Inverse problem for a singular differential operator - Semantic Scholar
Mathematical and Computer Modelling 47 (2008) 178–185 www.elsevier.com/locate/mcm
Inverse problem for a singular differential operator H. Koyunbakan ∗ , E.S. Panakhov Department of Mathematics, Firat University, 23119 Elazig, Turkey Received 5 September 2006; accepted 7 February 2007
Abstract In this paper, we give the solution of the inverse Sturm–Liouville problem on two partially coinciding spectra. In particular, in this case we obtain Hochstadt’s theorem concerning the structure of the difference q(x) − q(x) ˜ for the singular Sturm Liouville problem defined on the finite interval (0, π) having the singularity type 1 2 at the points 0 and π. 4 sin x c 2007 Elsevier Ltd. All rights reserved.
Keywords: Sturm–Liouville operator; Spectrum; Green function
1. Introduction The inverse problem of spectral analysis implies the reconstruction of a linear operator from some or other of its spectral characteristics. Such characteristics are spectra (for different boundary conditions), normalizing constants, spectral functions, nodal points, scattering data, etc. The starting point of the inverse problems seems to be a paper by Ambartsumyan [1]. In 1946, Borg [2] proved that two spectra uniquely determined the potential q(x). Marchenko [3] proved that two various spectra of the one singular Sturm–Liouville equation determine this equation uniquely. Later, Levitan and Gasimov [4] gave the effective method of construction for the singular Sturm–Liouville equation on two spectra. In the case of the regular equation, this problem was solved in [5,6] etc. The inverse problem for potentials having a peculiarity in both side points of the interval (0, π) was solved in [7]. We note that the inverse problem for the singular Sturm–Liouville equation is investigated in [8–10] etc. In later years, the problems of the regular and singular differential equations are solved by [11,12,10,13,14]. When we solve the inverse problem from the spectral data, the obvious question occurs: what if only partial spectral values are given? This was answered by Hochstadt [5], who showed that if a complete spectrum was prescribed but the second spectrum was missing information from a certain index set Λ, then q(x) could only be recovered modulo a sum over that index set of eigenfunctions of a q(x). In this paper, by using the Hochstadt method, we give the the structure concerning the difference q(x) − q(x) ˜ for the differential operators having the singularity type 12 . sin x
∗ Corresponding author.
E-mail addresses: [email protected] (H. Koyunbakan), [email protected] (E.S. Panakhov). c 2007 Elsevier Ltd. All rights reserved. 0895-7177/$ - see front matter doi:10.1016/j.mcm.2007.02.013
H. Koyunbakan, E.S. Panakhov / Mathematical and Computer Modelling 47 (2008) 178–185
179
We consider an operator d2 1 L = − 2 + q(x) − dx 4 sin2 x in which q(x) is a summable real function. Let λ0 < λ1 · · · be the spectrum of the problem L [u] = λu,
(1.1)
u(0) = 0, u(π − ε, λ) cos β + u 0 (π − ε, λ) sin β = 0
(1.2) (1.3)
where 0 < ε < x ≤ π − ε < π [15]. It is well known that the sequence {λn } satisfies the asymptotic [16] p 1 a 1 λn = n + + + O 4 n n2 and it is known that the solution of the problem (1.1)–(1.3) can be expressed as Z x u(x, λ) = Pλ (x) + K (x, t)Pλ (t)dt, 0
where r Pλ (x) ∼ =
2 cos πn
1 π n+ x− 2 4
is the solution of the equation Lu = λu when q(x) = 0 and K (x, t) is the solution of the problem ∂ 2 K (x, t) 1 ∂ 2 K (x, t) 1 − q(x) K t) = K (x, t) , + + (x, ∂x2 ∂t 2 4 sin2 x 4 sin2 t dK (x, x) = q(x). K (x, 0) = 0, 2 dx This problem can be solved by using the Riemann method [9]. 2. A Hochstadt theorem for legendre operators Theorem 2.1. Consider the problem 1 Lu = −u 00 + q(x) − u = λu, 4 sin2 x u(0) = 0,
0<x ) , (2.22) W (λ) where x< = min(x, y), y> = max(x, y). When the asymptotic formulas of above solution are considered, both the numerator and denominator of G(x, y) are entire functions of λ of 1/2. Then, for large λ, G(x, y) is bounded. Let Cn be a sequence of circles about the origin which intersect the positive λ-axis between λn and λn+1 . Hence, Z G(x, y) lim dµ = 0. (2.23) n→∞ C λ − µ n G(x, y) =
By using residue integration, it follows that Z n X 1 wk (x< )vk (x> ) G(x, y) dµ = −G(x, y) + . 0 (λ) (λ − λ ) 2π i Cn λ − µ W n k=0
(2.24)
From (2.23) and (2.24), the Mittak–Leffler expansion for G(x, y) as a function of λ is G(x, y) =
∞ X wn (x< )vn (x> ) , 0 (λ) (λ − λ ) W n k=0
(2.25)
where wn and vn are eigenfunctions corresponding to the simple eigenvalues λn . Therefore these functions are linearly dependent, that is rn wn = vn . For x = π, this is obtained as − sin β , β= 6 0 w n (π − ε) rn = 1 , β = 0, wn (π − ε)
(2.26)
(2.27)
H. Koyunbakan, E.S. Panakhov / Mathematical and Computer Modelling 47 (2008) 178–185
183
so that (λ − L)
−1
π
Z f =
G(x, y) f (y)dy =
0
X rn wn (x) Λ
Rπ
0 wn (y) f (y)dy . 0 W (λn ) (λ − λn )
(2.28)
Comparing (2.16), (2.17) and (2.28), we see that W 0 (λn ) . rn
kwn k2 =
Seperately, from (2.28) we obtain Rx Rπ v(x) 0 w(y) f (y)dy + w(x) x v(y) f (y)dy −1 . (λ − L) f = W (λn )
(2.29)
Our next claim is T (λ − L)
−1
f =
X v˜n (x) Λ
Rx 0
Rπ wn (y) f (y)dy + w˜ n (x) x vn (y) f (y)dy , W 0 (λn )(λ − λn )
(2.30)
where w˜ n (x) and v˜n (x) are solutions of (2.12) and (2.19), respectively. To prove this, we have to show that the righthand sides of (2.30) and (2.100 ) coincide. Now we deal with vn (x) = rn wn (x)
(2.31)
v˜n (x) = r˜n w˜n (x).
(2.310 )
and
Hence, the right-hand side of (2.30) is R R X r˜n w˜n (x) x wn (y) f (y)dy + rn w˜ n (x) π wn (y) f (y)dy 0 x . 0 (λ )(λ − λ ) W n n Λ If rn = r˜n , then (2.32) is reduced to R R X rn w˜ n (x) π wn (y) f (y)dy X w˜ n (x) π wn (y) f (y)dy 0 0 = . 0 (λ )(λ − λ ) W kw k2 (λ − λn ) n n n Λ Λ
(2.32)
(2.33)
Then, we have to show that rn = r˜n . Because λn are eigenvalues of the problem (2.1)–(2.3) and satisfy (2.13), it follows that wn (π − ε, λn ) cos β + wn0 (π − ε, λn ) sin β = 0, wn (π − ε, λn ) cos γ + wn0 (π − ε, λn ) sin γ = χ (λn ). When this is solved for wn (π − ε), we find that wn (π − ε) =
sin βχ (λn ) , sin (β − γ )
(2.34)
and, by using (2.27), we obtain rn =
sin (γ − β) . χ (λn )
(2.35)
In the same way, r˜n =
sin (γ − β) . χ (λn )
(2.350 )
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H. Koyunbakan, E.S. Panakhov / Mathematical and Computer Modelling 47 (2008) 178–185
Asymptotic formulas for wn and w˜n show that χ (λ) and χ˜ (λ) have the same asymtotic forms. Then we have rn = r˜n for n ∈ N . Therefore the statement (2.30) holds. From the formulas (2.18) and (2.30), we get X v˜n (x) R x wn (y) f (y)dy + w˜ n (x) R π vn (y) f (y)dy 0 x ˜ λ−L = T f. (2.36) 0 (λ )(λ − λ ) W n n Λ We calculate g(x) =
v(x) ˜
Rx 0
w(y) f (y)dy + w(x) ˜ W (λn )
Rπ x
v(y) f (y)dy
.
(2.37)
The Mittak–Leffler expansion of g(x) is R R R R X u˜n (x) x wn (y) f (y)dy + z˜ n (x) π vn (y) f (y)dy X v˜n (x) x wn (y) f (y)dy + w˜ n (x) π vn (y) f (y)dy 0 x 0 x + . 0 (λ )(λ − λ ) 0 (λ )(λ − λ ) W W n n n n Λ0 Λ ˜ and z˜ n (x) The second summation is in fact T (λ − L)−1 f , as given in (2.30). In the first term, u˜n (x) represents v(x) represents w(x) ˜ evaluated at λn . Hence, R R −1 X u˜n (x) x wn (y) f (y)dy + z˜ n (x) π vn (y) f (y)dy 0 x ˜ λ−L T f = g(x) − . (2.38) 0 (λ )(λ − λ ) W n n Λ 0
˜ and accordingly so is the right-hand side. Therefore, it is The left-hand side of (2.38) is in the domain (λ − L) continuous and we have the first derivative. Using the formulas (2.37) and differentiation of the right-hand side of (2.38), we obtain Rx Rπ Rx Rπ v˜ 0 (x) 0 w(y) f (y)dy + w˜ 0 (x) x v(y) f (y)dy X u˜ 0n (x) 0 wn (y) f (y)dy + z˜ n0 (x) x vn (y) f (y)dy − W (λ) W 0 (λn )(λ − λn ) Λ0 X u˜n (x)wn (x) − z˜ n (x)vn (x) v(x)w(x) ˜ − w(x)v(x) ˜ f (x). − + W (λ) W 0 (λn )(λ − λn ) Λ 0
An inspection of the term in the second set of braces shows that it vanishes identically. To do that, one merely computes the residue at each λn and observes that it becomes zero. One can differentiate the expression in the braces in the last expression and then we obtain, from (2.38), 0 (x)w(x) − w 0 (x)v(x) X u˜ 0 (x)wn (x) − z˜ 0 (x)vn (x) v ˜ ˜ n n f (x) Tf = − 0 (λ )(λ − λ ) W (λ) W n n Λ0 R R X x u˜ n (x)wn (y) f (y)dy + z˜ n (x) π vn (y) f (y)dy 0 x − . (2.39) 0 (λ ) W n Λ 0
The operator T must be independent of λ. To deduce the value of the expression in the braces in (2.39) we let λ → ∞. Using the asymptotic formulas, we seeRthat the term in the braces must reduce to unity. To simplify the second term π in (2.39) we recall that vn = rn wn and 0 wn (y) f (y)dy = 0. Then, Z x 1X Tf = f − y˜n (x) wn (y) f (y)dy, (2.40) 2 Λ 0 0
where 1 u˜n (x) − rn z˜ n (x) y˜n (x) = . 2 W 0 (λn )
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H. Koyunbakan, E.S. Panakhov / Mathematical and Computer Modelling 47 (2008) 178–185
Now, from (2.11), we conclude that L˜ T f = T L f . If we insert (2.40) in this equality and some straightforward computations, this yields X ( y˜n .wn )0 , q(x) − q(x) ˜ = Λ0
˜ Hence q = q. and, if Λ0 is empty, then T is a unitary operator and L = L. ˜ This completes the proof.
Corollary. We take the hypothesis of Theorem 2.1. Furthermore, suppose that sin β 6= 0 and that the finite index set Λ0 contains only {0}, that is, λn = λ˜ n , n ≥ 0. Then q(x) = q(x). ˜ References ¨ [1] V.A. Ambartsumyan, Uber eine frage der eigenwerttheorie, Z. Phys. 53 (1929) 690–695. [2] G. Borg, Eine umkehrung der Sturm–Liouvillesehen eigenwertaufgabe, Acta Math. 78 (1945) 1–96. [3] V.A. Marchenko, Certain problems of the theory of one dimensional linear differential operators of the second order I, Tr. Mosk. Mat. Obs. 1 (1952) 327–340. [4] B.M. Levitan, M.G. Gasimov, Determination of a differential equation from two spectra, Uspekhi Mat. Nauk 19, 2 (1964). [5] H. Hochstadt, The inverse Sturm–Liouville problem, Comm. Pure Appl. Math. XXVI (1973) 715–729. [6] B.M. Levitan, On the determination of the Sturm–Liouville operator from one and two spectra, Math. USSR Izvestija 12 (1978) 179–193. [7] M.M. Crum, Associated Sturm–Liouville Systems, Quar. J. Math. 6, 2 (1955) 121–128. [8] R. Carlson, A Borg–Levinson theorem for Bessel operator, Pacific J. Math. 177 (1) (1997) 1–26. [9] H. Koyunbakan, E.S. Panakhov, Transformation operator for singular Sturm–Liouville equations, Int. J. Pure Appl. Math. 14 (2) (2003) 135–143. [10] W Rundell, P.E. Sacks, Reconstruction of a radially symmetric potential from two spectral sequences, J. Math. Anal. Appl. 264 (2001) 354–381. [11] R.P. Agarwal, D. O’ Regan, Infinite Interval Problems for Differential, Difference and Integral Equations, Kluwer Academic, 2001. [12] R.P. Gilbert, A method of ascent for solving boundary value problems, Bull. Amer. Math. Soc. 75 (1969) 1286–1289. [13] J. P¨oschel, E. Trubowitz, Inverse Spectral Theory, Academic Press, Orlando, 1987. [14] J.C. Guillot, J.V. Raltson, Inverse spectral theory for a singular Sturm–Liouville operator on [0, 1], J. Differential Equations 76 (1988) 353–373. [15] H. Hochstadt, The Functions of Mathematical Physics, Wiley-Intersicience, New York, 1971. [16] H. Koyunbakan, Spectral properties of singular Sturm–Liouville problems, Ph.D. Thesis, Elazig, 2002. [17] J.A. Levin, Disribution of Zeros of Entire Function, AMS, 1964.
Inverse problem for a singular differential operator H. Koyunbakan ∗ , E.S. Panakhov Department of Mathematics, Firat University, 23119 Elazig, Turkey Received 5 September 2006; accepted 7 February 2007
Abstract In this paper, we give the solution of the inverse Sturm–Liouville problem on two partially coinciding spectra. In particular, in this case we obtain Hochstadt’s theorem concerning the structure of the difference q(x) − q(x) ˜ for the singular Sturm Liouville problem defined on the finite interval (0, π) having the singularity type 1 2 at the points 0 and π. 4 sin x c 2007 Elsevier Ltd. All rights reserved.
Keywords: Sturm–Liouville operator; Spectrum; Green function
1. Introduction The inverse problem of spectral analysis implies the reconstruction of a linear operator from some or other of its spectral characteristics. Such characteristics are spectra (for different boundary conditions), normalizing constants, spectral functions, nodal points, scattering data, etc. The starting point of the inverse problems seems to be a paper by Ambartsumyan [1]. In 1946, Borg [2] proved that two spectra uniquely determined the potential q(x). Marchenko [3] proved that two various spectra of the one singular Sturm–Liouville equation determine this equation uniquely. Later, Levitan and Gasimov [4] gave the effective method of construction for the singular Sturm–Liouville equation on two spectra. In the case of the regular equation, this problem was solved in [5,6] etc. The inverse problem for potentials having a peculiarity in both side points of the interval (0, π) was solved in [7]. We note that the inverse problem for the singular Sturm–Liouville equation is investigated in [8–10] etc. In later years, the problems of the regular and singular differential equations are solved by [11,12,10,13,14]. When we solve the inverse problem from the spectral data, the obvious question occurs: what if only partial spectral values are given? This was answered by Hochstadt [5], who showed that if a complete spectrum was prescribed but the second spectrum was missing information from a certain index set Λ, then q(x) could only be recovered modulo a sum over that index set of eigenfunctions of a q(x). In this paper, by using the Hochstadt method, we give the the structure concerning the difference q(x) − q(x) ˜ for the differential operators having the singularity type 12 . sin x
∗ Corresponding author.
E-mail addresses: [email protected] (H. Koyunbakan), [email protected] (E.S. Panakhov). c 2007 Elsevier Ltd. All rights reserved. 0895-7177/$ - see front matter doi:10.1016/j.mcm.2007.02.013
H. Koyunbakan, E.S. Panakhov / Mathematical and Computer Modelling 47 (2008) 178–185
179
We consider an operator d2 1 L = − 2 + q(x) − dx 4 sin2 x in which q(x) is a summable real function. Let λ0 < λ1 · · · be the spectrum of the problem L [u] = λu,
(1.1)
u(0) = 0, u(π − ε, λ) cos β + u 0 (π − ε, λ) sin β = 0
(1.2) (1.3)
where 0 < ε < x ≤ π − ε < π [15]. It is well known that the sequence {λn } satisfies the asymptotic [16] p 1 a 1 λn = n + + + O 4 n n2 and it is known that the solution of the problem (1.1)–(1.3) can be expressed as Z x u(x, λ) = Pλ (x) + K (x, t)Pλ (t)dt, 0
where r Pλ (x) ∼ =
2 cos πn
1 π n+ x− 2 4
is the solution of the equation Lu = λu when q(x) = 0 and K (x, t) is the solution of the problem ∂ 2 K (x, t) 1 ∂ 2 K (x, t) 1 − q(x) K t) = K (x, t) , + + (x, ∂x2 ∂t 2 4 sin2 x 4 sin2 t dK (x, x) = q(x). K (x, 0) = 0, 2 dx This problem can be solved by using the Riemann method [9]. 2. A Hochstadt theorem for legendre operators Theorem 2.1. Consider the problem 1 Lu = −u 00 + q(x) − u = λu, 4 sin2 x u(0) = 0,
0<x ) , (2.22) W (λ) where x< = min(x, y), y> = max(x, y). When the asymptotic formulas of above solution are considered, both the numerator and denominator of G(x, y) are entire functions of λ of 1/2. Then, for large λ, G(x, y) is bounded. Let Cn be a sequence of circles about the origin which intersect the positive λ-axis between λn and λn+1 . Hence, Z G(x, y) lim dµ = 0. (2.23) n→∞ C λ − µ n G(x, y) =
By using residue integration, it follows that Z n X 1 wk (x< )vk (x> ) G(x, y) dµ = −G(x, y) + . 0 (λ) (λ − λ ) 2π i Cn λ − µ W n k=0
(2.24)
From (2.23) and (2.24), the Mittak–Leffler expansion for G(x, y) as a function of λ is G(x, y) =
∞ X wn (x< )vn (x> ) , 0 (λ) (λ − λ ) W n k=0
(2.25)
where wn and vn are eigenfunctions corresponding to the simple eigenvalues λn . Therefore these functions are linearly dependent, that is rn wn = vn . For x = π, this is obtained as − sin β , β= 6 0 w n (π − ε) rn = 1 , β = 0, wn (π − ε)
(2.26)
(2.27)
H. Koyunbakan, E.S. Panakhov / Mathematical and Computer Modelling 47 (2008) 178–185
183
so that (λ − L)
−1
π
Z f =
G(x, y) f (y)dy =
0
X rn wn (x) Λ
Rπ
0 wn (y) f (y)dy . 0 W (λn ) (λ − λn )
(2.28)
Comparing (2.16), (2.17) and (2.28), we see that W 0 (λn ) . rn
kwn k2 =
Seperately, from (2.28) we obtain Rx Rπ v(x) 0 w(y) f (y)dy + w(x) x v(y) f (y)dy −1 . (λ − L) f = W (λn )
(2.29)
Our next claim is T (λ − L)
−1
f =
X v˜n (x) Λ
Rx 0
Rπ wn (y) f (y)dy + w˜ n (x) x vn (y) f (y)dy , W 0 (λn )(λ − λn )
(2.30)
where w˜ n (x) and v˜n (x) are solutions of (2.12) and (2.19), respectively. To prove this, we have to show that the righthand sides of (2.30) and (2.100 ) coincide. Now we deal with vn (x) = rn wn (x)
(2.31)
v˜n (x) = r˜n w˜n (x).
(2.310 )
and
Hence, the right-hand side of (2.30) is R R X r˜n w˜n (x) x wn (y) f (y)dy + rn w˜ n (x) π wn (y) f (y)dy 0 x . 0 (λ )(λ − λ ) W n n Λ If rn = r˜n , then (2.32) is reduced to R R X rn w˜ n (x) π wn (y) f (y)dy X w˜ n (x) π wn (y) f (y)dy 0 0 = . 0 (λ )(λ − λ ) W kw k2 (λ − λn ) n n n Λ Λ
(2.32)
(2.33)
Then, we have to show that rn = r˜n . Because λn are eigenvalues of the problem (2.1)–(2.3) and satisfy (2.13), it follows that wn (π − ε, λn ) cos β + wn0 (π − ε, λn ) sin β = 0, wn (π − ε, λn ) cos γ + wn0 (π − ε, λn ) sin γ = χ (λn ). When this is solved for wn (π − ε), we find that wn (π − ε) =
sin βχ (λn ) , sin (β − γ )
(2.34)
and, by using (2.27), we obtain rn =
sin (γ − β) . χ (λn )
(2.35)
In the same way, r˜n =
sin (γ − β) . χ (λn )
(2.350 )
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H. Koyunbakan, E.S. Panakhov / Mathematical and Computer Modelling 47 (2008) 178–185
Asymptotic formulas for wn and w˜n show that χ (λ) and χ˜ (λ) have the same asymtotic forms. Then we have rn = r˜n for n ∈ N . Therefore the statement (2.30) holds. From the formulas (2.18) and (2.30), we get X v˜n (x) R x wn (y) f (y)dy + w˜ n (x) R π vn (y) f (y)dy 0 x ˜ λ−L = T f. (2.36) 0 (λ )(λ − λ ) W n n Λ We calculate g(x) =
v(x) ˜
Rx 0
w(y) f (y)dy + w(x) ˜ W (λn )
Rπ x
v(y) f (y)dy
.
(2.37)
The Mittak–Leffler expansion of g(x) is R R R R X u˜n (x) x wn (y) f (y)dy + z˜ n (x) π vn (y) f (y)dy X v˜n (x) x wn (y) f (y)dy + w˜ n (x) π vn (y) f (y)dy 0 x 0 x + . 0 (λ )(λ − λ ) 0 (λ )(λ − λ ) W W n n n n Λ0 Λ ˜ and z˜ n (x) The second summation is in fact T (λ − L)−1 f , as given in (2.30). In the first term, u˜n (x) represents v(x) represents w(x) ˜ evaluated at λn . Hence, R R −1 X u˜n (x) x wn (y) f (y)dy + z˜ n (x) π vn (y) f (y)dy 0 x ˜ λ−L T f = g(x) − . (2.38) 0 (λ )(λ − λ ) W n n Λ 0
˜ and accordingly so is the right-hand side. Therefore, it is The left-hand side of (2.38) is in the domain (λ − L) continuous and we have the first derivative. Using the formulas (2.37) and differentiation of the right-hand side of (2.38), we obtain Rx Rπ Rx Rπ v˜ 0 (x) 0 w(y) f (y)dy + w˜ 0 (x) x v(y) f (y)dy X u˜ 0n (x) 0 wn (y) f (y)dy + z˜ n0 (x) x vn (y) f (y)dy − W (λ) W 0 (λn )(λ − λn ) Λ0 X u˜n (x)wn (x) − z˜ n (x)vn (x) v(x)w(x) ˜ − w(x)v(x) ˜ f (x). − + W (λ) W 0 (λn )(λ − λn ) Λ 0
An inspection of the term in the second set of braces shows that it vanishes identically. To do that, one merely computes the residue at each λn and observes that it becomes zero. One can differentiate the expression in the braces in the last expression and then we obtain, from (2.38), 0 (x)w(x) − w 0 (x)v(x) X u˜ 0 (x)wn (x) − z˜ 0 (x)vn (x) v ˜ ˜ n n f (x) Tf = − 0 (λ )(λ − λ ) W (λ) W n n Λ0 R R X x u˜ n (x)wn (y) f (y)dy + z˜ n (x) π vn (y) f (y)dy 0 x − . (2.39) 0 (λ ) W n Λ 0
The operator T must be independent of λ. To deduce the value of the expression in the braces in (2.39) we let λ → ∞. Using the asymptotic formulas, we seeRthat the term in the braces must reduce to unity. To simplify the second term π in (2.39) we recall that vn = rn wn and 0 wn (y) f (y)dy = 0. Then, Z x 1X Tf = f − y˜n (x) wn (y) f (y)dy, (2.40) 2 Λ 0 0
where 1 u˜n (x) − rn z˜ n (x) y˜n (x) = . 2 W 0 (λn )
185
H. Koyunbakan, E.S. Panakhov / Mathematical and Computer Modelling 47 (2008) 178–185
Now, from (2.11), we conclude that L˜ T f = T L f . If we insert (2.40) in this equality and some straightforward computations, this yields X ( y˜n .wn )0 , q(x) − q(x) ˜ = Λ0
˜ Hence q = q. and, if Λ0 is empty, then T is a unitary operator and L = L. ˜ This completes the proof.
Corollary. We take the hypothesis of Theorem 2.1. Furthermore, suppose that sin β 6= 0 and that the finite index set Λ0 contains only {0}, that is, λn = λ˜ n , n ≥ 0. Then q(x) = q(x). ˜ References ¨ [1] V.A. Ambartsumyan, Uber eine frage der eigenwerttheorie, Z. Phys. 53 (1929) 690–695. [2] G. Borg, Eine umkehrung der Sturm–Liouvillesehen eigenwertaufgabe, Acta Math. 78 (1945) 1–96. [3] V.A. Marchenko, Certain problems of the theory of one dimensional linear differential operators of the second order I, Tr. Mosk. Mat. Obs. 1 (1952) 327–340. [4] B.M. Levitan, M.G. Gasimov, Determination of a differential equation from two spectra, Uspekhi Mat. Nauk 19, 2 (1964). [5] H. Hochstadt, The inverse Sturm–Liouville problem, Comm. Pure Appl. Math. XXVI (1973) 715–729. [6] B.M. Levitan, On the determination of the Sturm–Liouville operator from one and two spectra, Math. USSR Izvestija 12 (1978) 179–193. [7] M.M. Crum, Associated Sturm–Liouville Systems, Quar. J. Math. 6, 2 (1955) 121–128. [8] R. Carlson, A Borg–Levinson theorem for Bessel operator, Pacific J. Math. 177 (1) (1997) 1–26. [9] H. Koyunbakan, E.S. Panakhov, Transformation operator for singular Sturm–Liouville equations, Int. J. Pure Appl. Math. 14 (2) (2003) 135–143. [10] W Rundell, P.E. Sacks, Reconstruction of a radially symmetric potential from two spectral sequences, J. Math. Anal. Appl. 264 (2001) 354–381. [11] R.P. Agarwal, D. O’ Regan, Infinite Interval Problems for Differential, Difference and Integral Equations, Kluwer Academic, 2001. [12] R.P. Gilbert, A method of ascent for solving boundary value problems, Bull. Amer. Math. Soc. 75 (1969) 1286–1289. [13] J. P¨oschel, E. Trubowitz, Inverse Spectral Theory, Academic Press, Orlando, 1987. [14] J.C. Guillot, J.V. Raltson, Inverse spectral theory for a singular Sturm–Liouville operator on [0, 1], J. Differential Equations 76 (1988) 353–373. [15] H. Hochstadt, The Functions of Mathematical Physics, Wiley-Intersicience, New York, 1971. [16] H. Koyunbakan, Spectral properties of singular Sturm–Liouville problems, Ph.D. Thesis, Elazig, 2002. [17] J.A. Levin, Disribution of Zeros of Entire Function, AMS, 1964.
