# Determination of differential pencils with spectral parameter dependent boundary conditions from interior spectral dataAcademic research paper on "Mathematics"

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## Academic research paper on topic "Determination of differential pencils with spectral parameter dependent boundary conditions from interior spectral data"

﻿Research Article

Received 2 January 2013 Published online in Wiley Online Library

(wileyonlinelibrary.com) DOI: 10.1002/mma.2844 MOS subject classification: 34A55; 34B24; 35K57; 47E05

Determination of differential pencils with spectral parameter dependent boundary conditions from interior spectral data

Chuan-Fu Yang*+ and Xiu-Juan Yu

Communicated by I. Stratis

Second-order differential pencils L(p,q,h0,hi,H0,Hi) on a finite interval with spectral parameter dependent boundary conditions are considered. We prove the following: (i) a set of values of eigenfunctions at the mid-point of the interval [0, n] and one full spectrum suffice to determine differential pencils L(p, q, h0, h1, H0, H1); and (ii) some information on eigenfunctions at some an internal point b € (y, n) and parts of two spectra suffice to determine differential pencils L(p, q, h0, hi, H0, Hi). Copyright © 2013 The Authors. Mathematical Methods in the Applied Sciences published by John Wiley & Sons, Ltd.

Keywords: differential pencils with spectral parameter dependent boundary conditions; interior spectral data; eigenvalue; inverse problem

1. Introduction

We consider the boundary value problem L = L(p, q, h0, h1, H0, H1) of the form:

lxy(x, A) d=f(x) C [A2 - 2Ap(x) - q(x)]y(x) = 0, 0 < x < n - y'(0) c (iAhi C h0)y(0) = 0 O.'O

y'(n) C (iAHi C H0)y(n) = 0,

where i = sj—1, A is a spectral parameter; p 2 W2 [0,n],q 2 W[0,n] are complex-valued functions; hj, Hj 2 C, j = 0,1, h1 / ±1, H1 /±1.The latter conditions exclude from consideration Redge-type problems [1], which require separate investigation. Differential equations with a nonlinear dependence on the spectral parameter frequently appear in mathematics as well as in applications. Inverse spectral problems consist in recovering operators from given their spectral characteristics [2-11]. Some aspects of spectral problems for second-order differential pencils were studied in [12-24] and other papers.

Inverse problem for interior spectral data of the differential operator lies in reconstructing this operator by some eigenvalues and information on eigenfunctions at some an internal point in the interval considered. The similar problems for the Sturm-Liouville operators [25,26] and differential pencils [27] were studied.

As far as I know, the inverse problem of interior spectral data for differential pencils L with spectral parameter dependent boundary conditions has not been considered before. The aim of this paper is to give two uniqueness theorems from some eigenvalues and information on eigenfunctions at some an internal point in the interval [0, n] provided that the spectrum is simple. The results obtained are new and a generalization of the well-known one for the classical Sturm-Liouville operator, which was studied in [25], for a special case that p(x) = 0, h1 = H1 = 0, and the parameters h and H are fixed. The results in this paper are also a generalization of theorems because of Yang and Guo [27], where authors consider a special case that h1 = H1 = 0 and assume either p(x) or q(x) is a prior known.

Department of Applied Mathematics, Nanjing University of Science and Technology, Nanjing 210094, Jiangsu, China

Correspondence to: Chuan-Fu Yang, Department of Applied Mathematics, Nanjing University of Science and Technology, Nanjing 210094, Jiangsu, China. ^ E-mail: chuanfuyang@mail.njust.edu.cn

This is an open access article under the terms of the Creative Commons Attribution Non-Commercial License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited and is not used for commercial purposes.

In fact, the last restriction is unnecessary. The proof presented here follows Mochizuki and Trooshin's proof in outline, but the results obtained are more general.

2. Main results

The eigenvalue asymptotics for pencils L are studied in [15], and it is shown that eigenvalues are not necessarily real, and all but a finite number of the eigenvalues are algebraically simple. It is well-known that the eigenvalues of pencils L consist of the sequence

Xn, n 2 A ={±0, ±1, ±2, •••}, and large eigenvalue Xn satisfies the classical asymptotic form

Xn = n + ! + 0(- ), (2.1)

1 Í" , 1 i (hi + 1)(Hi -1)

! = — I p(x)dx +--In —-—--.

n Jo () 2ni (h1 - 1)(H1 + 1)

To simplify the calculations later, we will assume that the eigenvalues Xn are all simple. Denote by Xn and yn(x), n e A, eigenvalues and corresponding eigenfunctions of the differential pencils L, respectively. We introduce the quantities(derivatives of the logarithm of |yn(x)|)

„(x) = ^. yn(x)

Considera second differential pencils

' 7xy(x, X) =fy"(x) + [X2 - 2Xp(x) - q(x)] y(x) = 0, 0 < x < n y'(0) + (iXH + ho) y(0) = 0 (2.2)

y'(n) + (iXH1 + H0) y(n) = 0,

where hk, H k,lq(x) and e(x) have the same properties of hk, Hk, q(x) and p(x), k = 0,1. We agree that if a certain symbol 1 denotes an object related to L(p,q,h0,h1,H0,H1),then 1 will denote an analogous object related to L (p,e,h0, h 1,H0, H^. Let l(n),r(n) be sequences of natural numbers with properties

l(n) = n(1 + em,), 0 < < 1, 61,n ! 0,

n (2.3)

r(n) = — (1 + e2,n), 0 < ff2 < 1, e2,n ! 0; ff2

and let be the eigenvalues of the pencils L(p, q, h0, h1, H0, H2), H1 / H2 e C. Now, we state the main results of this work.

Theorem 2.1 If for any n e A,

Xn = Xn, Kn(2) = K2) ,

then L(p,q,ho,hi,Ho,Hi) = L (p,q,ho,hi,Wo,W^. Remark 2.2

(1) Equation L(p,q,h0,h1,H0,H1) = L(p,Pq,h0, h 1, H0, H^ means that p(x) = p.x) on [0,n], q.x) = lq(x) a.e. on [0,n], and hk =

hk, Hk = Hk(k = 0,1).

(2) The solution of inverse problem in Theorem 2.1 is not unique without condition Kn = Hn (§), because single spectrum can not determine the pencils L. In particular, when yn(y) = 0, equation Kn = Hn {2) is replaced by yn(^)pn(§) = y'„( j)pn( j).

Theorem 2.3

Let l(n), r(n) and b 2 (§, n) be such that a1 > J - 1,02 > 2 - J. If for any n 2 A,

Xn = A n, ^l(n) = H l(n), Kr(n)(b) = Hr(n) then L(p,q,h0,h1,H0,H1) = L (p,Hp,h0,h1, H0, H^ and H2 = H2.

3. Integral representation of products

In this section, we shall study two products of the eigenfunctions of two differential pencils, see (3.6). The result will be used to derive an integral equation for the functions p(x) — 'p(x) and q(x) — Tq(x), which does not involve the parameter A.

Let yi (x, A) be the solution of equation ¡xy1 (x, A) = 0 satisfying the initial conditions yi (0) = land y (0) = —h0,then

yi (x, A) = cos [Ax - a(x)] + f Ai (x, t) cos(At)dt + f Bi (x, t) sin(At)dt,

where a(x) = f^ p(t)dt, and kernels Ai (x, t) and Bi (x, t) are solutions of the following problem [19]:

^ - 2p(x)^ - q(x)Ai(x, t) = ^, + 2p(x) - q(x)Bi (x, t) = ^, Ai (0,0) = -h0, Bi (x,0) = 0, lt=0 = 0

d 2 2 — [A1 (x,x) cos a(x) + B1 (x,x) sin a(x)] = q(x) + p2(x). dx

For each A € C, A / 0, let y2(x, A) be the solution of equation ¡xy2(x, A) = 0 satisfying the initial conditions y2(0) = 0and y2(0) = 1, then

y2(x, A) = A jsin[Ax - a(x)] cj A2(x, t) cos(Af)df C^ B2(x, t) sin(Af)df where the kernels A2(x, t), B2(x, t) are the solution of the problem

^ - 2p(x)^ - q(x)A2(x, t) = ,

C 2p(x)^ — q(x№(x, t) = ,

A2(0,0) = p(0), B2(x,0) = 0, Ï^J*11 lt=0 = 0,

with a(x) = /0p(t)dt. Moreover, there holds

d2 2 — [A2(x,x) sin a(x) + B2(x,x) cos a(x)] = q(x) + p2(x). dx

Let y(x, A) be the solution to equation ¡xy(x, A) = 0 with the initial conditions y(0) = 1 andy'(0) = -(/'Ah1 + h0), then from (3.1) and (3.2), we have

y(x, A) = y1 (x, A) — iXh1y2(x, A).

Moreover, we obtain

y(x, A) = cos[Ax — a(x)] — ih1 sin[Ax — a(x)] C I A(x, t) cos(At)dt

C f B(x, t) sin(At)dt,

A(x, t) = A1(x, t) - ih1A2(x, t), B(x, t) = B1(x, t) - ih1B2(x, t).

Simple calculations show that the characteristic equation of the pencils L can be reduced to !(A) = 0, where

«(A) = y (n, A) C (iAH1 C H0)y(n, A)

= A[(hH - 1) sin An C i(H1 - h1) cos An] C O (e™)

= A^(1 - h2) (1 - H2) sin(A - «)n C O (e™), r = |SA|.

Denote Gg = {A : |A - k- «|> 1, k = 0, ±1, ±2,•••}, 1 > 0. Using the known method (see, e.g. [4]), one can prove the following estimates for sufficiently large |A|:

|«(A)|>Ci|A|e™, A 2 GS.

Math. Meth. Appl. Sci. 2013

Similarly, for the solution p(x, X) of equation pp(x, X) = 0 with the initial conditions p(0) = 1 andpH(0) = — iXh+ h0), there has the following analogous result:

p(x,X) = cos [Xx — H(x)] — ih1 sin [Xx — H(x)] + I p(x, t) cos(At)dt

+ / B(x, t) sin(Xt)dt, 0

A(x, t) = At (x, t) — ihtA2(x, t), p(x, t) = B1 (x, t) — hB^^, t).

Next, using (3.3) and (3.5), and by extending the range of A(x, t),p (x, t) evenly with respect to the argument t and B(x, t), B(x, t) oddly with respect to the argument t and some straightforward calculations, for brevity denoting 9_(t) = a(t) — p(t) and 9+(t) = a(t) + c?(t), we can infer that

yy = cos[Xx — a(x)] cos [Xx — px)] — ih1 sin[Xx — a(x)] cos [Xx — px)]

— ihcos[Xx — a(x)] sin [Xx — c?(x)]—h1h sin[Xx—a(x)] sin [Xx — "a(x)]

+ f Hc(x, t) cos[2Xt — 9+(t)]dt + f Hs(x, t) sin[2Xt — 9+(t)]dt 00

= cos 9_(x) + 2 h — hp) sin 9_ (x) (3.6)

+ 1 + h h1 cos[2Xx — 9+(x)] — 2 (h1 + p sin[2Xx — 9+(x)]

+ f Hc(x, t) cos[2Xt — 9+(t)]dt + f Hs(x, t) sin[2Xt — 9+(t)]dt, 00

where Hc(x, t), Hs(x, t) 2 Wf([0,n] x [0,n]).

4. Completion of proofs

In this section, we shall complete the proofs of Theorems 2.1 and 2.3. The basic idea is to translate the integral equation (4.1) into an inhomogeneous integral equations that are independent of X and then show, step by step, that the inhomogeneous terms must vanish. Now, we can give the proofs of theorems in this work.

Proof of Theorem 2.1

If we multiply the equation in (1.1) byp(x) and the equation in (2.2) by y(x) and subtract, after integrating on [0, J-], we obtain

py — p 02 + [(p — q) + 2X (p — p)] ypdx = 0. (4.1)

Using the initial conditions at 0, then it yields

[y(2, x) y (2, ^ — y(2, ^y (2, ^] + iX (h1 — h1) + h0 — h0 + JQ 2 [(p — q) + 2X (p — p)] yydx = 0. (4.2)

Denote

Q(x) = q — q, P(x)=p — p,

H(A) = iX (ht - h!) + ho - ho W 2 [Q(x)+2AP(x)]yydx.

Because y(x,X) and y(x,X) are entire functions in variable X, H(A) is an entire function in variable X. For X = An, by the given assumptions, it follows that

pn (2) yn (2)—yn (n) pn (n) - o,

that is, the first term in (4.2) vanishes and hence

H(Xn/ = 0, n 2 A,

which implies that the eigenvalues An of pencils L are contained in the set of zeros of H(A). The eigenvalues An with account of multiplicity coincide with the zeros of its characteristic function «(A). From (3.6) and (4.3), we find that for all complex number A

|H(A)|<(Ci + C21 A | )e'T I* (4.4)

for some positive constants C1 and C2. Define

*(A) = !A), (4.5)

which is an entire function from the earlier arguments and it follows from (3.4) and (4.4) that

\$(A) = 0(1)

for sufficiently large | A |, A e G\$. Then by the maximum modulus principle for all complex number A, we obtain that for all complex number A

\$(A) = C,

where C is a constant.

Let us show that the constant C = 0. We can rewrite equation H(A) = C«(A) in the form

iA (hi - hi) + ho - h0 +j 2 [Q(x) + 2AP(x)]yydx = CA^j(1 - h])(l - H2) sin(A - a)it + 0 (e™),

i (hi - y) + +fQ 12 + 2P(x)j yydx = Cyj (1 - h2) (1 - H2) sin(A - co)n + .

By use of the Riemann-Lebesgue Lemma, we see that the limit of the left-hand side of the earlier equality exists as A n, A e R.Thus, we obtain that the constant C = 0. So, we have proved

H(A) = 0 for all complex number A. (4.6)

Substituting (3.6) into (4.3), we obtain

1 - h1h1 /"i

1 _hihi f

H(A) = iX (hi - hi) + ho - ho +-Q(x) cos 6-(x)dx

+ ^ (hi - hi) f 2 Q(x) sin 6-(x)dx 2 Jo

i + hihi /"t + Q(x) cos[2Ax - 6+(x)]dx

- - (hi + hi^ /"2 Q(x) sin[2Ax- 6+(x)]dx 2o

+ f 2 Q(x) i Hc(x, t) cos[2Af - 6+(t)]dtdx oo

+ f2 Q(x) i Hs(x, t) sin[2At - 6+(t)]dtdx oo

+ A (i - hihi) I 2 P(x) cos 6-(x)dx o

+ iA (hi - /"2 P(x) sin 6-(x)dx o

+ A (i + hihi) I 2 P(x) cos[2Ax- 6+(x)]dx o

- iA(hi + hi) /"2 P(x) sin[2Ax - 6+(x)]dx

- f 2 P(x) f

+ 2A i P(x) Hc(x, t) cos[2At - 6+(t)]dtdx

+ 2A f 2 P(x) /" Hs(x, t) sin[2At - 6+(t)]dtdx. oo

Moreover, from H(A) = 0 for all complex number X and by use of the Riemann-Lebesgue Lemma as X !i, X 2 R, we obtain that

— —

i (h- - h-) + (1 - h-h-) i 2 P(x) cos 8-(x)dx + i (h- - h-) i 2 P(x) sin 0_(x)dx = 0. ./0 Jo

Thus, we have

~ 1 - h1h 1 /"t

H(A) = ho - ho +-h-1 Q(x) cos 0_ (x)dx

+ (h1 - h 1) f 2 Q(x) sin 0_(x)dx 2 Jo

1 + h-ihi f — + + / Q(x) cos[2Ax - 0+ (x)]dx 2 Jo

• /* —

- - (h1 + h-W 2 Q(x) sin[2Ax- 0+(x)]dx 2 Jo

+ f2 Q(x) i Hc (x, f) cos[2Af - 0+(f)]dfdx oo

+ f2 Q(x) i Hs(x, f) sin[2Af - 0+(f)]dfdx oo

+ A (1 + h1eiW 2 P(x) cos[2Ax - 0+ (x)]dx

- iA (h1 + h1) f 2 P(x) sin[2Ax - 0+(x)]dx

+ 2A /"2 P(x^ Hc(x, f) cos[2Af - 0+(f)]dfdx oo

+ 2A /"2 P(x) /" Hs(x, f) sin[2Af - 0+(f)]dfdx. oo

Introduce

Q1 (f) = (1 + h1h 1) Q(f) + J 2 Q(x)Hc(x, f)dx,

Q2(f) = -i (h1 + hi) Q(f) ^ 2 Q(x)Hs(x, f)dx, —

Pi (f) = (1 + hi fr,) P(f) + ^ 2 2P(x)Hc(x, f)dx,

P2(f) = -i(hi + h 1) P(f) + J 2 2P(x)Hs(x, f)dx. By changing the order of integration, (4.9) can be rewritten as

H(A) = ho - ho + 2 Q(x) cos 0_(x)dx

2o • /• — + i (h1 - hew 2 Q(x) sin 0_(x)dx 2o

+ - 2 R1(f)e2iXfdf + - 2 R2(f)e_2iXfdf 2 o 2 o

Jo 2 Jo

+ A (ü 2 t, (f)e2iXfdf + Jo 2 T2(t)e_2iXfd^ ,

Ri (f) = Qi(f)_;iQ2(f) e_i0+(f), R2(f) = Qi (f)-2''Q2(f) ei0+(f) Ti (f) = Pi(f)_iP2(f) e_i8+(f)F T2(f) = Pi(f)-2iP2(f) ei0+(f).

(4.io)

(4.11)

By integration by parts in (4.11), we have

H(A) = ho — ho +

1 — h1 h1 f 2

+2 (hi — M I

i' Q(x) cos 6_(x)dx o

Q(x) sin 9_(x)dx

1 f ~ 1 f 7 + - 2 R1(t)e2iX'dt + - 2 R2(t)e_2iX'dt 2 o 2 o

(4.12)

+ -P (2) sin [A* — 6+(*/2)] — 2P2(0)

+ if T T- (t)e2iXtdt — if 7 T2(t)e_2iXt 2 o 2 o

2J0 1 ^0

Moreover, from H(A) = 0 for all complex number A and by use of the Riemann-Lebesgue Lemma as A !i, A e R, we obtain that

ho — ho +

p( y)=0,

1 h he C7 ' C7 1

— 1 1 / 2 Q(x) cos 6_(x)dx + - h — h^W 2 Q(x) sin 6_(x)dx — 1 P2(0) = 0,

2 o 2 o 2

f 2 [«1 (t) + iT1 (t)] e2iXtdt + f 2 [«2(t) - /T2(t)] e~2iXtdt = 0. 00

Because the exponential system |(e2'At,e~2lXt)T : A e ffij is complete in (L2(0,^/2))2, consequently,

«1 (t) + T1 (t) = 0 = R2(t) - iT2(t) on (0,^/2). From the definitions of R1 (t),R2(t), T1 (t), T2(t) by (4.11), we can infer

Q1 (t) + P1 (t)e;(t) + P2(t) = 0 = Q2(t) + P2(tW'+(t) - P1 (t). Substituting (4.10) into (4.16), together with P(^/2) = 0, it follows that

(1 + h^) Q(t) + [(1 + h1y1) 6';(t) - 2Hs(t, t)] P(t) -i (h1 + hy,) P(t) + J*'2 Hc(x, t)Q(x)dx + /t"/2 [2e;(t)Hc(x, t) + 2@tHs(x, t)] P(x)dx = 0,

-i (h1 + y) Q(t) + [-i (h1 + £) e';(t) + 2Hc(t, t)] P(t) - (1 + h1y) P(t) + /t"/2 Hs(x, t)Q(x)dx + /*'2 [2e;(t)Hs(x, t) - 2iHc(x, t)] P(x)dx = 0,

P(t) + I*'2 P'(x)dx = 0.

Introduce

F(t) = (Q(t), P(t), P (t))T,

^1(t) =

( 1 + h1 /T| (1 + h^) 6+(t) — 2Hs(t, t) —i (hn + e \ —i (h1 + h1) — i (h1 + h^) 6+ (t) + 2Hc(t, t) — (1 + h1 £) 0 1 0

K2(x, t) =

( Hc(x, t) 20+(t)Hc(x, t) + 2@tHs(x, t) 0\ , t) 26

Hs(x, t) 26+(t)Hs(x, t) — 2 ^Hc(x, t) 0

(4.13)

(4.14)

(4.15)

(4.16)

(4.17)

Math. Meth. Appl. Sci. 2013

Equation (4.17) can readily be reduced to a vector form

f 2/2 n

K1 (t)F(t) + K2 (x, t)F(x)dx = 0 for 0 < t < -. (4.18)

Because detK1(t) = — (1 — h2) (1 — h!2) / 0 under the assumption that h1 ^±1 and h1 /±1,(4.18) can be rewritten as

r'2 t n

F(t) + Kft (t)K2 (x, t)F(x)dx = 0 for 0 < t < -. t t 2

(4.t9)

But this is a homogeneous Volterra integral equation, and its solution is identically zero a.e. Thus, we have obtained

F(t) = 0 a.e. on [0, n/2],

which yields that

Q(t) = P(t) = 0 a.e. on [0,n/2].

Therefore, we have proven

p(x) = px) on [0,n/2] and q(x) = px) a.e. on [0,n/2].

Moreover, from (4.8) and (4.14), it is obvious that hk = hk (k = 0,1). To prove that

p(x) = px) on [n/2,n] and q(x) = px) a.e. on [n/2,n] (4.20)

and Hk = Hk (k = 0,1), we should repeat the earlier argument for the supplementary problem

y0(x) + [X2 — 2Ap1 (x) — q1 (x)] y(x) = 0, x 2 [0, n], y (0) + (—iXH1 — H0)y(0) = 0, y (n) + (—iXh1 — h0)y(n) = 0,

where q1 (x) = q(n — x) and p1 (x) = p(n — x). Then, we obtain P(n — t) = 0 = Q(n — t) a.e. on [0, J-], that is, (4.20) holds and Hk = Hk (k = 0,1). The proof of theorem is finished. □

To prove Theorem 2.3 in this paper, we first give a Lemma. Let m(n) be a sequence of natural numbers such that

m(n) = n (1 + en), 0 <o < 1, en ! 0. (4.21)

Lemma4.1 1. Let m(n) and b 2 (0, f) be such that o> J .If for any n 2 A,

Xm(n) = Xm(n), Km(n) (b) = 'pm(n)(b),

p(x) = px) on [0, b] and q(x) = 'H(x) a.e. on [0, b],

and hk = hk (k = 0,1). 2. Let m(n) and b 2 (§, n) be such that o > 2 — 2b. If for any n 2 A,

Xm(n) = Xm(n), Km(n) (b) = 'pm(n)(b),

p(x) = px) on [b,n] and q(x) = 'H(x) a.e. on [b,n],

and Hk = H (k = 0,1).

(1) Let y(x, A) be the solution to

- y1 '(x, A) + [q(x) + 2Ap(x)]y(x, A) = A2y(x, A) (4.22) with the initial conditions y(0) = 1 and y'(0) = — (iAh1 + h0). Similarly, let"y(x, A) be the solution of

— 7 '(x, A) + [q(x) + 2Apx)]"y(x, A) = A2y(x, A) (4.23)

with the initial conditions y(0) = 1 and y7(0) = — (iAh1 + h0).

If we multiply (4.22) by"y(x, A) and (4.23) by y(x, A) and subtract, after integrating on [0, b],we obtain

G(A) = fb [e(x)-q(x)] y(x, A)y(x, A)dx o

(4.24)

+ 2X f [p(x)—p(x)] y(x,X)p(x,X)dx + iX (h1 — h1) + h0 — h0

= p(x, X)y(x, X) — p(x, X)y (x, A)] |x=b. By the assumption ^m(n)(b) = нm(n)(b), it follows that

G (Xm(n)) = 0, n 2 A.

Using the same method in [27] (pp. 291-292), we can show that G(A) = 0 on the whole complex plane. As we already mentioned, if G(A) = 0, then the conclusion of Lemma is true. (2) Note that the interval [b, n] can be converted to an interval [0, n — b] by a transformation of variable x ! n — x. To prove (2), we should repeat arguments in part (1) for the supplementary problem L:

' by](x) d=70(x) + [X2 — 2Ap1 (x) — q1 (x)]y(x) = 0, x 2 [0, n], y (0) + (—iXH1 — H0)y(0) = 0, y (n) + (—iXh1 — h0)y(n) = 0,

where q1 (x) = q(n — x) and p1 (x) = p(n — x). A direct calculation implies thatyn(x) = yn(n — x) is a solution to the supplementary problem L and yn(n — b) = yn(b). Note that n — b 2 (0, J). Thus, for the supplementary problem L, the assumption in the case (1) is satisfied still. If we repeat the earlier arguments, we can obtain the proof of Lemma. ^

Proof of Theorem 2.3 Because

Xr(n) = Xr(n), Kr(n)(b) = 'Hr(n) (b),

where r(n) satisfies (2.3) and o2 > 2 — J, by Lemma 4.1, we obtain that

p(x) = px) on [b,n] and q(x) = px) a.e. on [b,n], and Hk = Hk (k = 0,1). Thus, we only need to prove that

p(x) = px) on [0, b] and q(x) = px) a.e. on [0, b]

and hk = hp (k = 0,1), H2 = W2. Similar to (4.24), in the case b 2 (J-, n), we have

G(X) 'i f' №)-,(x)ly(x,XM,X,dx

tb ^ - ~ ~ (4.25)

+ 2X / [p(x)—p(x)] y(x, X)p(x, X)dx + iX(h1 — h1) + h0 — h0

= p(x, X)y(x, X) — ^(x, X)/ (x, X^ |x=fa. Using the same method in [27] (pp. 292-293), we can prove that G(A) = 0 on the whole X-plane. This implies that

p(x) = px) on [0, b] and q(x) = px) a.e. on [0, b],

and hk = hk (k = 0,1).

Because p,n are eigenvalues of differential pencils L(p,q,h0,h1,H0,H2) and p'n are eigenvalues of differential pencils

L pp,h0,hi,H0,H2), by the relation (p,h1,pl(n)) = (p,hi,Th(n)) and the asymptotic expression (2.1), we obtain H2 = H2. The proof is complete. □

Acknowledgements

The author would like to thank the referees for valuable comments. This work was supported by the National Natural Science Foundation of China (11171152/A010602), Natural Science Foundation of Jiangsu Province of China (BK2010489),and the Outstanding Plan-Zijin Star Foundation of NUST (AB 41366).

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Math. Meth. Appl. Sci. 2013