lemma 3,they have the form (39) with some functions w α ( x ) ∈ L (0 ,
1) obeying condition (43). Accordingto the Paley–Wiener theorem, condition (49) and representation (50) imply w ( x ) + w ( x ) = 0 a . e . on (1 − a, . (51)Choose any function q ( x ) ∈ L (0 , a ) that obeys the functional equation w ( x ) = q ( a + 1 − x ) + q ( a − x ) x ∈ (1 − a, . (52)If a = 1 / , then we consider also the linear system w ( x ) = q (1 + a − x ) + q ( a + x ) , w ( x ) = − q (1 + a − x ) + q ( a + x ) , x ∈ ( a, − a ) , (53)which has the unique solution q ( x ) = w ( x − a ) + w ( x − a )2 , x ∈ (2 a, . (54)Thus, we constructed some function q ( x ) . Note that (43) and (51)–(53) imply (40).For α = 0 , , consider the boundary value problems L α ( q ( x ) , a ) with this q ( x ) . Denote by˜∆ α ( λ ) their characteristic functions. According to Lemma 1, they have the form (46) with thefunctions ˜ w α ( x ) determined by the formulae˜ w α ( x ) = q ( a + x ) + q ( a − x ) , x ∈ (0 , a ) , ( − α q (1 + a − x ) + q ( a + x ) , x ∈ ( a, − a ) , ( − α ( q (1 + a − x ) + q ( x − a )) , x ∈ (1 − a, , α = 0 , . Comparing this with (40) and using (39) and (46), we arrive at ˜∆ α ( λ ) ≡ ∆ α ( λ ) , α = 0 , , i.e.the spectra of the constructed problems coincide with the given sequences, respectively. (cid:3) Remark 5.
According to (43) and (51), Inverse Problem 1 is equivalent to Inverse Problem 2as soon as a = 1 / . Indeed, in this case, specification of the spectrum { λ n, } n ≥ is equivalentto the specification of { λ n, } n ≥ . By virtue of Remarks 1 and 5, it is sufficient to provide an algorithm for solving InverseProblem 2 only for a ∈ (0 , / . Algorithm 4.
For α = 0 , , let the spectra { λ n,α } n ≥ of the problems L α ( q ( x ) , a ) with a ∈ (0 , /
2) and q ( x ) ∈ L (0 ,
1) be given as well as the operator P in (47) with bijective I + P. Then the potential q ( x ) can be found by the following steps.1. Calculate the functions w ( x ) and w ( x ) , implementing the first two steps of Algorithm 3.2. Find q ( x ) on the interval (0 , a ) by the formula q ( a − x ) = P (( I + P ) − ( w ( x ))) , < x < a.
3. Find q ( x ) on the interval ( a, a ) by the formula q ( x + a ) = w ( x ) − q ( a − x ) , < x < a,
4. Finally, construct q ( x ) on (2 a,
1) by formula (54).13his algorithm allows one to describe the set of iso-bispectral potentials q ( x ) , i.e. of thosefor which the corresponding problems L α ( q ( x ) , a ) , α = 0 , , have one and the same pair ofthe spectra { λ n,α } n ≥ , α = 0 , . For definiteness, let a ∈ (0 , / . Then on the second step ofAlgorithm 4 one should use the constant operator P ( f ( x )) ≡ p ( x ) , (55)where p ( x ) ∈ L (0 , a ) is fixed. Theorem 8.
Let a ∈ (0 , / . If the function p ( x ) ranges over L (0 , a ) , then the corre-sponding functions q ( x ) constructed by Algorithm 4 form the set of all iso-bispectral potentialsfor the given pair of spectra { λ n, } n ≥ and { λ n, } n ≥ . Proof.
It is clear that for the operator P of the form (55) and for any p ( x ) ∈ L (0 , a ) , Algorithm 4 gives iso-bispectral potentials q ( x ) . On the other hand, by virtue of Theorem 6,no other iso-bispectral potentials exist. (cid:3)
Appendix A
Here we prove the following auxiliary assertion.
Lemma A1.
For α / ∈ Z , the system Λ α := { sin(2 n + α ) x } n ∈ Z is a Riesz basis in L (0 , π ) . For this purpose, we need the following theorem of Levin and Ljubarski˘ı (see [26]).
Theorem A1.
The system of functions { exp( iz k x ) } k ∈ N , where { z k } k ∈ N is the set of zerosof a sine-type function and inf k = n | z k − z n | > , is a Riesz basis in L ( − π, π ) . We remind that an entire function of exponential type S ( ρ ) is called a sine-type function if,for some positive constants c, C and K, the inequalities c < | S ( ρ ) | exp( −| Im ρ | π ) < C hold assoon as | Im ρ | > K. Proof of Lemma A1.
For real α / ∈ Z , this assertion was proved in [27]. Here we adapt itsproof to cover also the case Im α = 0 . Firstly note, that Λ α is complete in L (0 , π ) . Indeed,since for any f ( x ) ∈ L (0 , π ) we have1 σ ( ρ ) Z π f ( x ) sin ρx dx = o (1) , ρ → ∞ , ρ ∈ G δ, , where G δ, = { ρ : | ρ − n − α | ≥ δ, n ∈ Z } , δ > σ ( ρ ) = sin ρ + α π · sin ρ − α π, then f ( x ) = 0 a.e. on (0 , π ) as soon as f ( x ) is orthogonal to all elements of Λ α . Thus, it remainsto show the existence of positive constants A and A such that for any sequence { c n } ∈ l , thefollowing two-sided estimate holds: A ∞ X n = −∞ | c n | ≤ (cid:13)(cid:13)(cid:13) ∞ X n = −∞ c n sin(2 n + α ) x (cid:13)(cid:13)(cid:13) L (0 ,π ) ≤ A ∞ X n = −∞ | c n | . (56)Since ρ ± n = 2 n ± α, n ∈ Z , are all zeros of the sine-type function σ ( ρ ) , Theorem A1 impliesthat the system { exp( i (2 n ± α ) x ) } n ∈ Z is a Riesz basis in L ( − π, π ) . Hence, there exist positiveconstants B and B such that for any { b n } ∈ l , the following two-sided estimate holds: B ∞ X n = −∞ | b n | ≤ (cid:13)(cid:13)(cid:13) X ν =0 ∞ X n = −∞ b n + ν exp( i (2 n + ( − ν α ) x ) (cid:13)(cid:13)(cid:13) L ( − π,π ) ≤ B ∞ X n = −∞ | b n | . (57)14n particular, with b n = − ic n / b n +1 = ic − n / n ∈ Z , estimates (57) imply B ∞ X n = −∞ | c n | ≤ (cid:13)(cid:13)(cid:13) ∞ X n = −∞ c n sin(2 n + α ) x (cid:13)(cid:13)(cid:13) L ( − π,π ) ≤ B ∞ X n = −∞ | c n | . (58)Since k f k L ( − π,π ) = √ k f k L (0 ,π ) for any odd (as well as any even) function f ( x ) ∈ L ( − π, π ) , estimates (58) imply (56) with A j = B j / , j = 1 , , which finishes the proof. (cid:3) Acknowledgement.
The authors are grateful to Natalia Bondarenko for sharing the ideaof proving Lemma A1.
Funding.
The first author was supported by Grant 20-31-70005 of the Russian Foundationfor Basic Research. The second author was supported by National Natural Science Foundationof China (11871031).
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