00 Boundary Value Problems
a SpringerOpen Journal
RESEARCH
Open Access
Inverse eigenvalue problems for a discontinuous Sturm-Liouville operator with two discontinuities
Yalçin Güldü*
"Correspondence: yguldu@cumhuriyet.edu.tr Department of Mathematics, Faculty of Sciences, Cumhuriyet University Sivas, 58140, Turkey
Abstract
In this paper, we consider a discontinuous Sturm-Liouville operator with parameter-dependent boundary conditions and two interior discontinuities. We obtain eigenvalues and eigenfunctions together with their asymptotic approximate formulas. Then, we give some uniqueness theorems by using Weyl function and spectral data, which are called eigenvalues and normalizing constants for solution of inverse problem.
MSC: Primary 34A55; secondary 34B24; 34L05
Keywords: Sturm-Liouville problem; eigenvalues; eigenfunctions; transmission conditions; Weyl function
1 Introduction
It is well known that the theory of Sturm-Liouville problems is one of the most actual and extensively developing fields of theoretical and applied mathematics, since it is an important tool in solving many problems in mathematical physics (see [1-4]). In recent years, there has been increasing interest in spectral analysis of discontinuous Sturm-Liouville problems with eigenvalue-linearly and nonlinearly dependent boundary conditions [1, 5-12]. Various physics applications of such problems can be found in [1, 3, 4, 13-19] and corresponding bibliography cited therein.
Some boundary value problems with discontinuity conditions arise in heat and mass transfer problems, mechanics, electronics, geophysics and other natural sciences (see [3] also [20-29]). For instance, discontinuous inverse problems appear in electronics for building parameters of heterogeneous electronic lines with attractive technical characteristics [20, 30, 31]. Such discontinuity problems also appear in geophysical forms for oscillations of the earth [32, 33]. Furthermore, discontinuous inverse problems appear in mathematics for exploring spectral properties of some classes of differential and integral operators.
Inverse problems of spectral analysis form recovering operators by their spectral data. The inverse problem for the classical Sturm-Liouville operator was studied first by Ambar-sumian in 1929 [34] and then by Borg in 1945 [35]. After that, direct and inverse problems for Sturm-Liouville operator have been extended to so many different areas.
Springer
© 2013 Guldu; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We consider a discontinuous Sturm-Liouville problem L with function p (x)
ly := -p(x)y"(x) + q(x)y(x) = Xy(x), x e [a, 5i) U (5i, S2) U (S2, b] = (1)
P (x) =
^ 2 a < x < 51,
p22, 51 < x < S2, p-2, 52 < x < b,
and p1, p2, and p3 are given positive real numbers; q(x) e L2[^, R]; X e C is a complex spectral parameter; boundary conditions at the endpoints
ky := X(6[y(a) - d^y'(a)) - (0iy(a) - d2y'(a)) = 0,
hy := M^i'y(^) - Y2y'(b)) + (yiy(b) - y^y'(b)) = 0
with discontinuity conditions at two points x = 51) x = 52
fey := y(5i + 0) - 03y(Si - 0) - y3y'(5i - 0) = 0, hy := y'(5i + 0) - 04y№ - 0) - y4y'(<$i - 0) = 0, ky := y(&2 + 0) - d5y(&2 - 0) - y5y'(S2 - 0) = 0, ley := y'(52 + 0) - 06y(S2 - 0) - ye/(«2 - 0) = 0,
where 0i, yi and 0-, yj (i = i, 6, j = i, 2) are real numbers and
(2) (3)
(6) (7)
03 Y3 >0, a2 = 05 Y5 >0, Pi = 0i 0i > 0 and
04 Y4 06 Y6 02 02
In the present paper, we construct a linear operator T in a suitable Hilbert space such that problem (i)-(7) and the eigenvalue problem for operator T coincide. We investigate eigenvalues and eigenfunctions together with their asymptotic behaviors of operator T. Besides, we study some uniqueness theorems according to Weyl function and spectral data, which are called eigenvalues and normalizing constants.
2 Operator formulation and spectral properties
We make known the inner product in the Hilbert space H := Hi ® C2, where Hi = (L2(^), (•, •>i), C denotes the Hilbert space of complex numbers and a self-adjoint operator T defined on H such that (i)-(7) can be dealt with as the eigenvalue problem of operator T. We define an inner product in H by
P&i P&2 fb
(F, G> := aia2p^ I f (x)g(x) dx + a2p^ I f (x)g(x) dx + p" / f (x)g(x) dx
J a J ¿>i J
aia2 _ i _
+ i + 7~f2,g2
(f (x)\ A \f2
e H, G =
(g(x)\ g1 \g2 /
Consider the operator T defined by the domain
D(T) = {F e H: f (x),f'(x) e ACi«(n), If e Hi, /3 = /4 = /5 = l6 = O,f = Of (a) - Of(a) /2 = kj!/ - y-if '(b))
such that TF := (//, Oj/ (a) - fa), -(yf (b) - y/'(b))) for F = (/, Of (a) - Of (a), X];/(b) -y/'(b)) e D(T) and also /3-/g are satisfied for/. Thus, we can rewrite the considered problem (l)-(7) in the operator form as TF = XF.
Theorem 1 The operator T is symmetric in H.
Proof Let F, G e D(T). By two partial integrations, we get
(TF, G) = (F, TG) + a^W f,g, Si - O) - W f,g, a))
+ a-(W(/,g, S2 - O) - Wf,g, Si + O)) + W(/,g, b) - Wf,g, S2 + O)
«1«2 7T
(0!f (a)-0f '(a)) 0g (a)-02 ¿'(a))
- 7T (Ktf (b) - Yf'(b)) (y/g(b) - Y2'S?(b))
«1«2 7T
(0^ (a)-02g'(a)) (0f (a)-0f '(a))
+ 7T (Y^(b) - Y2gg'(b)) {yf (b) - yf'(b)),
where by W f, g; x), we denote the Wronskian of the functions f and g as f (x)g'(x) -f(x)g(x).
Since f and g satisfy the boundary conditions (2)-(3) and transmission conditions (4)-(7), we obtain
«1«2
[{01f (a) - 02f'(a))(0{g(a) - 02g») - (a) - 02g»){0f (a) - 02f (a))]
= aa W(f, g, a)
x [(y^(b) - Y2g'(b))(yif(b) - y^'(b)) - (ylf(b) - yf'(b))(Wg(b) - y2g'(b))] P2
= - W (f, g, b) W (f, g, + 0) = a,W f, g, 5, - 0).
Thus, we have {TF, G) = (F, TG), i.e., T is symmetric.
Lemma 1 Problem (l)-(7) can be considered as the eigenvalue problem of the symmetric operator T.
Corollary 1 All eigenvalues and eigenfunctions of problem (l)-(7) are real, and two eigen-functions y (x, A4) and y (x, A2), corresponding to different eigenvalues Al and A2, are orthogonal in the sense of
Çäl räi
ala2p'2 I y(x,Al)^(x, A2) dx + a2p| I y(x, Al)^(x, A2) dx
Ja J äl
+ P2 I y(x, Al)y(x, A2) dx
+ —— (û[y(a, Ai)-e2 y (a, Al)) (û[y(a, Ai)-9ii y'(a, A2)) Pi
+ -1 (yMb, Ai) - KiV(b, Ai))(yiV(b, Ai) - Yi1 <p'(b, Ai)) = 0.
We define the solutions
y(x, A) =
yl(x, A), x e [a, äl), yi(x, A), x e (äl, äi), y3(x, A), x e (äi,b],
ty (x, A) =
tyl(x, A), x e [a, äl), tyi(x, A), x e (äl, äi), ty'(x, A), x e (äi, b]
of equation (l) by the initial conditions
yl(a, A)= AOi - 02, yl(a, A) = AOl - 0l,
yi(äl, A) = 03yl(äl, A) + Y3yl (äl, A), yi (äl, A) = 04yl(äl, A) + y4yl (äl, A), (9)
y3(äi, A) =05yi(äi, A) + y5yi (äi, A), y3 (äi, A) = 06yi(äi, A) + y6yi (äi, A),
and similarly,
f3(b, A) = AyI + Yi, ty' (b, A) = Ayl + yl, Y6ty'(äi, A) - Y5^3(äi, A)
tyi(äi,A) = tyl(äl,A) =
y4tyi(äl, A) - Y'tyi(äl, A) al ,
06f'(äi, A)-05f3 (äi, A) tyi (äi, A) =---, (l0)
tyl (äl, A) =
04tyi(äl, A) - P'tyi(äl, A) -al ,
respectively.
These solutions are entire functions of A for each fixed x e [a, b] and satisfy the relation ^ (x, An) = Kny(x, An)
for each eigenvalue An, where
ty'(a, An) - Oty(a, An) Kn = ß '
Lemma 2 Let X = k2, k = a + ia>. Then the following integral equations and also asymptotic behaviors hold for v = 0,1:
dv , x t , x dv —<m*, X) = (X02 - O2) — coskpi(x - a)
1 t , x dv
+ kP' X0i - dxv sin kPi (x - a)
+ — I —— sin kp1(x - t)q(t)^1(t, X) dt k J a dxv
= (X02 - 02) —— cos kp1(x - a) v ' dxv
+ — (X01 - 00 sinkp1(x - a) + O(|k|v+1elImk|(x-a)p1), kpi dxv
dv dv dxv^2(x, X) = (03<M<$1, X) + (51,X)) — coskp2(x - 51)
+ kp^4^1(51, X) + (51, X«dxv sinkP2(x - 51)
+ p2 i sinkp2(x - t)q(t)<p2(t, X) dt k j 51 dxv
= - (X02 - 02)Y3kp1 sinkp1(51 - a)—j~v coskp2(x - 51)
+ O( |k| v+2el Imk|((51-a)p1+(x-51)p2^,
dv dv dxv^3(x, X) = (05^2, X) + Y5^2 (52, X)) — cos kp3(x - 52)
+ -.-(05 ^2(52, X) + Y6^2 (52, X))— sin kp3(x - 52)
kp3 dxv
p^ /• x dv
+ — I ——sinkp3(x - t)q(t)y3(t, X) dt k t/52 dxv
= (X02 - 02) Y3Y5k2pp sin kp1(51 - a) sin kpsfe - 51) — cos kp3(x - 52)
O( |k|
v+3el Imk|((51 -a)p1+(52-51 )p2+(x-52)p3)x
Lemma 3 Let X = k2, k = a + ia>. Then the following integral equations and also asymptotic behaviors hold for v = 0,1:
dv , , t , x dv , , ,,
^3(x, X) =XY2 + Y2) dxv cos kp3(x - b)
1 t , x dv , , „
+ kp3 (XY1 +n) dxsinkp3(x - b)
p3 f b dv
—— / ——sin kp3(x - t)q(t)^3(t, X) dt k x dx
= (Xy2 + Y2) dxv cos kp3(x - b)
+ 1— (XYi' + Y1) sinkp3(x - b) + O(|k|v+1e|Imk|(x-b)p3), kp3 1 dx
dv . , .. Y6^3(52,(52,A) dv
— fa (x, A) = -— cos kp2 (x - 52)
dxv a2 dxv
1 06^3(52,A)-05^3(52,A) dv
sin kp2(x - 52)
kp2 -a2 dx
p2 i sinkp2(x - t)q(t)^2(t, A) dt k Jx dxv
aY2 + ' kp3Y5 sinkp3(52 - b)coskp2(x - 52)
a2 /\ dxv
+ 0( |k| v+2glImk|((52-b)p3+(x-52)p2^,
d' .. Y4^2(5i,A)-Y3^2(5i,A) dv
— fa(x, A) =-— cos kpi(x - 5i)
dxv a1 dxv
1 04^2(5i,A)-03^2(5i,A) dv ;
sin kpi(x - 5i)
kpi -ai dx
pi f 5i dv
— I —— sin kpi (x - t)q(t) fa (t, A) dt k Jx dxv
ay2 + Y2
* (k2P2P3Y3Y5 sinkp3(52 - b) sinkp2(5i - 52)d-v cos kpi(x - 5i)
v+3g| Imk|((52-b)p3+(5i-52)P2+(x-5i)pi)\
+ O( |k|
The function A (A) is called the characteristic function, and numbers |^«}«>i are called the normalizing constants of problem (i)-(7) such that
A (A) = A(Yi'y(b, A) - y2 q>'(b, A)) + {ywb A) - Y2' v'(b, A)), (ii)
/»5i /»52 pb
¡in := aia2pi I ty2(x,An)dx + a2p2 I ty2(x, An)dx + p3 / ^2(x, An)dx
J a J 5i J 52
aa 2 i 2
+ —(0i^(a, An) - 02 v'(a, An)) + — (YiV(b, A„)-Y2v'(b, An)). (i2)
Lemma 4 The following equality holds for each eigenvalue An
aia2A (An) = Knin. Proof Since
-p(x)^ "(x, A) + q(x)^ (x, A) = A'ft (x, A), -p(x)q>"(x, An) + q(x)y(x, An) = Ay(x, An), we get
y'(x,An)^(x, A) - f '(x, A)y(x, An)(|* + |52 + |b2) r 5i
= (A - An)p2 I ^ (x, A)y(x, An) dx
52 b + (A - An)p2 I ^ (x, A)y(x, An) dx + (A - An)p| I ^ (x, A)y(x, An) dx.
J 5i J 52
After that, add and subtract A(X) on the left-hand side of the last equality, and by using conditions (2)-(7), we obtain
A(X) + (X - Xn)(0ty(a,X) - $2ty'(a,X)) -(X - Xn)(y{q>(b,Xn) - y21q>'(b,Xn)) + (1 -aO(ty(5-,X)y'(5-,Xn) - ^(5-,Xn) ty '(5-,X)) + (1 -a2)(ty(5-,X)q)'(5-,X^ - ^(52,X^ty'(52,X))
= (X - Xn)p2 I ty(x, X)q>(x, Xn) dx + (X - Xn)p22 / ty(x, X)q>(x, Xn) dx
J a J 5i
+ (X - Xn)p2 / ty(x, X)^(x, Xn) dx,
aia2A(X) 2 f51 , , , w 2 f52 , ^ , -= aia2p1 I ty(x, X)^(x, Xn) dx + a2p2 I ty(x, X)^(x, Xn) dx
X - Xn J a J 51
+ p"2 I ty(x, X)^(x, Xn) dx
+ a1a2 ($1ty (a, X) -0'2 ty '(a, X))(0^(a, Xn) -02 q>'(a, Xn)) p1
+ IT (^Mb, Xn) - y2<p'(b, Xn))(yity(b, X) - y2ty'(b, X)). P2
For X ^ Xn, a1a2AA(Xn) = Knjin is obtained by using the equality ty (x, Xn) = Knp(x, Xn)
and (12). □
Corollary 2 The eigenvalues of problem L are simple.
Lemma 5 [36] Let {ai}p,=1 be the set of real numbers satisfying the inequalities a0 > a0 > ••• > ap-1 > 0, and let {ai}p=1 be the set of complex numbers. Ifap = 0, then the roots of the equation
eaoX + a1ea1X + ••• + ap-ieaoX + ap = 0
have the form 2n ni
Xn =-+ ^ (n) (n = 0,±1,...),
where ^ (n) is a bounded sequence. Now, from Lemma 2 and (11), we can write
A(X) - A0(X) = O(k6eImkl((5l2a)pl+(5225l)p2+(b252)pз^, where A0(X) = k702y2'y3y5P1P2P3 sinkp^ - a) sinkp2(52 - 51) sinkp3(b - 52).
We can see that non-zero roots, namely A°n of the equation A0(A) = 0, are real and analytically simple. Furthermore, it can be proved by using Lemma 5 that
A/A0 = 7-:-:-:-T-,-;— + *n, sup |%| < TO. (i3)
V (5i- a)pi + (52- 5i)p2 + (b - 52)p3 n
Theorem 2 The eigenvalues {An}n>0 have the following asymptotic behavior for sufficiently large n:
:=VAL4+o(i). (i4)
Proof Denote
Gn := {A : k2 = A, |k|< |*0|+ 5}, where kn = ^/A0 and 5 is a sufficiently small number. The relations |Ao(A)| > C5 |k|7el Jmk|((5i-a)pi+(52-5i)p2+(b-52)p3)
A (A) - A0(A) = O (k6 e | k|((5i-a)pi+(52-5i)p2+(&-52)p3)^ are valid for A e d Gn.
Then, by Rouche's theorem that the number of zeros of An(A) coincides with the number of zeros of A (A) in Gn, namely n + 4 zeros, An, Ai, A2,..., An+3. In the annulus, between Gn and Gn+i, A (A) has accurately one positive zero, namely k1: kn = yA + 5n, for n > i. So, it follows that An+4 = Applying to Rouche's theorem in qe = {k: |k - k° | < e} for sufficiently small e and sufficiently large n, we get 5n = o(i). Finally, we obtain the asymptotic formula
= yAn-4+o(i).
Denote
Ai(A) := W(yh fi, x) := Pifi - x e ^ (i = i, 3),
which are independent of x e and are entire functions such that ^ = [a, 5i), = (5i, 52), ^3 = (52, b]. It can be easily seen that
A (A) := A3 (A) = a2 A2(A) = aia2 Ai(A). □
Example Let q = 0, a = 0, b = n, 5i = 4, 52 = f, 03 = y4 = 0, y3 = i, 04 = -i, 05 = Y6 = 0, Y5 = i, 06 = -i, y2 = i, Y6 = -i, Yi = Y2 = 0, 02 = i 0i = -i, 0{ = 02 = 0.
A(X) = p1p2p3k7 sinkp151 sinkp2(52 - 51) sinkp3(n - 52)
+ O(k6eI !mk|(51P1+(52-51)p2+(n-52)p3^,
the eigenvalues of the boundary value problem (1)-(7) satisfy the following asymptotic formulae:
4(n -4) 4(n -4)
n1 = kn1 = + &n, V Xn2 = kn2 = + &n,
1 1 p1 2 2 p2
_ 2(n -4)
n3 = kn3 = + &n,
3 3 p3
where en = O(n-1). 3 Inverse problems
In this section, we study the inverse problems for the reconstruction of the boundary value problem (1)-(7) by Weyl function and spectral data.
We consider the boundary value problem L with the same form of L but with different coefficients q(x), 0i, ■pi, 5 j, 0, y j, i = 1 - 6,j = 1,2.
If a certain symbol a denotes an object related to L, then the symbol a denotes the corresponding object related to L.
The Weyl function Let $(x, X) be a solution of equation (1), which satisfies the condition (X01 - 01)$(a, X) - (X02 - 02)$(a, X) = 1 and transmissions (4)-(7).
Assume that the function ft(x, X) is the solution of equation (1) that satisfies the conditions ft(a, X) = Pf^, ft' (a, X) = Pf^! and the transmission conditions (4)-(7).
Since W[ft, y] = 1, the functions ft and y are linearly independent. Therefore, the function ty (x, X) can be represented by
02 ty '(a, X)-01'ty (a, X) „ N „
ty(x,X) = 2Y y '-l^—Ly(x,X) + A(X)x(x,X)
$(x, X) = ^ = x (x, X) + 02 ty '(a, X)-01ty (a,X) y(x, x) (15)
v ' A(X) v ' p1A(X) ^ ' '
that is called the Weyl solution, and
= M(X) (6)
02 ty' (a, X) -01ty (a, X)
P1A(X) is called the Weyl function.
Theorem 3 IfM(X) = M(X), then L = L, i.e., q(x) = q(x), a.e. and 0i = 0i, yi = Y, i = 1-6, 5j = 5, 0/ = 0/, Yj = jj',7 = 1,2.
Proof We introduce a matrix P(x, A) = [Pkj(x, A)]kj=i,2 by the formula
P(x,A)(<k j< $\
, \<p' $'J
<Pii(x, A) Pi2(x, A)\ = /-<$' + $<k' p$ - $<k \ \Pn(x, A) P22(x, A)l l-p'<t' + <k'$' - <k$') ,
where $(x, A) =
Pii(x, A) = x(x, A)<k'(x, A) -p(x, A)x'(x, A) + (M(A) -M(A))p(x, A)<k'(x, A), Pi2(x, A) = <(x, A)x(x, A) - x(x, A)<k(x, A) + (M(A) -M(A))<(x, A)<k(x, A).
Thus, if M(A) = MM (A), then the functions Pii(x, A) and Pi2(x, A) are entire in A for each fixed x.
Denote Gw = {A : A = k2, |k - k51 > w, 5 = i, 2,...} and Gw = {A : A = k2, |k - k51 > w, 5 = i, 2,...}, where w is sufficiently small number, k5 and k5 are square roots of the eigenvalues of the problem L and L, respectively. It is easily shown that
$(v)(x, A) < Cw|k|V-(2+i)e-| I^v/A|((5i-a)pi+(52-5i)p2+(b-52)p3), x e Qi (i = i, 2,3), v = 0, i (i8)
are valid for sufficiently large |A|, where Qi = [a, 5i), Q2 = (5i, 52) and Q3 = (52, b]. Hence, Lemma 2 and (i8) yield that
|Pii(x, A) | < Cw, |Pi2(x, A) | < Cw|k|_i for A e Q and for A e Gw n Gw. (i9)
According to (i9), and Liouville's theorem, Pii(x, A) = C(x) and Pi2(x, A) = 0 for x e [a, b]\{5i, 52, 5^ 52}. Byvirtue of (i7), we get
<(x, A) = C(x)<k (x, A), $(x, A) = C(x)$ (x, A). (20)
It is obvious that
W$ (x, A), < (x, A) ] = $ (a, A) (0i A - 0^ - $' (a, A) (0!1 A - 02) _ f (a, A)(0iA - 0i)-f '(a, A)(02A - 02)
= a(A)
and similarly, W[$(x, A), <k(x, A)] = i. Thus, we have C2(x) = i.
Otherwise, the following asymptotic expressions hold
yi(x, X) = <pi(x, X) = X e-ik(x-a)n (1 + o(l)) for x < 5i and x < 5i,
y2(x, X) = ^(x, X) = -—e-ik((5i-a)pi+(x-5i)p2)(i + 0(1))
for 5i < x < S2 and ^ < x < 52, (2i)
X2 ( )
y3(x, X) = y3(x, X) = — e-ik((Si-a)Pi+(S2-Si)P2+(x-S2p)(i + o(i)) for 52 < x and 52 < x.
Without loss of generality, we assume that Si < 5i and S2 < 52. From (20)-(2i), we get C(x) = i for x e [a, 5i) U (5i, S2) U (52, b] and also
(i + o(i))C(x) = -e(x-5i)p2 (i + o(i)) for x e (5i, 5i), Xi (22)
__ e(x-5i)p2(i + o(i)) C(x) = -e(52-5i)p2+(x-52)p3(i + o(i)) for x e (52,
As |X| —> ^ in (22), we contradict C2(x) = i. Therefore, 5i = 5i, 52 = 52. Thus, y(x,X) = y(x,X), $(x,X) = $(x,X) and JxX = JxX). Hence, from equation (i) and transmission conditions (4)-(7), q(x) = q(x), a.e., 0i = 0i, yi = Y, i = 3-6, and from (ii) and (i2), 0i = 0i,
0! = 0j, yj = y! ,j=i, 2. □
Lemma 6 The following representation holds
M(X) = £
n=0 pn(Xn - X)
Proof Weyl function M(X) is a meromorphic function with respect to X, which has simple poles at Xn. Therefore, we calculate
Re sM(X) =
02 j '(a, Xn) - 0j (a, Xn)
P1A (Xn)
Since Kn = 02ty,(a,XnP20lty(a,Xn) and A(Xn) = ,
Re sM(X) = a1a2. (3)
X=Xn //n
Let rN = {X : X = k2, |k| = *JxN + e}, where e is a sufficiently small number. Consider the contour integral IN(X) = f^ MX d/, X e int rN. For X e Gw,
A(X) > IX17/2 CweI !mkI((51 -a)p1+(52-51 )p2+(b-52)p3) satisfies. Using this equality and (16), we get
|M(X)| < Cw for X e Gw. |X|
Thus, IN(A) = 0. As a result, the residue theorem and (23) yield
«i«2
M(X) = £
n=0 ln(An - A) □
Theorem 4 IfAn = An and in = |nfor all n, then L = L, i.e., q(x) = q(x), a.e., 0i = 0i, yi = Y, i = i - 6, 5j = 5j, 0j = &., yj = j j = i, 2. Hence, problem (i)-(7) is uniquely determined by spectral data {An, in}.
Proof If An = An and in = ln for all n, then M(A) = M(A) by Lemma 6. Therefore, we get L = L by Theorem 3.
Let us consider the boundary value problem Li that we get the condition 02y'(a, A) -0iy(a, A) = 0 instead of condition (2) inL. Let {Tn}n>0 be the eigenvalues of the problem Li. It is clear that xn are zeros of
Ai(t ):=#2 f '(a, t )-6[_f (a, t ). □
Theorem 5 I/Xn = Xn and Tn = tnfor a// n, then L(q, Sj, , yi, y!) = L(q, Sj, , Y, yj), i = 1-6, k = 3-6,j =1,2.
Hence, the problem L is uniquely determined by the sequences {Xn} and {Tn}, except coefficients 0j and 0.
Proof Since the characteristic functions A(X) and A1(t) are entire of order 2, functions A(X) and A1(t) are uniquely determined up to multiplicative constant with their zeros by Hadamard's factorization theorem [37]
A(X) = C 1-f),
n=o ^ Xn'
œ , s
Ai(T) = C^ 1- - ,
n=0 ^ Tn'
where C and Ci are constants dependent on {An} and {rn}, respectively. Therefore, when An = An and Tn = Tn for all n, A (A) = A(A) and Ai(r) = Ai(r). Hence, 02 ty '(a, t ) -0if (a, t) = 02f '(a, t) - 0$(a, t). As a result, we get M(A) = M(A) by (i6). So, the proof is completed by Theorem 3. □
Competing interests
The author declares that he has no competing interests.
Received: 18 June 2013 Accepted: 26 August 2013 Published: 11 September 2013 References
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doi:10.1186/1687-2770-2013-209
Cite this article as: Guldu: Inverse eigenvalue problems for a discontinuous Sturm-Liouville operator with two discontinuities. Boundary Value Problems 2013 2013:209.
