Anal.Math.Phys. DOI 10.1007/s13324-015-0120-6
Recovering differential operators with nonlocal boundary conditions Vjacheslav Anatoljevich Yurko1 · Chuan-Fu Yang2
Received: 25 April 2015 / Revised: 27 October 2015 / Accepted: 2 November 2015 © Springer Basel 2015
Abstract Inverse spectral problems for Sturm–Liouville operators with nonlocal boundary conditions are studied. As the main spectral characteristics we introduce the so-called Weyl-type function and two spectra, which are generalizations of the wellknown Weyl function and Borg’s inverse problem for the classical Sturm–Liouville operator. Two uniqueness theorems of inverse problems from the Weyl-type function and two spectra are presented and proved, respectively. Keywords problems
Differential operators · Nonlocal boundary conditions · Inverse spectral
Mathematics Subject Classification
34A55 · 34L05 · 47E05
1 Introduction Consider the differential equation − y (x) + q(x)y(x) = λy(x), x ∈ (0, T ),
B
Vjacheslav Anatoljevich Yurko
[email protected] Chuan-Fu Yang
[email protected]
1
Department of Mathematics, Saratov University, Astrakhanskaya 83, Saratov 410012, Russia
2
Department of Applied Mathematics, School of Science, Nanjing University of Science and Technology, Nanjing 210094, Jiangsu, China
(1)
V.A. Yurko, C. Yang
and the linear forms
T
U j (y) :=
y(t)dσ j (t),
j = 1, 2.
(2)
0
Here q ∈ L(0, T ) is a complex-valued function, σ j (t) are complex-valued functions of bounded variations and are continuous from the right for t > 0. There exist finite limits H j := σ j (+0) − σ j (0). Linear forms (2) can be written in the form U j (y) := H j y(0) +
T
y(t)dσ j0 (t),
j = 1, 2,
(3)
0
where σ j0 (t) in (3) are complex-valued functions of bounded variations and are continuous from the right for t ≥ 0. Throughout this paper we assume H1 = 0. A natural assumption that U1 = kU2 holds for all k ∈ C, that is, two linear forms U j ( j = 1, 2) are linearly independent. Boundary problems with nonlocal conditions are a part of fast developing differential equations theory. Samarskii and Bitsadze first formulated and investigated nonlocal boundary problem for elliptic equation [1]. Afterwards the number of differential problems with nonlocal boundary conditions had increased. Note that such nonlocal operators appear not only in quantum mechanics but also in other areas such as the theory of diffusion processes, see the related references in [2–6]. Boundary value problems with nonlocal boundary conditions also arise in various fields of physics [7– 9], biology, biotechnology [10,11] and other fields. Nonlocal boundary conditions come up when value of the function on the boundary is connected with values inside the domain. Quite new area, related to problems of this type, deals with investigation of the spectrum of differential equations with nonlocal conditions. In this paper we study inverse spectral problems for Sturm–Liouville operators with nonlocal boundary conditions, which are defined by linear forms (2) or (3). Classical inverse problems for Eq. (1) with two-point separated boundary conditions have been studied fairly completely in many works (see the monographs [12–15] and the references therein). Some aspects of the inverse problem theory for different nonlocal operators can be found in [16–25]. In this paper we prove uniqueness theorems for the solution of the inverse spectral problems for Eq. (1) with nonlocal boundary conditions. In Sect. 1 we suggest statements of the inverse problems and formulate our main results (Theorems 1 and 2). Section 2 introduces important notions and properties of spectral characteristics. The proofs of Theorems 1 and 2 are given in Sect. 3. For this purpose we use the ideas of the method of spectral mappings [15]. In Sect. 4 we present counterexamples related to the statements of the inverse problems (see also [18]). Additional spectral data are introduced in Sect. 5. In Sect. 6, as an example, we consider an inverse problem of recovering the potential q from the given three spectra. Let X k (x, λ) and Z k (x, λ), k = 1, 2, be the solutions of Eq. (1) under the initial conditions X 1 (0, λ) = X 2 (0, λ) = Z 1 (T, λ) = Z 2 (T, λ) = 1,
Recovering differential operators with nonlocal. . .
X 1 (0, λ) = X 2 (0, λ) = Z 1 (T, λ) = Z 2 (T, λ) = 0. Consider the boundary value problem (BVP) L 0 for Eq. (1) with the conditions U1 (y) = U2 (y) = 0. Denote ω(λ) := det[U j (X k )] j,k=1,2 , and assume that ω(λ) ≡ 0. The function ω(λ) is entire in λ of order 1/2, and its zeros = {ξn }n≥1 (counting multiplicity) coincide with the eigenvalues of L 0 . The function ω(λ) is called the characteristic function for L 0. Denote V j (y) := y ( j−1) (T ), j = 1, 2. Consider the BVP L j , j = 1, 2, for Eq. (1) with the conditions U j (y) = V1 (y) = 0. The eigenvalue set j = {λn j }n≥1 (counting multiplicity) of the BVP L j coincide with the zeros of the characteristic function j (λ) := det[U j (X k ), V1 (X k )]k=1,2 . For λ = λn1 , let (x, λ) be the solution of Eq. (1) under the conditions U1 ( ) = 1, V1 ( ) = 0. Denote M(λ) := U2 ( ). The function M(λ) is called the Weyltype function. It is known [14] that for Sturm–Liouville operators with classical twopoint separated boundary conditions, the specification of the Weyl function uniquely determines the potential q(x). In this case with nonlocal boundary conditions, it is not true; the specification of the Weyl-type function M(λ) does not uniquely determine the potential (see counterexamples in Sect. 4). In this case the inverse problem is formulated as follows. Inverse Problem 1 Given M(λ) and ω(λ), construct the potential q(x). We note that the functions σ j (t) are known a priori, and only the potential q(x) has to be constructed. Let us formulate a uniqueness theorem for inverse problem 1. For this purpose, together with q we consider another potential q, ˜ and we agree that if a certain symbol α denotes an object related to q, then α˜ will denote an analogous object related to q. ˜ ˜ Theorem 1 Let 1 ∩ = ∅. If M(λ) = M(λ) and ω(λ) = ω(λ), ˜ then q(x) = q(x) ˜ a.e. on (0, T ). Thus, under condition S, the specification of M(λ) and ω(λ) uniquely determines the potential. The proof of Theorem 1 is given in Sect. 3. We note that if condition S does not hold, then the specification of M(λ) and ω(λ) does not uniquely determine the potential (see counterexamples in Sect. 4). In this case we have to specify an additional spectral information (see Sect. 5). Consider the BVP L 11 for Eq. (1) with the conditions U1 (y) = V2 (y) = 0. The eigenvalue set 11 := {λ1n1 }n≥1 of the BVP L 11 coincide with the zeros of the characteristic function 11 (λ) := det[U1 (X k ), V2 (X k )]k=1,2 . Clearly, {λn1 }n≥1 ∩ {λ1n,1 }n≥1 = ∅. Inverse Problem 2 Given {λn1 , λ1n1 }n≥1 , construct q(x). This inverse problem is a generalization of the well-known Borg’s inverse problem [26] for Sturm–Liouville operators with classical two-point separated boundary conditions, and coincides with it when U1 (y) = y(0). We note that in Inverse Problem 2 there are no restrictions on behavior of the spectra.
V.A. Yurko, C. Yang
Theorem 2 If λn1 = λ˜ n1 , λ1n1 = λ˜ 1n1 , n ≥ 1, then q(x) = q(x) ˜ a.e. on (0, T ). The proof of Theorem 2 is given in Sect. 3.
2 Auxiliary propositions Let λ = ρ 2 , τ := Im ρ ≥ 0. It is known (see, for example, [14]) that there exists a fundamental system of solutions {Yk (x, ρ)}k=1,2 of Eq. (1) such that for |ρ| → ∞: Y1(ν) (x, ρ) = (iρ)ν exp(iρx) 1 + O(ρ −1 ) , (ν) Y2 (x, ρ) = (−iρ)ν exp(−iρx) 1 + O(ρ −1 )
(4)
for ν = 0, 1, and (ν−1) det Yk (x, ρ)
k,ν=1,2
= −2iρ 1 + O(ρ −1 ) .
(5)
Lemma 1 Let {Wk (x, λ)}k=1,2 be a fundamental system of solutions of Eq. (1), and let Q j (y), j = 1, 2, be linear forms. Then (ν−1) (x, λ) det[Q j (Wk )]k, j=1,2 = det[Q j (X k )]k, j=1,2 det Wk
k,ν=1,2
.
(6)
Proof One has, for ν = 1, 2, Wν (x, λ) =
2
Aνk (λ)X k (x, λ),
k=1
where the coefficients Aνk (λ) do not depend on x. This yields det[Q j (Wk )]k, j=1,2 = det[Q j (X k )]k, j=1,2 det[Aνk (λ)]k, j=1,2 , and det Wk(ν−1) (x, λ) (ν−1)
Since det[X k
k,ν=1,2
= det X k(ν−1) (x, λ)
k,ν=1,2
det[Aνk (λ)]k, j=1,2 .
(x, λ)]k,ν=1,2 = 1, we arrive at (6).
It follows from (5) to (6) that det[Q j (Z k )]k, j=1,2 = det[Q j (X k )]k, j=1,2 , det[Q j (Yk )]k, j=1,2 = −2iρ 1 + O(ρ −1 ) det[Q j (X k )]k, j=1,2 .
(7) (8)
Recovering differential operators with nonlocal. . .
Consider the functions ϕ(x, λ) = − det[X k (x, λ), U1 (X k )]k=1,2 , θ (x, λ) = det[X k (x, λ), U2 (X k )]k=1,2 , ψ(x, λ) = det[X k (x, λ), V1 (X k )]k=1,2 . Clearly, U1 (ϕ) = 0, U2 (ϕ) = ω(λ), V1 (ϕ) = 1 (λ), V2 (ϕ) = 11 (λ), U1 (θ ) = ω(λ), U2 (θ ) = 0, V1 (θ ) = −2 (λ), U j (ψ) = j (λ), V1 (ψ) = 0, V2 (ψ) = −1. Moreover, using (6), (7), we calculate det[θ (ν−1) (x, λ), ϕ (ν−1) (x, λ)]ν=1,2 = ω(λ), det[ψ (ν−1) (x, λ), ϕ (ν−1) (x, λ)]ν=1,2 = 1 (λ), 1 (λ) = −U1 (Z 2 ), 2 (λ) = −U2 (Z 2 ), 11 (λ) = U1 (Z 1 ).
(9) (10)
Considering boundary conditions of , ψ, ϕ and θ, we obtain ψ(x, λ) , 1 (λ) 1 2 (λ) (x, λ) = ϕ(x, λ) . θ (x, λ) + ω(λ) 1 (λ)
(x, λ) =
(11) (12)
Hence, 2 (λ) , M(λ) := U2 ( ) = 1 (λ) det (ν−1) (x, λ), ϕ (ν−1) (x, λ)
(13) ν=1,2
= 1.
(14)
Let v1 (x, λ) and v2 (x, λ) be the solutions of Eq. (1) under the conditions v1 (T, λ) = v2 (T, λ) = 1, v1 (T, λ) = 0, U1 (v2 ) = 0. Obviously, v1 (x, λ) = Z 1 (x, λ), v2 (x, λ) = Z 2 (x, λ) + N (λ)Z 1 (x, λ), (ν−1) det vk (x, λ) = 1, k,ν=1,2
(15)
where N (λ) =
1 (λ) U1 (Z 2 ) =− . 11 (λ) U1 (Z 1 )
(16)
V.A. Yurko, C. Yang
Denote U1a (y) :=
a
y(t)dσ1 (t), a ∈ (0, T ].
0
Clearly, U1 = U1T , and if σ1 (t) ≡ C (constant) for t ≥ a, then U1 = U1a . Let λn1 = ρn2 . For sufficiently small δ > 0, we denote δ := {ρ : arg ρ ∈ [δ, π − δ]}, G δ := {ρ : |ρ − ρn | ≥ δ, ∀n ≥ 1}. Lemma 2 For |ρ| → ∞, ρ ∈ δ , we have (iρ)ν exp(iρx)(1 + o(1)), x ∈ [0, T ), H1 (iρ)ν (ν) exp(−iρ(T − x)) 1 + O(ρ −1 ) , x ∈ [0, T ), v1 (x, λ) = 2 H1 exp(−iρT )(1 + o(1)), 1 (λ) = − 2iρ H1 exp(−iρT )(1 + o(1)). 11 (λ) = 2
(ν) (x, λ) =
(17) (18)
(19)
Let σ1 (t) ≡ C (constant) for t ≥ a (i.e. U1 = U1a ). Then for |ρ| → ∞, ρ ∈ δ , it yields ϕ (ν) (x, λ) =
H1 (−iρ)ν−1 exp(−iρx)(1 + o(1) + O(exp(iρ(2x − a)))), 2 x ∈ (0, T ], (20)
(ν)
v2 (x, λ) = (−iρ)ν−1 exp(iρ(T − x))(1 + o(1) + O(exp(iρ(2x − a)))), x ∈ [0, T ). (21) Proof One has (x, λ) = A1 (λ)Y1 (x, ρ) + A2 (λ)Y2 (x, ρ).
(22)
Since U1 ( ) = 1, V1 ( ) = 0, it follows from (22) that A1 (λ)U1 (Y1 ) + A2 (λ)U1 (Y2 ) = 1, A1 (λ)V1 (Y1 ) + A2 (λ)V1 (Y2 ) = 0.
(23)
By virtue of (4), we have for |ρ| → ∞, ρ ∈ δ : U1 (Y1 ) = H1 (1 + o(1)), U1 (Y2 ) = O(exp(−iρT )), (24) V1 (Y1 ) = exp(iρT ) 1 + O(ρ −1 ) , V1 (Y2 ) = exp(−iρT )(1 + O(ρ −1 )). (25)
Recovering differential operators with nonlocal. . .
Solving linear algebraic system (23) and using (24), (25), we calculate A1 (λ) = H1−1 (1 + o(1)), A2 (λ) = O(exp(2iρT )). Substituting these relations into (22), we arrive at (17). Formulas (18)–(21) are proved similarly. By the well-known method (see, for example, [14]) one can also obtain the following estimates for x ∈ (0, T ), τ ≥ 0 :
v1(ν) (x, λ) = O ρ ν exp(−iρ(T − x)) ,
(ν) (x, λ) = O ρ ν exp(iρx) , ρ ∈ G δ .
(26) (27)
Moreover, if σ1 (t) ≡ C (constant) for t ≥ a (i.e. U1 = U1a ), then for x ≥ a/2, τ ≥ 0:
ϕ (ν) (x, λ) = O ρ ν−1 exp(−iρx) ,
(ν) v2 (x, λ) = O ρ ν−1 exp(iρ(T − x)) , ρ ∈ G δ ,
(28) (29)
where G δ := {ρ : |ρ − ρn | ≥ δ, ∀n ≥ 1}, δ > 0, and ρn2 = λ1n1 .
3 Proofs of Theorems 1, 2 Firstly we prove Theorem 2. Let λn1 = λ˜ n1 , λ1n1 = λ˜ 1n1 , n ≥ 1. The characteristic function 1 (λ) of the BVP L 1 is entire in λ of order 1/2. Therefore, by Hadamard’s factorization theorem, 1 (λ) is uniquely determined up to a multiplicative constant ˜ 1 (λ) ≡ C (constant). Taking (19) into account, we calculate by its zeros, i.e., 1 (λ)/ ˜ 1 (λ). Analogously, we get 11 (λ) ≡ ˜ 11 (λ). C = 1, and consequently, 1 (λ) ≡ By virtue of (16), this yields N (λ) ≡ N˜ (λ).
(30)
P1 (x, λ) = v1 (x, λ)v˜2 (x, λ) − v˜1 (x, λ)v2 (x, λ), P2 (x, λ) = v2 (x, λ)v˜1 (x, λ) − v˜2 (x, λ)v1 (x, λ).
(31)
Consider the functions
In view of (15) and (30), one gets P1 (x, λ) = (Z 1 (x, λ) Z˜ 2 (x, λ) − Z˜ 1 (x, λ)Z 2 (x, λ)) + ( N˜ (λ) − N (λ))Z 1 (x, λ) Z˜ 1 (x, λ) = Z 1 (x, λ) Z˜ (x, λ) − Z˜ (x, λ)Z 2 (x, λ), 2
1
P2 (x, λ) = Z 2 (x, λ) Z˜ 1 (x, λ) − Z˜ 2 (x, λ)Z 1 (x, λ) + (N (λ) − N˜ (λ))Z 1 (x, λ) Z˜ 1 (x, λ) = Z 2 (x, λ) Z˜ 1 (x, λ) − Z˜ 2 (x, λ)Z 1 (x, λ).
V.A. Yurko, C. Yang
Thus, for each fixed x, the functions Pk (x, λ), k = 1, 2, are entire in λ. On the other hand, taking (18) and (21) into account we calculate for each fixed x ≥ T /2 and k = 1, 2 : Pk (x, λ) − δ1k = o(1), |ρ| → ∞, ρ ∈ δ , where δ1k is the Kronecker symbol. Moreover, in view of (26) and (29), we get for k = 1, 2 Pk (x, λ) = O(1), |ρ| → ∞, ρ ∈ G δ . Using the maximum modulus principle and Liouville’s theorem for entire functions, we conclude that P1 (x, λ) ≡ 1,
P2 (x, λ) ≡ 0, x ≥ T /2.
Together with (31) this yields vk (x, λ) = v˜k (x, λ), Z k (x, λ) = Z˜ k (x, λ), q(x) = q(x), ˜ x ≥ T /2.
(32)
Let us now consider the BVPs L a1 and L a11 for Eq. (1) on the interval (0, T ) with the conditions U1a (y) = V1 (y) = 0 and U1a (y) = V2 (y) = 0, respectively. Then, according to (10), the functions a1 (λ) := −U1a (Z 2 ) and a11 (λ) := U1a (Z 1 ) are the characteristic functions of L a1 and L a11 , respectively. One has a a/2 U1 (Z k ) = U1a (Z k ) − Z k (t, λ)dσ1 (t), k = 1, 2, a/2
hence a/2
1 (λ) = a1 (λ) + a/2
11 (λ) = a11 (λ) −
a
Z 2 (t, λ)dσ1 (t),
a/2 a
Z 1 (t, λ)dσ1 (t).
(33)
a/2
T (λ) = (λ), it follows from Let us use (33) for a = T. Since 1T (λ) = 1 (λ), 11 11 (32), (33) that T /2 ˜ T /2 (λ), T /2 (λ) = ˜ T /2 (λ). 1 (λ) = 1 11 11
Repeating preceding arguments subsequently for a = T /2, T /4, T /8, . . . , we conclude that q(x) = q(x) ˜ a.e. on (0, T ). Theorem 2 is proved. ˜ ω(λ) = ω(λ). ˜ Now we will prove Theorem 1. Let 1 ∩ = ∅, and M(λ) = M(λ), Consider the functions ˜ (x, λ)ϕ(x, λ), R1 (x, λ) = (x, λ)ϕ˜ (x, λ) − ˜ R2 (x, λ) = ϕ(x, λ) (x, λ) − ϕ(x, ˜ λ) (x, λ).
(34)
Recovering differential operators with nonlocal. . .
It follows from (11) and (34) that 1 ψ(x, λ)ϕ˜ (x, λ) − ψ˜ (x, λ)ϕ(x, λ) , 1 (λ) 1 ˜ ϕ(x, λ)ψ(x, λ) − ϕ(x, ˜ λ)ψ(x, λ) . R2 (x, λ) = 1 (λ) R1 (x, λ) =
This yields that for each fixed x, the functions Rk (x, λ) are meromorphic in λ with possible poles only at λ = λn1 . On the other hand, taking (12) into account we calculate 1 θ (x, λ)ϕ˜ (x, λ) − θ˜ (x, λ)ϕ(x, λ) , ω(λ) 1 ϕ(x, λ)θ˜ (x, λ) − ϕ(x, ˜ λ)θ (x, λ) . R2 (x, λ) = ω(λ) R1 (x, λ) =
(35) (36)
Hence the functions Rk (x, λ) are regular at λ = λn1 . Thus, for each fixed x, the functions Rk (x, λ) are entire in λ. Furthermore, by virtue of (17) and (20), we obtain for x ≥ T /2 : Rk (x, λ) − δ1k = o(1), |ρ| → ∞, ρ ∈ δ . Moreover, using (27), (28), we get for x ≥ T /2 : Rk (x, λ) = O(1), |ρ| → ∞, ρ ∈ G δ . Therefore, R1 (x, λ) ≡ 1, R2 (x, λ) ≡ 0. Together with (14) and (34), this yields ˜ ϕ(x, λ) = ϕ(x, ˜ λ), ψ(x, λ) = ψ(x, λ), q(x) = q(x), ˜ x ≥ T /2. In particular, we obtain Z k (x, λ) = Z˜ k (x, λ), k = 1, 2, x ≥ T /2. Since ϕ(x, λ) = U1 (Z 1 )Z 2 (x, λ) − U1 (Z 2 )Z 1 (x, λ), it follows that ˜ 1 (λ), 11 (λ) = ˜ 11 (λ). 1 (λ) = Using Theorem 2, we conclude that q(x) = q(x) ˜ a.e. on (0, T ). Theorem 1 is proved.
V.A. Yurko, C. Yang
4 Counterexamples (1) Let T = π, U1 (y) = y(0), U2 (y) = y(π/2), q(x) = q(x +π/2), x ∈ (0, π/2), and q(x) ≡ q(π − x). Take q(x) ˜ = q(π − x), x ∈ (0, π ). We see that the BVP L˜ 1 is that Eq. (1) with q(x) ˜ = q(π − x) and the conditions U1 (y) = V1 (y) = 0; the BVP L˜ 2 is that Eq. (1) with q(x) ˜ = q(π − x) and the conditions U2 (y) = V1 (y) = 0; and the ˜ = q(π/2−x) and the conditions U1 (y) = U2 (y) = 0. BVP L˜ 0 is that Eq. (1) with q(x) Then ˜ 1 (λ), 2 (λ) = ˜ 2 (λ), ω(λ) = ω(λ), ˜ 1 (λ) = ˜ and, in view of (13), M(λ) = M(λ). Condition S does not hold. This means, that the specification of M(λ) and ω(λ) does not uniquely determine the potential q. (2) Let T = π, U1 (y) = y(0), U2 (y) = y(π − α), where α ∈ (0, π/2). Then 1 (λ) = X 2 (π, λ), ω(λ) = X 2 (π − α, λ), and 2 (λ) = X 2 (π, λ)X 1 (π − α, λ) − X 2 (π − α, λ)X 1 (π, λ). Obviously, if 1 (λ∗ )2 (λ∗ )ω(λ∗ ) = 0 for a certain λ∗ , then either 1 (λ∗ ) = 2 (λ∗ ) = ω(λ∗ ) = 0 (i.e. λ∗ is an eigenvalue for all boundary value problems L 0 , L 1 , L 2 ), or λ∗ is an eigenvalue for only one problem from L 0 , L 1 , L 2 . In other words, it is impossible that λ∗ is an eigenvalue for only two problems from L 0 , L 1 , L 2 . Let q(x) ≡ q(π − x), and let q(x) ≡ 0 for x ∈ [0, α0 ] ∪ [π − α0 , π ], where α0 ∈ (0, π/2). If α < α0 , then λn2 = (π n/α)2 , n ≥ 1. Choose a sufficiently small α < α0 such that 1 ∩2 = ∅. Clearly, such choice is possible. Then 1 ∩ = ∅, i.e. ˜ 1 (λ), 2 (λ) = ˜ 2 (λ), condition S holds. Take q(x) ˜ := q(π − x). Then 1 (λ) = ˜ and consequently, M(λ) = M(λ). Thus, condition S holds, but the specification of M(λ) does not uniquely determine the potential q.
5 Additional spectral data If condition S does not hold, then the specification of M(λ) and ω(λ) does not uniquely determine the potential q. We introduce additional spectral data. For simplicity, we confine ourselves to the case when zeros of ω(λ) are all simple. By virtue of (9),
det θ (ν−1) (x, λ), ϕ (ν−1) (x, λ) ν=1,2 = ω(λ). Then the functions ϕ(x, ξn ) and θ (x, ξn ) are linearly dependent, i.e., there exist numbers An and Bn (|An | + |Bn | > 0) such that An ϕ(x, ξn ) = Bn θ (x, ξn ). Consider the sequence (so-called a sequence of norming constants) D = {dn }n≥1 , where dn := Bn /An (dn := ∞ if An = 0). The inverse problem is formulated as follows. Inverse Problem 3 Given M(λ), ω(λ) and D, construct q(x).
Recovering differential operators with nonlocal. . .
We note that if condition S holds (i.e., 1 ∩ = ∅), then by virtue of (11), (12), ˜ ˜ and we arrive at implies that D = D, dn = −M −1 (ξn ). In this case M(λ) = M(λ) Inverse Problem 1. ˜ ˜ then q(x) = q(x) Theorem 3 If M(λ) = M(λ), ω(λ) = ω(λ) ˜ and D = D, ˜ a.e. on (0, T ). Proof Fix n ≥ 1, and consider the functions Rk (x, λ), k = 1, 2, in a neighborhood of the point λ = ξn . If dn = d˜n = ∞, then, in view of (35), (36), Rk (x, λ) are regular at λ = ξn . If dn = d˜n = ∞, then θ (x, ξn ) = θ˜ (x, ξn ) = 0. By virtue of (35), (36), this yields that Rk (x, λ) are regular at λ = ξn . Thus, the functions Rk (x, λ), k = 1, 2, are entire in λ. Repeating the arguments from the proof of Theorem 1, we conclude that q(x) = q(x) ˜ a.e. on (0, T ).
6 Example (inverse problem from three spectra) Fix a ∈ (0, T ). Consider Inverse Problem 1 in the particular case, when U1 (y) = y(0), U2 (y) = y(a). Then the boundary value problems L 0 , L 1 , L 2 take Eq. (1) and the linear forms L 0 :
y(0) = y(a) = 0,
: :
y(0) = y(T ) = 0, y(a) = y(T ) = 0.
L 1 L 2
Denote by j and j = {λn j } the characteristic functions and spectrum of L j ( j = 0, 1, 2), respectively, and assume that 0 ∩ 1 = ∅ (condition S ). Inverse Problem 4 Given three spectra 0 , 1 and 2 , construct q(x). The following theorem is a consequence of Theorem 1: condition S implies that ˜ ˜ ( j = 1, 2) ensure that M(λ) = M(λ) in 1 ∩ = ∅ in Theorem 1; j = j ˜ Theorem 1; and 0 = 0 means ω(λ) = ω(λ) ˜ in Theorem 1. Note that by Hadamard’s factorization theorem j is uniquely determined by j , respectively, since j are all entire in λ of order 1/2. ˜ , j = 0, 1, 2, then q(x) = q(x) ˜ a.e. Theorem 3 Let condition S hold. If j = j on (0, T ). We note that Inverse Problem 4 was studied by many authors (see, for example, [27,28]). Acknowledgments The authors would like to thank the referees for valuable comments. The research work of Yurko was supported in part by Grant 13-01-00134 of Russian Foundation for Basic Research. Yang was supported in part by the National Natural Science Foundation of China (11171152) and Natural Science Foundation of Jiangsu Province of China (BK 20141392).
V.A. Yurko, C. Yang
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