INVERSE STURM-LIOUVILLE PROBLEMS WITH A SPECTRAL ...

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Abstract. In this manuscript, we study the inverse problem for non self-adjoint Sturm–Liouville operator −D2 + q with eigenpa- rameter dependent boundary and ...
Sahand Communications in Mathematical Analysis (SCMA) Vol. 3 No. 2 (2016), 75-89 http://scma.maragheh.ac.ir

INVERSE STURM-LIOUVILLE PROBLEMS WITH A SPECTRAL PARAMETER IN THE BOUNDARY AND TRANSMISSION CONDITIONS MOHAMMAD SHAHRIARI

Abstract. In this manuscript, we study the inverse problem for non self-adjoint Sturm–Liouville operator −D2 + q with eigenparameter dependent boundary and discontinuity conditions inside a finite closed interval. By defining a new Hilbert space and using its spectral data of a kind, it is shown that the potential function can be uniquely determined by part of a set of values of eigenfunctions at some interior point and parts of two sets of eigenvalues.

1. Introduction Let us consider the boundary value problem (1.1)

ℓy := −y ′′ + qy = λy, U (y) := λ(y ′ (0) + h1 y(0)) − h2 y ′ (0) − h3 y(0) = 0,

(1.2)

V (y) := λ(y ′ (π) + H1 y(π)) − H2 y ′ (π) − H3 y(π) = 0,

with the jump conditions U1 (y) := y(a + 0) − a1 y(a − 0) = 0, (1.3)

U2 (y) := y ′ (a + 0) − a2 y ′ (a − 0) − a3 y(a − 0) = 0,

where q(x) is a real function in ∈ L2 (0, π), hi , Hi , and ai (i = 1, 2, 3), are real with a ∈ (0, π), a1 a2 > 0, r1 := h3 − h1 h2 > 0 and r2 := H1 H2 − H3 > 0. 2010 Mathematics Subject Classification. Primary: 34B24, 47A05; Secondary: 34B20, 47A10. Key words and phrases. Inverse Sturm-Liouville problem, Jump conditions, Non self-adjoint operator, Parameter dependent condition Received: 12 May 2015, Accepted: 21 September 2015. 75

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M. SHAHRIARI

For simplicity we use the notation L = L(q(x); h1 ; h2 ; h3 ; H1 ; H2 ; H3 ; a) for the problem (1.1)–(1.3). Here λ is the spectral parameter. In the present paper, we are interested in the inverse problem, that is, the question of existence, uniqueness and reconstruction of the potential function q from given spectral data. These problems originated in the work of Ambarzumian(1929)[3], were continued by Borg(1945)[6], and have been gradually elucidated over the past seventy years. Here we want to look at the question of uniqueness for the above problem using two sets of spectra, or one spectrum plus a part of a set of value of eigenfunctions at some interior point. Such kind of problems have a long tradition and we refer the reader to [20]–[7], [34, 14, 25], and the references therein. In particular, the operator ℓ plays an important role as the one-dimensional Schr¨odinger operator in quantum mechanics and our transmission conditions include the case of point interactions (see e.g. the monographs [29] and [2]). In section 2 we define a new Hilbert space for a non-self-adjoint Sturm–Liouville operator by using similar techniques as in [23, 1], to obtain the asymptotic form of solutions and eigenvalues. In section 3 we formulate a novel inverse spectral problem which is a generalization of [10, 28] and [22]–[11]. 2. Asymptotic form of solutions and eigenvalues In this section, we introduce a special inner product in the Hilbert space (L2 (0, a) ⊕ L2 (a, π)) ⊕ C2 and define a linear operator A in it such that the problem (1.1)-(1.3) can be interpreted as the eigenvalue problem of A. To this end we introduce the function { 1, 0 ≤ x < a, (2.1) w(x) = 1 a < x ≤ π. a1 a2 , Now we define a new Hilbert space inner product on H := (L2 (0, a) ⊕ L2 (a, π)) ⊕ C2 , by



π

w(0) w(π) f1 g¯1 + f2 g¯2 , r1 r2 0     f (x) g(x) where F =  f1  and G =  g1  ∈ H and we let f2 g2

(2.2)

⟨F, G⟩H :=

f g¯w +

R1 (u) := u′ (0) + h1 u(0),

R1′ (u) := h2 u′ (0) + h3 u(0),

R2 (u) := u′ (π) + H1 u(π),

R2′ (u) := H2 u′ (π) + H3 u(π).

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In this Hilbert space we construct the operator (2.3) with domain (2.4)

A : H → H,  

  f (x), f ′ (x) ∈ AC[0, a) ∪ (a, π] and, f (a ± 0), f (x)  ′ D(A) = F =  f1  f (a ± 0) is defined, ℓf ∈ L2 (0, π), f1 = R1 (f ),   f2 = R2 (f ), U (f ) = U1 (f ) = U2 (f ) = 0, f2 

and by action law   ℓf AF =  R1′ (f )  R2′ (f )

 f (x) F =  R1 (f )  ∈ D(A). R2 (f ) 

with

So, we can change the boundary value problem (1.1)-(1.3) as following form   y(x) (2.5) AY = λY Y :=  R1 (y)  ∈ D(A), R2 (y) in the Hilbert space H. It is easy to verify that the eigenvalues of the operator A coincide with those of the problem (1.1)-(1.3). Theorem 2.1. The operator A is self-adjoint. Proof. We omit the proof, since the arguments are the same as in [23]. □ Suppose that the functions φ(x, λ) and ψ(x, λ) are solutions of (1.1) under the initial conditions (2.6)

φ(0, λ) = h2 − λ,

φ′ (0, λ) = λh1 − h3 ,

and (2.7)

ψ(π, λ) = H2 − λ,

ψ ′ (π, λ) = λH1 − H3 .

By attaching a subscript 1 or 2 to the functions φ and ψ, we mean to refer to the first subinterval [0, a) or to the second subinterval (a, π]. By virtue of [33], problem (1.1) under the initial conditions (2.6) or (2.7), has a unique solution φ1 (x, λ) or ψ2 (x, λ), which is an entire function of λ ∈ C for each fixed point x ∈ [0, a) or x ∈ (a, π]. From the linear differential equations we obtain that the modified Wronskian ( ) (2.8) W (u, v) = w(x) u(x)v ′ (x) − u′ (x)v(x) , is constant on x ∈ [0, a) ∪ (a, π] for two solutions ℓu = λu, ℓv = λv satisfying the transmission conditions (1.3). Moreover, we set (2.9)

∆(λ) := W (φ(λ), ψ(λ)).

Then ∆(λ) is an entire function whose roots coincide with the eigenvalues of problem L. Moreover, the eigenfunctions φi (x, λn ) and ψi (x, λn )

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M. SHAHRIARI

associated with a certain eigenvalue λn , satisfy the relation ψi (x, λn ) = βn φi (x, λn ), thus from (2.6), ψ ′ (0, λn ) + h1 ψ(0, λn ) . r1 We define the norming constant by (2.10)

βn =

γn := ⟨Φ(x, λn ), Φ(x, λn )⟩H , where Φ(x, λ) = (φ1 (x, λ), φ2 (x, λ), R1 (φ), R2 (φ))T . ˙ n ) = −γn βn holds for each eigenLemma 2.2 ([27]). The equality ∆(λ value λn . Lemma 2.3 ([27]). The eigenvalues of the problem L are real and simple. Theorem 2.4. Let λ = ρ2 and τ := Imρ. For equation (1.1) with boundary conditions (1.2) and jump conditions (1.3) as |λ| → ∞, the following asymptotic formulas hold: (2.11) φ(x; λ) =

(2.12) ′

φ (x; λ) =

 2 ( ρ cos ρx + ρ −h1 +   

1 2

∫x 0

) q(t)dt sin ρx + O(exp(|τ |x)), x < a,

ρ2 (b1 cos ρx + b2 cos ρ(2a − x)) + ρ (f1 (x) sin ρx    +f2 (x) sin ρ(2a − x)) + O(exp(|τ |x)),  ( −ρ3 sin ρx + ρ2 −h1 +   

1 2

∫x 0

) q(t)dt cos ρx + O(ρ exp(|τ |x))

ρ3 (−b1 sin ρx + b2 sin ρ(2a − x)) + ρ2 (f1 (x) cos ρx    −f2 (x) cos ρ(2a − x)) + O(ρ exp(|τ |x)),

x > a, x < a,

x > a,

where (2.13) and

b1 = (a1 + a2 )/2,

b2 = (a1 − a2 )/2,

( ) ∫ 1 x a3 f1 (x) = b1 −h1 + q(t)dt + , 2 0 2 ( ) ∫ x ∫ a 1 a3 f2 (x) = b2 −h1 − q(t)dt + q(t)dt + . 2 0 2 0

Then the characteristic function is ∆(λ) = ρ5 (−b1 sin ρπ + b2 sin ρ(2a − π)) + ρ4 [(f1 (π) + H1 b1 ) cos ρπ (2.14)

+(−f2 (π) + H1 b2 ) cos ρ(2a − π)] + O(ρ3 exp(|τ |π)).

Proof. We omit the proof, since the arguments are the same as in [27]. □

INVERSE STURM-LIOUVILLE SPECTRAL PARAMETER PROBLEMS ...

79

Define (2.15)

∆◦ (ρ) := ρ5 (−b1 sin ρπ + b2 sin ρ(2a − π)).

So we have the following Lemma: Lemma 2.5. The roots of characteristic function ∆◦ (ρ) are ρ◦n = n −

(2.16)

5 + ηn , 2

where ηn ∈ (0, 1) for n ∈ N. Proof. There exist a similar proof for obtaining ηn ∈ (0, 1) in Lemma 2.6 of [26]. □ Theorem 2.6. The corresponding eigenvalues {λn } of the boundary value problem L, admit the following asymptotic form as n → ∞: ( ) 1 5 , (2.17) ρn = n − + η n + O 2 n where ηn ∈ (0, 1). Proof. From (2.14) and (2.15), we see that ( ) (2.18) ∆(ρ) = ∆◦ (ρ) + O ρ4 exp(|τ |π) . For sufficiently large value of ρ, the functions ∆(ρ) and ∆◦ (ρ) have the same number of zeros counting multiplicities according to Rouche’s theorem. So if ρn and ρ◦n are roots of ∆(ρ) and ∆◦ (ρ) respectively, then we have ρn = ρ◦n + εn where limn→∞ εn = 0. Since numbers ρn are zeros of the characteristic function ∆(ρ), therefore by a simple calculation we (1) □ get εn = O n . 3. Main results We consider the boundary value problem ˜ 1; h ˜ 2; h ˜ 3 ; H1 ; H2 ; H3 ; a), ˜ = L(˜ L q (x); h which is the same as L with diferent q, h1 , h2 and h3 . Let φ(x, λ) and φ(x, ˜ λ) be the solutions of (3.1)

− φ′′ + qφ = λφ, φ(0, λ) = h2 − λ, φ′ (0, λ) = λh1 − h3 ,

and (3.2)

− φ˜′′ + q˜φ˜ = λφ, ˜ ˜ 2 − λ, φ′ (0, λ) = λh ˜1 − h ˜ 3. φ(0, λ) = h

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M. SHAHRIARI

Suppose b is an arbitrary and fixed point on [0, π2 ], then the following representation holds (see [19]) ] [ ∫ x k1 (x, t) cos(ρt)dt φ(x, λ) = (λ − h2 ) cos(ρx) + 0 [ ] ∫ x sin(ρx) sin(ρt) + (h3 − λh1 ) + k2 (x, t) dt , (3.3) ρ ρ 0 and

(3.4)

[ ] ∫ x ˜ ˜ φ(x, ˜ λ) = (λ − h2 ) cos(ρx) + k1 (x, t) cos(ρt)dt 0 [ ] ∫ x sin(ρx) sin(ρt) ˜ 3 − λh ˜ 1) + (h + k˜2 (x, t) dt , ρ ρ 0

for x ∈ (0, a), where k1 (x, t), k2 (x, t), k˜1 (x, t), and k˜2 (x, t) are continuous functions which does not depend on λ. Using (3.3) and (3.4) it is easy to verify that (

) ∫ x 1 (1 + cos(2ρx)) + v1 (x, t) cos(2ρt)dt 2 0 ( ) ∫ x 1 ˜ ˜ + (h3 − λh1 )(h3 − λh1 ) (1 + cos(2ρx)) + v2 (x, t) cos(2ρt)dt 2 0 ( ) ∫ x ˜ 3 − λh ˜ 1 ) 1 (sin(2ρx) + + (λ − h2 )(h v3 (x, t) sin(2ρt)dt) 2ρ 0 ( ) ∫ x 1 ˜ 2 )(h3 − λh1 ) + (λ − h (sin(2ρx) + v4 (x, t) sin(2ρt)dt) 2ρ 0 ( ) ∫ x 2 = λ B1 (1 + cos(2ρt)) + B1 R1 (x, t) cos(2ρt)dt 0 ( ) ∫ x + λ B2 (1 + cos(2ρt)) + B2 R2 (x, t) cos(2ρt)dt ∫ x 0 + B3 (1 + cos(2ρt)) + B3 R3 (x, t) cos(2ρt)dt,

˜2) φ(x, λ)φ(x, ˜ λ) = (λ − h2 )(λ − h

0

(3.5)

for x ∈ (0, a). Using a similar method of [34, 14] and a simple calculation we obtain ( ) ∫ x 2 φ(x, λ)φ(x, ˜ λ) = λ B1 A(x, ρ) + B1 R1 (x, t) cos(2ρt)dt 0 ( ) ∫ x + λ B2 A(x, ρ) + B2 R2 (x, t) cos(2ρt)dt ∫ x 0 + B3 A(x, ρ) + B3 R3 (x, t) cos(2ρt)dt, 0

(3.6)

INVERSE STURM-LIOUVILLE SPECTRAL PARAMETER PROBLEMS ...

81

for a ≤ x ≤ b; where A(x, ρ) = 2A1 + 2A2 cos(2ρx) + 2A3 cos 2ρ(x − a) + 2A4 cos 2ρ(x − 2a). Here A1 = A3 =

1 + 12 (a21 2a21 a21 −a22 4 ,

− a22 )φ(a − 0, λ)φ(a ˜ − 0, λ),

A2 = A4 =

(a1 +a2 )−2 , 8 2 (a1 −a2 ) , 8

and B1 =

˜1 1 + h1 h , 2

B2 = −

˜ 2 + h1 h ˜3 + h ˜ 1 h3 h2 + h , 2

B3 =

˜ 2 + h3 h ˜3 h2 h . 2

The functions Ri (x, t) (i = 1, 2, 3) are continuous which does not depend on λ. Let l(n) be a subsequence of natural numbers such that (3.7)

l(n) =

n (1 + ϵ1n ), σ1

0 < σ1 ≤ 1,

ϵ1n → 0.

Suppose ∆l (λ) is a restriction of ∆(λ) such that all roots of ∆l (λ) is {λl(n) }. Lemma 3.1. Let Gη = {ρ : |ρ − ρl(n) | > η} and fix η > 0, then there is a positive constant Cη such that ( |∆l (λ)| ≥ Cη |ρ|5 exp

|τ |π σ1

) .

Proof. The proof of this lemma is similar to the proof of Theorem 1.1.3 [8, P. 11]. □ Theorem 3.2. If a ∈ (0, π) be a jump point, fix point b ∈ [0, π2 ], and σ1 > 2b π ˜ l(n) , λl(n) = λ

⟨yl(n) , y˜l(n) ⟩(b) = 0,

for any n ∈ N, then q(x) = q˜(x) almost everywhere(a.e.) on [0, b] and ˜ i for i = 1, 2, 3. hi = h Proof. Define Q(x) := q˜(x) − q(x). Multiplying φ˜ by (3.1), φ by (3.2), and subtracting the result and integrating on [0, b] ∪ [b, a) ∪ (a, π], we

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M. SHAHRIARI

obtain

 ∫b ˜ 2 )(λh1 − h3 ) − (λ − h2 )(λh ˜1 − h ˜ 3 ),  Q(x)φ(x, λ)φ(x, ˜ λ)dx + (λ − h    0 ∫a ∫b G(λ) := |a1 | 0 Q(x)φ(x, λ)φ(x, ˜ λ)dx + |a12 | a Q(x)φ(x, λ)φ(x, ˜ λ)dx     ˜ 2 )(λh1 − h3 ) − (λ − h2 )(λh ˜1 − h ˜ 3 ), +|a1 |(λ − h (3.8)

=

b ≤ a,

a < b,

 b ˜ 2 )(λh1 − h3 ) [φ ˜′ (x, λ)φ(x, λ) − φ(x, ˜ λ)φ′ (x, λ)]0 + (λ − h    ˜ ˜  −(λ − h2 )(λh1 − h3 ) 

b ≤ a,

 a  |a1 | [φ ˜′ (x, λ)φ(x, λ) − φ(x, ˜ λ)φ′ (x, λ)]0 + |a12 | [φ ˜′ (x, λ)φ(x, λ)    ′ b ˜ 2 )(λh1 − h3 ) − (λ − h2 )(λh ˜1 − h ˜ 3 ), −φ(x, ˜ λ)φ (x, λ)]a + |a1 |(λ − h

a < b.

We now claim that G(λ) = 0 on the whole λ-plane. Using (3.8) we get G(λl(n) ) = 0. From the inequality (3.9)

| cos(ρx)| ≤ exp(x|τ |)

and

| sin(ρx)| ≤ exp(x|τ |),

we see that the entire function G(λ) satisfies (3.10)

 C1 |ρ|4 exp(2b|τ |),      ( |G(λ)| ≤ |ρ|4 2a12 + C2 exp(2a|τ |) + A2 exp(2b|τ |) + A3 exp(2(b − a)|τ |)   1 )    +A4 exp(2|b − 2a||τ |) + C3 exp(2b|τ |) ,

b ≤ a,

b > a,

for some positive constants C1 , C2 , C3 and |λ| large enough. It is easy to see that b = max{a, b, b − a, |b − 2a|} for the above case. From (3.8), (3.9), and (3.10) we see that the entire function G(λ) satisfies (3.11)

|G(λ)| ≤ M |ρ|4 exp(2|τ |b),

for some positive constant M and |λ| large enough. Define ϕ(λ) :=

G(λ) . ∆l (λ)

The definition of ∆l (λ) and G(λ) implies ϕ(λ) is an entire function. It follows from (3.11) and Lemma 3.1 that ( ) 1 ϕ(λ) = O , |ρ| exp(|τ |(2b − σπ1 )) for sufficiently large |λ|. From the condition of σ1 the relation 2b− σπ1 > 0 holds. Hence Liouville’s theorem implies (3.12)

ϕ(λ) = 0,

for all λ, therefore (3.13)

G(λ) = 0,

for all λ.

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83

Now we want prove that q(x) = q˜(x) a.e. on [0, b]. We have two cases for obtaining this result. Case (i): If b ≤ a, from (3.5), (3.8), and (3.13), we obtain [ 2

G(λ) = λ





b

Q(x)dx + B1

B1

0





b

+ λ B2

b

Q(x)dx + B2



0



b

+ B3

b

Q(x)dx + B3



0

)

b

]

Q(t)R2 (x, t)dx dt t



)

b

cos(2ρt) Q(t) +

0

]

Q(t)R1 (x, t)dx dt t

( (

)

b

cos(2ρt) Q(t) +

0

[



cos(2ρt) Q(t) +

0

[

(

b

]

Q(t)R3 (x, t)dx dt t

˜ 1 ) + λ(h3 − h ˜3 + h ˜ 1 h2 − h1 h ˜ 2 ) + h3 h ˜ 2 − h2 h ˜ 3 = 0, + λ2 (h1 − h (3.14)

on the whole λ-plane. Letting λ → ∞ for three section in (3.14) and real λ, we see that from the Riemann-Lebesgue lemma ∫ b ˜ 1 = 0, B1 Q(x)dx + h1 − h 0



b

B2

˜3 + h ˜ 1 h2 − h1 h ˜ 2 = 0, Q(x)dx + h3 − h

0

∫ B3

b

˜ 2 − h2 h ˜ 3 = 0, Q(x)dx + h3 h

0

and



b

Bi

) ( ∫ b Q(t)Ri (x, t)dx dt = 0, cos(2ρt) Q(t) +

i = 1, 2, 3.

t

0

Note that there is at least one Bi ̸= 0, (i = 1, 2, 3). From the completeness of the function cos(2ρt) on the interval [0, b], we have ∫ b Q(t) + Q(x)Ri (x, t)dx = 0, 0 < t < b. t

But this equation is a homogeneous Volterra integral equation and has only the zero solution. Thus Q(x) = 0 a.e. on 0 ≤ x ≤ b, that is, q(x) = q˜(x) a.e. on [0, b]. So, from the above equations we obtain ˜ 1 , h2 = h ˜ 2 , and h3 = h ˜ 3. h1 = h Case (ii): If a < b ≤ π2 , from (3.8), (3.13), and using the similar result of case (i) and changing the integral variables, we get

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M. SHAHRIARI

[ 2

G(λ) = B1 λ



∫ 2A2 b |a1 | Q(x)dx + Q(x)dx |a2 | a 0 ( ) ] ∫ b ∫ b + cos(2ρt) F (t) + Q1 (t)R1 (x, t)dx dt

[

a



+ B2 λ |a1 |

0 a



t b

2A2 Q(x)dx |a2 | a 0 ( ) ] ∫ b ∫ b + cos(2ρt) F (t) + Q1 (t)R2 (x, t)dx dt Q(x)dx +

[ ∫ + B3 |a1 |

0



a

t b

2A2 Q(x)dx |a2 | a 0 ( ) ] ∫ b ∫ b + cos(2ρt) F (t) + Q1 (t)R3 (x, t)dx dt Q(x)dx +

0

t

˜ 1 ) + λ(h3 − h ˜3 + h ˜ 1 h2 − h1 h ˜ 2 ) + h3 h ˜ 2 − h2 h ˜ 3 = 0, + λ (h1 − h 2

(3.15) for i = 1, 2, 3 and on the whole λ-plane. Letting ρ → ∞ in (3.15), by the Riemann-Lebesgue lemma ∫ a ∫ [ ] 2A1 b ˜ 1 = 0, B1 |a1 | Q(x)dx + Q(x)dx + h1 − h |a2 | a 0 ∫ a ∫ [ ] 2A1 b ˜3 + h ˜ 1 h2 − h1 h ˜ 2 = 0, B2 |a1 | Q(x)dx + Q(x)dx + h3 − h |a2 | a 0 ∫ a ∫ ] [ 2A1 b ˜ 2 − h2 h ˜ 3 = 0, Q(x)dx + Q(x)dx + h3 h B3 |a1 | |a2 | a 0 and (3.16)

∫ b(

∫ F (t) +

Bi 0

b

) Ri (x, t)Q1 (x)dx cos(2ρt)dt = 0,

t

for all ρ and i = 1, 2, 3, where { |a1 |Q(x), x < a, (3.17) Q1 (x) := 1 |a2 | Q(x), a < x ≤ b. Since the functions cos(2ρt) is complete on L2 [0, b], we have ∫ b F (t) + Ri (x, t)Q1 (x)dx = 0. (3.18) t

The form of F (t) will help us to obtain that Q(x) = 0 a.e. in [0, b]. First, we consider the terms with Ri (x, t) in (3.15). Since Ri (x, t) is bounded

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85

on (x, t) ∈ [0, π] × [0, π] and Q(x) is integrable on [0, π], by the Fubini’s Theorem, ∫ a ∫ x ∫ b ∫ x Q1 (x) Ri (x, t) cos(2ρt)dtdx + Q1 (x) Ri (x, t) cos(2ρt)dtdx 0

0

(3.19)

∫ b∫ =

a

0

b

Ri (x, t)Q1 (x)dx cos(2ρt)dt. 0

t

Second, we consider the remainder terms in (3.15). Specifically, we have (3.20) ∫ a ∫ b ∫ b b Q(x) cos(2ρx)dx, Q(x) cos(2ρx)dx + A2 Q(x) cos(2ρx)dx = 0

a

where b Q(x) = ∫ (3.21)

b

0

{

Q(t), t ∈ [0, a], 2A2 Q(t), t ∈ [a, b], ∫

A3 Q(x) cos 2ρ(x − a)dx =

a

b−a

A3 Q(x + a) cos(2ρx)dx, 0

and (3.22)  ∫ 2a−b  ∫ b A4 Q(2a − x) cos(2ρx)dx, b ≤ 2a,  a A4 Q(x) cos 2ρ(x−2a)dx =  a  ∫ b−2a A Q(x + 2a) cos(2ρx)dx, 2a < b. 4 −a The equations (3.20)-(3.22), imply that F (t) in (3.19) has the following form. We have four cases for F (t): Case I: If b < 32 a  |a1 |Q(t) +     |a |Q(t), 1 (3.23) F (t) = |a  1 |Q(t) +    2A2 Q(t), |a2 |

t ∈ [0, b − a], t ∈ [b − a, 2a − b], 2A4 Q(2a − t)), t ∈ [2a − b, a], |a2 | t ∈ [a, b].

2A3 |a2 | Q(t

+ a),

Case II: If 23 a ≤ b < 2a (3.24)  3 |a1 |Q(t) + 2A  |a2 | Q(t + a),    |a1 |Q(t) + 2 (A3 Q(t + a) + A4 Q(2a − t)), |a2 | F (t) = 2A4 |a |Q(t) +  1  |a2 | Q(2a − t),   2A2 |a2 | Q(t),

t ∈ [0, 2a − b], t ∈ [2a − b, b − a], t ∈ [b − a, a], t ∈ [a, b].

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Case III: If 2a ≤ b < 3a (3.25)  2A4   |a2 | Q(t + 2a),   2    |a1 |Q(t) + |a2 | (A3 Q(t + a) + A4 Q(2a + t)), 3 |a1 |Q(t) + 2A F (t) = |a2 | Q(t + a),  2    |a2 | (A2 Q(t) + A3 Q(t + a)),    2A2 Q(t), |a2 |

t ∈ [−a, 0], t ∈ [0, b − 2a], t ∈ [b − 2a, a], t ∈ [a, b − a], t ∈ [b − a, b].

Case IV: If b ≥ 3a (3.26)  2A4   |a2 | Q(t + 2a),   2    |a1 |Q(t) + |a2 | (A3 Q(t + a) + A4 Q(2a + t)), 2 F (t) = |a2 | (A2 Q(t) + A3 Q(t + a) + A4 Q(2a + t)),  2    |a2 | (A2 Q(t) + A3 Q(t + a)),    2A2 Q(t), |a2 |

t ∈ [−a, 0], t ∈ [0, a], t ∈ [a, b − 2a], t ∈ [b − 2a, b − a], t ∈ [b − a, b].

To prove that Q(x) = 0 in [0, b], we shall consider only case I. From (3.17), (3.18), and (3.24), we see that ( ) ∫ b 1 (3.27) 2A2 Q(t) + Q(x)Ri (x, t)dx = 0, t ∈ [a, b]. |a2 | t Note that |a12 | ̸= 0 and A2 ̸= 0. Since (3.27) is a homogenous Volterra integral equation, so Q(t) = 0 a.e. on [a,b]. From (3.14) and (3.15) we ˜ 1, get Q(x) = 0 a.e. on [0, b]. So we get q(x) = q˜(x) a.e. on [0, b], h1 = h ˜ ˜ h2 = h2 , and h3 = h3 . By considering a similar way we can prove other cases. □ Suppose m(n) and r(n) are subsequences of natural numbers such that n (3.28) m(n) = (1 + ϵ2n ), 0 < σ2 ≤ 1, ϵ2n → 0, σ2 (3.29)

r(n) =

n (1 + ϵ3n ), σ3

0 < σ3 ≤ 1,

ϵ3n → 0.

Corollary 3.3. Let b ∈ ( π2 , π), a ∈ (0, π), and fix σ3 > 2 − ˜ r(n) , λr(n) = λ

and

2b π.

If

⟨yr(n) , y˜r(n) ⟩(b) = 0,

˜ 1 , h2 = h ˜ 2 , and h3 = h ˜ 3 , then q(x) = q˜(x) for each n ∈ N, and h1 = h a.e. on [b, π].

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87

Proof. To prove that q(x) = q˜(x) a.e. on [b, π], we will consider the b by changing x by π − x. Let t = π − x, then supplementary problem L for problem (1.1) we have −y ′′ + q(π − t)y = λy. We define q1 (t) = q(π − t), then the above equation is the following form b := −y ′′ + q1 (t)y = λy, 0 < t < π ℓy λ(y ′ (0) − H1 y(0)) − H2 y ′ (0) + H3 y ′ (0) = 0, λ(y ′ (π) − h1 y(π)) − h2 y ′ (π) + h3 y(π) = 0, with the discontinuous conditions 1 y(π − a + 0) = y(π − a − 0), a1 1 a3 (3.30) y ′ (π − a + 0) = y ′ (π − a − 0) + y(π − a − 0). a2 a1 a2 By the similar proof of Theorem 3.2, we obtain q1 (t) = q˜1 (t) a.e. on [0, π − b]. By replacement x = π − t we have q(x) = q˜(x) a.e. on [b, π]. □ Let µn be the eigenvalues of problems (3.1) and (3.31) and µ ˜n be the eigenvalues of problems (3.2) and (3.31) with the jump conditions (1.3) such that (3.31)

λ(y ′ (π) + H1 y(π)) − H2 y ′ (π) − H3 y ′ (π) = 0,

where H1 ̸= H1 and Hi , (i = 1, 2, 3) are real and H1 H3 − H2 > 0 . Theorem 3.4. Let b ∈ ( π2 , π) and a ∈ (0, π), fix σ2 > σ3 > 2 − 2b π . If ˜n, λn = λ

µm(n) = µ ˜m(n) ,

2b π

− 1 and

⟨yr(n) , y˜r(n) ⟩(b) = 0,

for each n ∈ N, then q(x) = q˜(x) a.e. on [b, π]. ˜ corProof. Let yn (x, λn ) and y˜n (x, λn ) be the eigenfunctions of L and L responding to the eigenvalues λn , respectively. Since the eigenfunctions yn (x, λn ) and y˜n (x, λn ) have the same boundary condition at point π and from Corollary 3.3, q(x) = q˜(x) a.e. on [b, π], we obtain (3.32)

y˜n (x, λn ) = an yn (x, λn ),

x ∈ (b, π], n ∈ N,

where an is constant. From (3.8) and assumptions we get (3.33)

G(λn ) = 0,

G(µl(n) ) = 0.

We now show that G(λ) = 0, for all λ ∈ C. By applying the similar method of the proof of Theorem 3.2 we obtain q(x) = q˜(x) almost ˜ 1 , h2 = h ˜ 2 , and h3 = h ˜ 3 . Using Corollary everywhere on [0, b], h1 = h

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M. SHAHRIARI

3.3 and the assumption of this theorem we obtain q(x) = q˜(x) a.e. on [b, π]. □

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