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Three spectra inverse Sturm–Liouville problems with overlapping eigenvalues

Shouzhong Fu

B1

, Zhong Wang

1

and Guangsheng Wei

2

1College of Mathematics and Statistics, Zhaoqing University, Zhaoqing 526061, PR China

2College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, PR China

Received 28 January 2017, appeared 9 May 2017 Communicated by Miklós Horváth

Abstract. In the paper we show that the Dirichlet spectra of three Sturm–Liouville differential operators defined on the intervals [0, 1],[0,a] and[a, 1]for some a∈ (0, 1) fixed, together with the knowledge of the normalizing constants corresponding to the overlapped eigenvalues, uniquely determine the potentialq on[0, 1]. In this situation we also provide the algorithm for recovering the potential by using the given spectral data.

Keywords: eigenvalue, normalizing constant, inverse eigenvalue problem.

2010 Mathematics Subject Classification: 34A55, 34B24.

1 Introduction

In this paper we are concerned with the inverse problem for recovering the potential q on interval[0, 1]of a Sturm–Liouville equation

lu:=−u00+q(x)u=λu (1.1)

using the three spectra σ(L),σ(L)andσ(L+)corresponding to three Sturm–Liouville oper- atorsL,LandL+. These operators are generated inL2 spaces by the differential expressions ldefined on[0, 1],[0,a]and[a, 1], respectively, and the Dirichlet boundary conditions

u(0) =0=u(1), (1.2)

u(0) =0=u(a), (1.3)

u(a) =0=u(1). (1.4)

Here the potential q∈ L2[0, 1]is a real-valued function and a∈(0, 1)is fixed.

Recently there has been much interest in three spectra inverse problems (see [1–7,9,16]

and the references therein). This problem was first investigated by Pivovarchik [15] under condition thata=1/2 andσ(L)andσ(L±)are all Dirichlet spectra. Further investigation has

BCorresponding author. Email: shzhfu@sina.com

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been carried out by Gesztesy and Simon [9] under the more general situations of a ∈ (0, 1) and no-Dirichlet spectra. In particular, Gesztesy and Simon [9] proved uniqueness of the reconstructed qwhenever the three spectra do not overlap and suggested a counterexample to uniqueness otherwise. They also raised an interesting assertion: We believe the analysis of the situation whereσ(L)∩σ(L+)has k-points will yield k-parameter sets of potentials q consistent with the given sets of eigenvalues (see [9, p. 91]).

One of our purposes of this paper is to answer affirmatively the above assertion. More specifically, we shall show that the uniqueness result remains valid by using the three spectra provided that the normalizing constants αn corresponding to λn are employed as the addi- tional spectral data, whenλnσ(L)∩σ(L+). This implies that whenσ(L)∩σ(L+)hask- points, there then exists ak-parameter set of potentialsqcorresponding to the given three spec- tra. Our second purpose here is to reconstruct potentials in terms of the three spectra, together with the normalizing constants{αn}nΛ whereΛ ⊂Nsatisfying σ(L+)∩σ(L) = {λn}nΛ. The method used here is similar to that used solving half-inverse problem [15] and three spectra inverse problem without the overlapping eigenvalues [14].

Section 2 includes the uniqueness theorem and its proof. Section 3 gives the algorithm for the reconstruction ofqin terms of given spectral data.

2 Uniqueness

In this section we shall prove our uniqueness result concerning with three spectra inverse Sturm–Liouville problems with overlapping eigenvalues.

Denote byu(x,λ)andu+(x,λ)the solutions of Eq. (1.1) with the initial conditions u(0,λ) =u0(0,λ)−1=0, (2.1) u+(1,λ) =u0+(1,λ)−1=0. (2.2) It is known (see, e.g., [8] and [13, Lemma 3.4.2 and proof of Theorem 3.4.1]) that, for each x∈ [0, 1],u±(x,λ)are entire functions ofλand the eigenvalues{λn}n=1of the operator Lare precisely zeros of the function∆(λ)defined by

∆(λ) =

u(a,λ) u+(a,λ) u0(a,λ) u0+(a,λ)

=u(1,λ), (2.3)

where∆(λ)is called the characteristic function of L, which also has the following representa- tion:

∆(λ) =

k=1

λkλ k2π2

. (2.4)

Similarly, lettingσ(L+) ={µ+n}n=1andσ(L) ={µn}n=1, then we have

+(λ):=u+(a,λ) = (1−a)

k=1

µ+nλ k2π2/(1−a)2

, (2.5)

(λ):=u(a,λ) =a

k=1

µnλ k2π2/a2

, (2.6)

where∆±(λ)denote the characteristic functions of the operators L±. Moreover, for anyλnσ(L), the normalizing constant,αn, associated with theλnis defined by

αn =

Z 1

0 u2(x,λn)dx. (2.7)

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In virtue of the above preliminaries, we are now in a position to give the uniqueness result of this paper.

Theorem 2.1. Let the Sturm–Liouville differential operators L and L±be defined by(1.1)–(1.4). Let σ(L+)∩σ(L) ={λn}nΛ (2.8) whereΛ(⊂ N)is the associated index set. Then the potential q is uniquely determined almost every- where on[0, 1]by the three spectraσ(L),σ(L±)and the normalizing constants{αn}nΛ.

Proof. Let us consider another operators ˜L, ˜L and ˜L+ of the same form (1.1)–(1.4) but with different coefficient ˜q(x) on [0, 1]. Denote by ˜∆(λ)and ˜±(λ) the characteristic functions of operators ˜L and ˜L±. Then by the hypotheses of Theorem 2.1, these operators and L and L± have common spectra σ(L)and σ(L±)respectively; and normalizing constants αn for n ∈ Λ corresponding to the common eigenvalues λnσ(L)∩σ(L±).

From (2.3)–(2.6) we have u±(a,λ) = u˜±(a,λ)andu(1,λ) =u˜(1,λ). This together with (2.3) implies

u(a,λ) u+(a,λ) u0(a,λ)−u˜0(a,λ) u0+(a,λ)−u˜0+(a,λ)

=(λ)−˜(λ)≡0. (2.9) We first prove thatu0±(a,µ±k ) =u˜0±(a,µ±k )for allk∈N. Ifµk ∈/ σ(L)then by (2.5) and (2.6) we haveu+(a,µk )6=0,u(a,µk ) =0 and thereforeu0(a,µk ) =u˜0(a,µk ). Similar argument yields u0+(a,µ+k) =u˜0+(a,µ+k )for all µ+k ∈/σ(L).

On the other hand, if there existk1, k2 andk ∈ Nsuch that µk

1 = µ+k

2 = λkσ(L), then u(a,λk) =0=u+(a,λk). This together with (2.9) yields

˙

u(a,λk) u˙+(a,λk) u0(a,λk)−u˜0(a,λk) u0+(a,λk)−u˜0+(a,λk)

=0. (2.10)

Here ˙u±=∂u±/∂λ. Sinceλk is a simple eigenvalue of the operators LandL±, it follows that

˙

u±(a,λk)6=0. If

u0±(a,λk)−u˜0±(a,λk)6=0, (2.11) then we have the following equality

˙

u(a,λk) u˙+(a,λk) = u

0(a,λk)−u˜0(a,λk)

u0+(a,λk)−u˜0+(a,λk). (2.12) Note that ˙∆(λk) =−αkκk whereκk = u0+(a,λk)/u0(a,λk)(see [8]). Since αk =α˜k and∆(λ) =

∆˜(λ)for allλC, it follows thatκk =κ˜k. This combined with (2.11) gives u0(a,λk)

u0+(a,λk) = u˜

0(a,λk)

˜

u0+(a,λk) = u

0(a,λk)−u˜0(a,λk)

u0+(a,λk)−u˜0+(a,λk) = u˙(a,λk)

˙

u+(a,λk). (2.13) Moreover, because each zero of the characteristic function∆(λ)is simple, we have

∆˙(λk) =

˙

u(a,λk) u˙+(a,λk) u0(a,λk) u0+(a,λk)

6=0, (2.14)

which contradicts (2.13). Therefore,u0±(a,λk) =u˜0±(a,λk)for allk ∈Λ.

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By the discussion above, we see that u0±(a,µ±k )−u˜0±(a,µ±k ) = 0 for all k ∈ N. Let us consider the functionF(λ)defined as

F(λ) = u

0(a,λ)−u˜0(a,λ)

u(a,λ) . (2.15)

It is clear to check thatF(λ)is an entire function and it has the following representation F(λ) = u

0(a,λ) u(a,λ)−u˜

0(a,λ)

˜

u(a,λ) =: ˜m(a,λ)−m(a,λ). (2.16) Herem(a,λ)is called the Weylm-function associated with the Sturm–Liouville operatorL. It is known that m(a,λ) = i√

λ+o(1) as |λ| → in any sector ε < Arg(λ) < πε for ε >0, where√

λis the square root branch with Im(√

λ)≥0, which implies that m(a,iy)−

˜

m(a,iy) =o(1)asy →+∞. It together with (2.16) yieldsF(λ) =0 and thereforeu0(a,λ) =

˜

u0(a,λ)for allλC.

The above argument implies that m±(a,λ) ≡ m˜±(a,λ). By the uniqueness result of Marchenko [13], we have either q = q˜ on [0,a] by using m(a,λ) ≡ m˜(a,λ) or q = q˜ on [a, 1]by usingm+(a,λ)≡m˜+(a,λ). The proof is complete.

3 Reconstruction

By the uniqueness theorem above, we know that the solution of the inverse problem discussed by us, if exists, is unique. In this section we provide the way of recovering the potential q by using three Dirichlet spectra σ(L), σ(L±) and (if necessary) the normalizing constants {αn}nΛ corresponding to the overlapped eigenvalues λnσ(L), where Λ = {n : λnσ(L)∩σ(L+)}. The method used here is similar to that used solving half-inverse problem [15] and three spectra inverse problem [14] without the overlapping eigenvalues.

Without loss of generality, we always assume the eigenvalues{λk}k=1,{µ±k}k=1are all pos- itive. Denote bys2 =λandγ2 =µ. Under these notations, we shall use increasing sequences {sk}kZ\{0} sk =−sk,(sk)2 =λk fork ∈N and{γ±k }kZ\{0} γ±k =−γ±k , (γ±k )2= µ±k for k∈N, and{αn}nΛ to recoveringq.

Based on condition that the potentialqexists corresponding to our spectral data, according to [13, p. 32] we have

u(x,λ) = sin(sx)

s +

Z x

0 K1(x,t)sin(st)

s dt, (3.1)

u+(x,λ) = sin s(1−x)

s +

Z 1

x K2(x,t)sin s(1−t)

s dt. (3.2)

Using (3.1) and (3.2), we obtain

u0(a,λ) =cos(sa) +K1sin(sa)

s + ψ1(s)

s , (3.3)

u0+(a,λ) =cos s(1−a)−K2sin s(1−a)

s +ψ2(s)

s , (3.4)

whereψ1(0) =0,ψ2(0) =0, K1= 1

2 Z a

0 q(t)dt, K2= 1 2

Z 1

a q(t)dt,

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ψ1(s) =

Z a

0 K01,x(a,t)sin(st)dt, ψ2(s) =

Z 1

a K2,x0 (a,t)sin s(1−t)dt. (3.5) Moreover, theK1(x,t)andK2(x,t)possess partial derivatives of the first order, and belonging to L2(0,a) and L2(a, 1), respectively, as a function of each of its variables when the other variable is fixed. From [10, Theorem 8] we see that two functional sequences

sin γk t k=1 and

sin γk+(t−1)

k=1 are Riesz bases of L2[0,a] and L2[a, 1], respectively. This together with the fact thatK1,x0 (a,t)∈ L2(0,a)andK2,x0 (a,t)∈ L2(a, 1)implies

{ψ1(γk )}kZ\{0}, {ψ2(γ+k )}kZ\{0} ∈l2 (3.6) (see [13, Lemma 1.4.3]). On the other hand, letting La denote the class of entire functions of exponential type ≤ a, which belong to L2(−,∞) for real λ, then we have ψ1 ∈ La and ψ2∈ L(1a).

Now we are in a position to constructq. We begin by introducing the following lemma.

Lemma 3.1. We have

u0(a,µk ) =









(µk )

u+(a,µk ) ifµ

k ∈/σ(L),

αn˙(µk )

∆˙(µk )u˙(a,µk ) +αn+(a,µk ) ifµ

k =λnσ(L),

(3.7)

and

u0+(a,µ+k ) =









∆(µ+k )

u(a,µ+k ) ifµ

+

k ∈/σ(L),

∆˙(µ+k )2

∆˙(µ+k )u˙(a,µ+k ) +αn+(a,µ+k) ifµ

+

k =λnσ(L).

(3.8)

Proof. By (2.3), if λ = µk ∈/ σ(L) then we have u+(a,µk ) 6= 0 and therefore u0(a,µk) =

−4(µk)/u+(a,µk); moreover, if µk = λnσ(L) then we haveµkσ(L+). This together with (2.3) implies

∆˙(µk ) =

˙ u+

u0 u0+

(a,µk ). (3.9)

Note that ˙∆(µk ) = −αnκn and κn = u0+(a,µk )/u0(a,µk ), which together with (3.9) yields (3.7). Similar argument implies that (3.8) remains true. The proof is complete.

The above lemma helps us to solve the values of ψ1(γk) and ψ2(γk+) for k ∈ Z\{0} in terms of (3.3) and (3.4). Thus, fork ∈Nwe have

ψ1(γk ) =γk u0(a,µk)−cos(aγk )−K1sin(aγk ) γk

!

, (3.10)

ψ1(γk) =γk u0(a,µk )−cos(aγk)−K1sin(aγk) γk

!

;

ψ2(γ+k ) =γ+k u0+(a,µ+k) +cos (1−a)γ+k

+K2sin (1−a)γk+ γk+

!

, (3.11)

ψ2(γ+k) =γ+k

u0+(a,µ+k ) +cos

(1−a)γ+k

+K2 sin

(1−a)γ+k

γ+k

,

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where γk = −γk, γ+k = −γk+. Moreover, it is easy to check from (3.1) and (3.2) that both functionssu(a,λ) andsu+(a,λ) are of sine-type (see, e.g., [11]), that is, there exist positive numbersm,M andp such that

me|Ims|a ≤ |su(a,λ)| ≤ Me|Ims|a, me|Ims|(1a) ≤ |su+(a,λ)| ≤ Me|Ims|(1a) for|Ims|> p. We use Theorem A in [11] and find

ψ1(s) =su(a,s2)

k=−∞,k6=0

ψ1(γk )

dsu(a,s2) ds |s=γ

k (s−γk ), (3.12) ψ2(s) =su+(a,s2)

k=−∞,k6=0

ψ2(γ+k )

dsu+(a,s2) ds |s=γ+

k (s−γ+k ), (3.13) where ψ1(0) = ψ2(0) = 0. The series on the right-hand sides of (3.12) and (3.13) converge uniformly on any compact subdomain ofC and in L2(−∞,) for realλ to functions which belong toLa andL1a. Then we find the characteristic functionsu0(a,λ)andu0+(a,λ)of the Dirichlet-Neumann Sturm–Liouville problems on the intervals [0,a] and [a, 1], respectively.

Denote by{νk }∞,k6=0and{νk+}∞,k6=0the sets of zeros of the functionsu0(a,λ)andu0+(a,λ). We are able to construct the part of the potentialq∈ L2(0,a)by the two spectra{γk }∞,k6=0 and{νk},k6=0 and the part of the potential q∈ L2(a, 1)by the two spectra{γ+k },k6=0 and {νk+}∞,k6=0 via the procedure of [12,13].

The method of reconstruction of potentials is described by the following algorithm.

Algorithm 3.2. Suppose that the three spectraσ(L) = {λn}n=1, σ(L) = {µn}n=1 andσ(L+) = {µ+n}n=1 corresponding to Sturm–Liouville operators L, L and L+ defined by (1.1)–(1.4) and the normalizing constants{αn}nΛwhereΛ= {n:λnσ(L)∩σ(L+)}are given. Then the algorithm of constructing q from the given data follows.

(1) Construct∆(λ)and u±(a,λ)by(2.3)–(2.6)in Section 2.

(2) Findψ1(s)andψ2(s)by(3.12)–(3.13)and K1and K2by the asymptotics of{µ±n}n=1:

µ+n = a

2

+ K1

a +o(1), µ+n = nπ

1−a 2

+ K2

1−a+o(1); and then find u0(a,λ)and u0+(a,λ)by(3.3)and(3.4).

(3) Solve the zeros,{νk}∞,k6=0and{νk+}∞,k6=0, of functions u0(a,λ)and u0+(a,λ).

(4) Using the algorithm of Gelfand and Levitan (see [12,13]) to construct the (unique) potential q on [0,a]and[a, 1]by{γ±k }kZ\{0}and

νk± kZ\{0}.

Acknowledgements

The authors would like to thank the referees for their insightful suggestions and comments which improved and strengthened the presentation of this manuscript.

The research was supported in part by the NNSF of China (No. 11571212, 11171295).

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Nachr.289(2016), 1727–1738.MR3563902;url

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