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Volume 2010, Article ID 171967, 17 pagesdoi:10.1155/2010/171967 Research Article On an Inverse Scattering Problem for a Discontinuous Sturm-Liouville Equation with a Spectral Parameter i

Volume 2010, Article ID 171967, 17 pages

doi:10.1155/2010/171967

Research Article

On an Inverse Scattering Problem for

a Discontinuous Sturm-Liouville Equation with

a Spectral Parameter in the Boundary Condition

Khanlar R Mamedov

Mathematics Department, Science and Letters Faculty, Mersin University, 33343 Mersin, Turkey

Correspondence should be addressed to Khanlar R Mamedov,hanlar@mersin.edu.tr

Received 9 April 2010; Accepted 22 May 2010

Academic Editor: Michel C Chipot

Copyrightq 2010 Khanlar R Mamedov This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

An inverse scattering problem is considered for a discontinuous Sturm-Liouville equation on the half-line0, ∞ with a linear spectral parameter in the boundary condition The scattering data of

the problem are defined and a new fundamental equation is derived, which is different from the classical Marchenko equation With help of this fundamental equation, in terms of the scattering data, the potential is recovered uniquely

1 Introduction

We consider inverse scattering problem for the equation

−ψ qxψ  λ2ρ xψ 0 < x < ∞, 1.1 with the boundary condition

−α1ψ 0 − α2ψ0 λ2

β1ψ 0 − β2ψ0, 1.2

where λ is a spectral parameter, qx is a real-valued function satisfying the condition

∞

0

1  xq xdx < ∞, 1.3

ρ x is a positive piecewise-constant function with a finite number of points of discontinuity,

α i , β i i  1, 2 are real numbers, and γ  α1β2− α2β1> 0.

The aim of the present paper is to investigate the direct and inverse scattering problem

on the half-line 0, ∞ for the boundary value problem 1.1–1.3 In the case ρx ≡ 1,

the inverse problem of scattering theory for1.1 with boundary condition not containing spectral parameter was completely solved by Marchenko1,2, Levitan 3,4, Aktosun 5,

as well as Aktosun and Weder6 The discontinuous version was studied by Gasymov 7 and Darwish 8 In these papers, solution of inverse scattering problem on the half-line

0, ∞ by using the transformation operator was reduced to solution of two inverse problems

on the intervals 0, a and a, ∞ In the case ρx / 1, the inverse scattering problem was

solved by Guse˘ınov and Pashaev 9 by using the new nontriangular representation of Jost solution of1.1 It turns out that in this case the discontinuity of the function ρx strongly

influences the structure of representation of the Jost solution and the fundamental equation of the inverse problem We note that similar cases do not arise for the system of Dirac equations with discontinuous coefficients in 10 Uniqueness of the solution of the inverse problem and geophysical application of this problem for1.1 when qx ≡ 0 were given by Tihonov 11 and Alimov12 Inverse problem for a wave equation with a piecewise-constant coefficient was solved by Lavrent’ev13 Direct problem of scattering theory for the boundary value problem1.1–1.3 in the special case was studied in 14

When ρx ≡ 1 in 1.1 with the spectral parameter appearing in the boundary conditions, the inverse problem on the half-line was considered by Pocheykina-Fedotova15 according to spectral function, by Yurko16–18 according to Weyl function, and according

to scattering data in19,20 This type of boundary condition arises from a varied assortment

of physical problems and other applied problems such as the study of heat conduction by Cohen 21 and wave equation by Yurko 16, 17 Spectral analysis of the problem on the half-line was studied by Fulton22

Also, physical application of the problem with the linear spectral parameter appearing

in the boundary conditions on the finite interval was given by Fulton23 We recall that inverse spectral problems in finite interval for Sturm-Liouville operators with linear or nonlinear dependence on the spectral parameter in the boundary conditions were studied

by Chernozhukova and Freiling24, Chugunova 25, Rundell and Sacks 26, Guliyev 27, and other works cited therein

This paper is organized as follows InSection 2, the scattering data for the boundary value problem1.1–1.3 are defined InSection 3, the fundamental equation for the inverse problem is obtained and the continuity of the scattering function is showed Finally, the uniqueness of solution of the inverse problem is given inSection 4

For simplicity we assume that in1.1 the function ρx has a discontinuity point:

ρ x 

⎧

⎨

⎩

α2, 0 ≤ x < a,

where 0 < α / 1

The function

f0x, λ  1

2

1 1

ρ x

e iλμx1

2

1− 1

ρ x

e iλμ−x , 1.5

is the Jost solution of1.1 when qx ≡ 0, where μ±x  ±x ρ x  a1 ∓ ρ x.

It is well known9 that, for all λ from the closed upper half-plane, 1.1 has a unique

Jost solution f x, λ which satisfies the condition

lim

x→ ∞f x, λe −iλx 1 1.6 and it can be represented in the form

f x, λ  f0x, λ 

∞

μx

K x, te iλt dt, 1.7

where the kernel Kx, t satisfies the inequality

∞

μx |Kx, t|dt ≤ C

exp ∞

x

tq tdt , 0 < C  const, 1.8

and possesses the following properties:

dK

x, μx

dx  − 1

4 ρ x

1 1

ρ x

d

dx



K

x, μ−x  0− Kx, μ−x − 0 1

4 ρ x

1− 1

ρ x

q x. 1.10

In addition, if qx is differentiable, Kx, t satisfies a.e. the equation

ρ x ∂2K

∂t2 −∂2K

∂x2  qxK  0, 0 < x < ∞, t > μx. 1.11 Denote that

ϕ λ α2 β2λ2

f0, λ −α1 β1λ2

f 0, λ. 1.12

According toLemma 2.2inSection 2, the equation ϕλ  0 has only a finite number of simple roots in the half-plane Im λ > 0; all these roots lie in the imaginary axis The behavior of this

boundary value problem1.1–1.3 is expressed as a self-adjoint eigenvalue problem

We will call the function

S λ 



α2 β2λ2

f0, λ −α1 β1λ2

f 0, λ



α2 β2λ2

f0, λ −α1 β1λ2

f 0, λ 1.13

the scattering function for the boundary value problem1.1–1.3, where f0, λ denotes the complex conjugate of f 0, λ.

We denote by m−2k the normalized numbers for the boundary problem1.1–1.3:

m−2k ≡

∞

0

ρ xf x, iλ k2

dx 1

γβ2f0, iλ k  − β1f 0, iλ k2

, 1.14

where k  1, 2, , n It turns out that the potential qx in the boundary value problem 1.1–

1.3 is uniquely determined by specifying the set of values {Sλ, λ k , m k } The set of values

is called the scattering data of the boundary value problem1.1–1.3 The inverse scattering problem for boundary value problem1.1–1.3 consists in recovering the coefficient qx

from the scattering data

The potential qx is constructed by slightly varying the method of Marchenko Set

F0x  1

2π

∞

−∞S0λ − Sλe −iλx dλn

k1

m2k e −λ kx ,

F

x, y

 1

2

1 1

ρ x

F0



y  μx1

2

1− 1

ρ x

F0



y  μ−x,

1.15

where

S0λ 

⎧

⎪

⎪

⎨

⎪

⎪

⎩

f00, λ

f00, λ  e −2iλa

1 τe −2iλaα

e −2iλaα  τ , if β2  0,

f00, λ

f00, λ  −e −2iλa

1− τe −2iλaα

e −2iλaα − τ , if β2/  0,

1.16

and τ  α − 1/α  1.

We can write out the integral equation

F

x, y



∞

μx K x, tF0



t  ydt  Kx, y

1− ρ x

1 ρ x K



x, 2a − y 0, 1.17

for the unknown function Kx, t The integral equation is called the fundamental equation

of the inverse problem of scattering theory for the boundary problem 1.1–1.3 The fundamental equation is different from the classic equation of Marchenko and we call the

equation the modified Marchenko equation The discontinuity of the function ρx strongly

influences the structure of the fundamental equation of the boundary problem1.1–1.3 By

Theorem 4.1inSection 4, the integral equation has a unique solution for every x≥ 0 Solving

this equation, we find the kernel Kx, y of the special solution 1.7, and hence according to formula1.10 it is constructed the potential qx.

We show that formula1.7 is valid for 1.1 For this, let us give the algorithm of the proof in9 For fx, λ let us consider the integral equation

f x, λ  f0x, λ 

∞

x

Φx, t, λqtft, λdt, 1.18

Φx, t, λ  s0t, λc0x, λ − s0x, λc0t, λ, 1.19

while s0x, λ and c0x, λ are solutions of 1.1 when qx ≡ 0, satisfying the initial conditions

c00, λ  s

00, λ  1 and c

00, λ  s00, λ  0.

It is not hard to show that the functionΦx, t, λ satisfies the formula

Φx, t, λ 

σ x,t

where

K0x, t, z 

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

1

2α , |z| ≤ σx, t, x ≤ t ≤ a,

1 4

1 1

α , t − a − αa − x ≤ |z| ≤ σx, t, x ≤ a ≤ t,

1

2, |z| ≤ t − a − αa − x, x ≤ a ≤ t,

1

2, |z| ≤ σx, t, t ≥ x ≥ a,

σ x, t 

t

x



ρ sds 

⎧

⎪

⎨

⎪

⎩

α t − x, x ≤ t ≤ a,

α a − x  t − a, x ≤ a ≤ t,

t − x, a ≤ x ≤ t.

1.21

Substituting the expression1.7 for fx, λ in the integral equation 1.18 and using formula

1.20 for Φx, t, λ after elementary operations, the following integral equations for the kernel

K x, t are obtained:

K x, t  1

4α

11

α

a

αxαa−at/2α q zdz  1

4α

1−1

α

a

αxαaa−t/2α q zdz

1

4

1 1

α

∞

a

q zdz −1

4

1− 1

α

t−αxαaa/2

a

q zdz

 1

2α

mint,ααa−a/α

x

q z

t αz−x

t −αz−x K z, sds dz

−1

4

taαa−αx/2

a

q z

t −z−a−αaαx

t z−a−αaαx K z, sds dz,

1.22

for 0 < x < a, αx − αa  a < t < −αx  αa  a;

K x, t  1

4

1 1

α

∞

tαx−αaa/2 q zdz 1

4

1− 1

α

∞

t−αxαaa/2 q zdz

 1

2α

a

x

q z

t αz−x

t −az−x K z, sds dz

−1 4

1− 1

α

a αa−αx

a

q z

t −zaαa−αx

t z−a−αaαx K z, sds dz

1 4

1− 1

α

∞

a αa−αx q z

t z−a−αaαx

t −zaαa−αx K z, sds dz,

1.23

for 0 < x < a, t > −αx  αa  a;

K x, t  1

2

∞

xt/2 q zdz 1

2

∞

x

q zdz

t z−x

t −z−x K z, sds, 1.24

for t ≥ x ≥ a.

The solvability of these integral equations is obtained through the method of successive approximations By using integral equations 1.22–1.24 for Kx, t, equalities

1.9, 1.10 are obtained By substituting the expressions for the functions fx, λ and

fx, λ in 1.1, it can be shown that 1.11 holds

2 The Scattering Data

For real λ /  0, the functions fx, λ and fx, λ form a fundamental system of solutions of

1.1 and their Wronskian is computed as W{fx, λ, fx, λ}  2iλ Here the Wronskian is defined as W {f, g}  fg − fg.

Let ωx, λ be the solution of 1.1 satisfying the initial condition

ω 0, λ  α2 β2λ2, ω0, λ  α1 β1λ2. 2.1 The following assertion is valid

Lemma 2.1 The identity

2iλωx, λ



α2 β2λ2

f0, λ −α1 β1λ2

f 0, λ  fx, λ − Sλfx, λ 2.2

holds for all real λ /  0, where

S λ 



α2 β2λ2

f0, λ −α1 β1λ2

f 0, λ



α2 β2λ2

f0, λ −α1 β1λ2

f 0, λ 2.3

S λ  S−λ  S−λ−1. 2.4

The function Sλ is called the scattering function of the boundary value problem 1.1–

1.3

Lemma 2.2 The function ϕλ may have only a finite number of zeros in the half-plane Im λ > 0.

Moreover, all these zeros are simple and lie in the imaginary axis.

Proof Since ϕ λ / 0 for all real λ / 0, the point λ  0 is the possible real zero of the function

ϕ λ Using the analyticity of the function ϕλ in upper half-plane and the properties of

solution1.7 are obtained that zeros of ϕλ form at most countable and bounded set having

0 as the only possible limit point

Now let us show that all zeros of the function ϕλ lie on the imaginary axis Suppose that μ1and μ2are arbitrary zeros of the function ϕλ We consider the following relations:

−f

x, μ1

 qxfx, μ1

 μ2

1ρ xfx, μ1

,

−f

x, μ2



 qxfx, μ2



 μ2 ρ xfx, μ2



.

2.5

Multiplying the first of these relations by f x, μ2 and the second by fx, μ1, subtracting the second resulting relation from the first, and integrating the resulting difference from zero to infinity, we obtain



μ21− μ2

 ∞

0

ρ xfx, μ1



f

x, μ2



dx − Wf

x, μ1



, f

x, μ2



x0 0. 2.6

On the other hand, according to the definition of the function ϕλ, the following relation

holds:

ϕ

μ j

α2 β2μ2

j



f

0, μ j

−α1 β1μ2

j



f

0, μ j

 0, j  1, 2. 2.7 Therefore,

f

x, μ j

 1

γ



β2f

0, μ j

− β1f

0, μ j

ω

x, μ j

, j  1, 2. 2.8

This formula yields

W

f

x, μ1



, f

x, μ2



x0 1

γ



β2f

0, μ1



− β1f

0, μ1



×β2f

0, μ2

− β1f

0, μ2

μ2 − μ2 1



.

2.9

Thus, using2.6 and 2.9 we have



μ21− μ2

∞

0

ρ xfx, μ1



f

x, μ2



dx1

γ



β2f

0, μ1



− β1f

0, μ1



×β2f

0, μ2

− β1f

0, μ2

 0.

2.10

Here ρx > 0, γ > 0 In particular, the choice μ2  μ1at2.10 implies that μ2

1− μ1  0, or

μ1  iλ1, where λ1 ≥ 0 Therefore, zeros of the function ϕλ can lie only on the imaginary axis Now, let us now prove that function ϕλ has zeros in finite numbers This is obvious

if ϕ 0 / 0, because, under this assumption, the set of zeros cannot have limit points In the

general case, since we can give an estimate for the distance between the neighboring zeros of

the function ϕλ, it follows that the number of zeros is finite see 2, page 186

Let

m−2k ≡

∞

0

ρ xf x, iλ k2

dx 1

γβ2f0, iλ k  − β1f 0, iλ k2

 1

2iμ k γ ϕ

iλ kβ2f0, iλ k  − β1f 0, iλ k, k  1, 2, , n.

2.11

These numbers are called the normalized numbers for the boundary problem1.1–1.3 The collections{Sλ, −∞ < λ < ∞; λ k ; m k k  1, 2, , n} are called the scattering

data of the boundary value problem1.1–1.3 The inverse scattering problem consists in recovering the coefficient qx from the scattering data

3 Fundamental Equation or Modified Marchenko Equation

From1.9, 1.10, it is clear that in order to determine qx it is sufficient to know Kx, t To derive the fundamental equation for the kernel Kx, t of the solution 1.7, we use equality

2.2, which was obtained inLemma 2.1 Substituting expression1.7 for fx, λ into this

equality, we get

2iλωx, λ

ϕ λ − f0x, λ  S0λf0x, λ



∞

μx K x, te −iλt dt  S0λ − Sλf0x, λ



∞

μx K x, tS0λ − Sλe iλt dt − S0λ

∞

μx K x, te −iλt dt.

3.1

Multiplying both sides of relation3.1 by 1/2πe iλy and integrating over λ from−∞ to ∞,

for y > μx at the right-hand side we get

K

x, y

 1

2π

∞

−∞S0λ − Sλf0x, λe iλy dλ



∞

μx K x, t

 1

2π

∞

−∞S0λ − Sλe iλ ty dλ

dt

−

∞

μx K x, t

 1

2π

∞

−∞S0λe iλ ty dλ

dt.

3.2

Now we will compute the integral1/2π−∞∞S0λe iλ ty dλ By elementary transforms we

obtain

S0λ  e −2iλa



1− τ2

e 2iλaα

1 τe 2iλaα  τe −2iλa

 e −2iλa1−α

1− τ2∞

k0

−1k τ k e 2iλaαk  τe −2iλa ,

3.3

where β2 0 Thus we have

1

2π

∞

−∞S0λe iλ ty dλ1− τ2∞

k0

−1k τ k δ

t  y − 2a1 − α  2aαkτδt  y − 2a,

3.4

where δt is the Dirac delta function.

For β2/ 0, similarly we get

1

2π

∞

−∞S0λe iλ ty dλτ2− 1∞

k0

−1k τ k δ

t  y − 2a1 − α  2aαkτδt  y − 2a.

3.5 Consequently,3.2 can be written as

K

x, y

 F S



x, y



∞

μx K x, tF 0S



t  ydt − τKx, 2a − y

−1− τ2∞

k0

−1k τ k K

x, 2a 1 − α − 2aαk − y,

3.6

F 0S x ≡ 1

2π

∞

−∞S0λ − Sλe iλx dλ,

F S



x, y

≡ 1 2

1 1

ρ x

F 0S



μx  y

1 2

1− 1

ρ x

F 0S



μ−x  y.

3.7

Let us show that for y > μx the last expression in the sum equals zero We note that

K x, z  0 for z < x For y > μx we have

2a1 − α − 2aαk − y < μx, k  0, 1, 2, 3.8

If 0 < x < a, then μx  αx − αa  a, and hence

2a1 − α − 2aαk − y < 2a − 2aαk  1 − αx  αa − a

 a − aα − 2aαk − αx < a1 − α ≤ μx. 3.9

If x ≥ a, then μx  x, and hence, for this case, the inequality holds.

Therefore, for y > μx 3.2 takes the form

K

x, y

 F S



x, y



∞

μx K x, tF 0S



t  ydt1− ρ x

1 ρ x K



x, 2a − y. 3.10

On the left-hand side of3.1 with help of Jordan’s lemma and the residue theorem and by takingLemma 2.2into account for y > μx, we obtain

−n

k1

2iλ k ω x, iλ k

ϕiλ k e −λ ky . 3.11

From the definition of normalized numbers m k k  1, 2, , n in 2.11 we have

−n

k1

2iλ k ω x, iλ k e −λ ky

ϕiλ k  −

n



k1

2iλ k e −λ ky f x, iλ k



β2f0, iλ k  − β1f 0, iλ kϕiλ k

 −n

k1

m2k f x, iλ k e −λ ky

 −n

k1

m2

k

f0x, iλ k e −λ k xy

∞

μx

K x, te −λ k ty dt

.

3.12

equation the modified Marchenko equation The discontinuity... values

is called the scattering data of the boundary value problem< /i>1.1–1.3 The inverse scattering problem for boundary value problem 1.1–1.3 consists in recovering the. .. −z? ?a? ??? ?a αx

t z? ?a? ??? ?a αx K z, sds dz,

1.22