C.Ü. Fen-Edebiyat Fakültesi
Fen Bilimleri Dergisi (2008)Cilt 29 Say 1
On the solutions of the quadratic pencil of the Sturm-Liouville equation with steplike potential
A. Adilo lu Nabiev
Cumhuriyet University, Faculty of Education, 58140, Sivas, Turkey E-mail: aadiloglu@cumhuriyet.edu.tr
Received: 12.02.2008, Accepted: 18.02.2008
Abstract: The Jost solutions of the quadratic pencil of the Sturm-Liouville equation with steplike potential are investigated on the real line. The Jost solutions are defined when the potential is asymptotic to different constants as x , . The integral representations are obtained for the Jost solutions and some spectral properties are investigated.
Keywords: Sturm-Liouville equation with steplike potentials, energy-dependent Sturm-Liouville equation, transformation operators, the Jost solution, spectral analysis of differential operators
Basamak türünden potansiyele sahip Sturm-Liouville denkleminin ikinci dereceden demetinin çözümleri üzerine
Özet: Bu çal mada reel eksende basamak türünden potansiyele sahip Sturm-Liouville denkleminin ikinci dereceden demetinin Jost çözümleri incelenmektedir. Potansiyelin x , . iken farkl sabitlere yakla mas durumu için Jost çözümleri belirlenmektedir. Jost çözümleri için integral gösterilimler elde edilmekte ve baz spektral özellikler incelenmektedir.
Sturm-1. Introduction
This article deals with the investigation of the Jost solutions of the quadratic pencil of the Sturm-Liouville equation when the potential is asymptotic to different constants as x , . It is well known that the Jost solutions play an important role in the investigation direct and inverse scattering problems for Sturm-Liouville operators. The direct and inverse problems for the classical Sturm-Liouville operators with steplike potentials was first studied by Buslayev and Fomin [1]. After Buslayev and Fomin's work Hruslov [2] applied their results to the solution of the Cauchy problem for the Korteweg-de Vries equation with steplike initial profile. Later A.Cohen [5] constructed a counterexample asserted some incorrects in [1]. The direct and inverse scattering for steplike potentials for the Sturm-Liouville equation was developed and completely solved in [6,7].
The full-line direct and inverse scattering problems for the quadratic pencil of the Sturm-Liouville equation with potentials having zero limit at infinity, as a generalization of the Marchenko method, was investigated in [3,4].
It is natural to set and solve the analogous problems for the quadratic pencil of the Sturm-Liouville equation with steplike potentials.. But in this case there are some difficulties with the construction of the Jost solutions which play a basic role in solving the direct and inverse scattering problems. The aim of this paper is to obtain the integral representation for the Jost solutions.
2. The Jost Solutions
On the entire real line let us consider the generalized Sturm-Liouville equation , , ) , ( '' V x y 2y x y (1) with the energy dependent potential V(x, ) q(x) 2 p(x), where q(x),p(x) are real valued functions defined on the entire real line and is a complex parameter. Assume that the following conditions are satisfied:
(i) the function q(x) is locally summable and has different limit values: , ) ( lim , ) ( lim q x c12 q x c22 x x where 0 c12 c22;
(ii) the function p(x) is absolutely continuous on each finite segment ) ; ( ] ; [ and ; 2 ; 1 , ] ) ( ' ) ( 2 ) ( )[ 1 (x q x cj2 cj p x p x dx j (iii) Denote by x j j j x j j j dt t p c c t q x t x dt t p t p c c t q x , ] ) ( ) 1 2 ( ) ( [ ) ( , ] ) ( ' ) ( 2 ) ( [ ) ( 2 2 where j 1;2.
Let j is the of Riemann surface of the function kj( ) cj , j 2 2
is the upper sheet of this surface and the "boundary" of j is j which is the set of points of the upper and the lower lips of the -plane cut along the rays cj ( j 1;2).
For j we define the Jost solutions fj(x, )of the equation (1) with the asymptotic conditions , 1 ) , ( lim 1 ( ) 1 x ik x f x e (2) . 1 ) , ( lim 2 ik2( )x x f x e (3) Rewrite the equation (1) in the form
, ) ( 2 ) ) ( ( '' q x c 2 y k 2 c 2p x y k 2y y j j j j (1’) where kj kj( ).It is easy to see that the Jost solution fj(x, )satisfies the integral equation , ) , ( )] ( ) ( 2 ) ( 2 ) ( [ ) ( sin ) , ( 2 2 2 dt t f t p k c k t p k c t q k x t k e x f j j j j j j x j j x ik j j (4)
where "+" and "-" corresponds to j 1and j 2consequently.
Theorem 1: If the conditions (i)-(ii) are satisfied then for all j j the solution
) , (x
, ) , ( ) ( ) , (x R x e ( ) A x t e ( ) dt f x t ik j x ik j j j j (5) where , ) ( exp ) ( x j x i p t dt R (6) and the kernel Aj( tx, ) satisfies the inequality
)). ( exp( 2 2 1 ) , (x t x t x Aj j j (7) In addition, 2 1 1 ( , ) ( ( ) ) ( ) ( ) ( ) ( , ) ( ) . 2 2 j j j j j x x A x x q t c R t dt p x R x A t t p t dt i (8)
Proof: Suppose that the integral equation (4) has the solution of the form (5). Substituting the expression (5) of fj(x, ) in (4) we have
2 2 2 ( ( ) 1) ( , ) sin ( ) [ ( ) 2 ( ) 2( ) ( )] ( ) ( , ) . j j j j ik x ik t j j x j j j j j j j x ik t ik j j t R x e A x t e dt k t x q t c k p t k c k p t k R t e A t e d dt (9)
Rewrite (9) in the form
(2 ) 2 ( ( ) 1) ( , ) ( ( ) ) ( ) 2 j j j j ik t x ik x ik x ik t j j j j j x x e e R x e A x t e dt q t c R t dt ik (2 ) (2 ) 2 2 ( ) ( ) 2 ( ) ( ) ( ) 2 ( ) ( ) ( ) 2 ( ( ) ) ( , ) 2 j j j j j j ik t x ik x ik t x ik x j j j j j j x x ik t x ik t x j j j x t e e i e e p t R t dt k c k p t R t dt ik e e q t c dt A t d ik ( ) ( ) ( ) ikj t x ikj t x ( , ) j x t i p t dt e e A t d t j j j j j x t ik x t ik x d t A k c k ik e e dt t p j j ) , ( ) ( 2 ) ( 2 2 ) ( ) ( (9’) Now let us require Rj(x) to satisfy
x j
j x i p t R t dt
R ( ) 1 ( ) ( )
From here we have
x
j x i p t dt
R ( ) exp ( ) Further, using the formulas (see [8])
, 2 ) ( ) ( ds e ik e e t x x t s ik j x t ik x t ik j j j t j j x t ik j j j j j k d c I dt e ic ik k c k j 0 1 2 2 , 0 Im , ) (
where I1(.) is the Bessel function, after some simple transformations we have the following equality: 2 2 ) ( ) ) ( ( 2 1 ) , ( t x j j x t ik t ik x j x t e dt e dt q s c R s ds A j j 2 0 1 ) ( ) ( ) ( 2 2 2 1 t x j x t j j j d p s R s ds c I ic t x R t x p i ds x s t s A s p i d s A ds c s q j x x s t x s t j x j ) ( , ) ( ) ( , ) ) ( ( 2 1 2 , ) , ( ) ( ) ( ) , ( ) ( 0 1 2 t s x x s t j x x t j j j t x x d s A ds s p d c I ic ds x s t s A s p i (12)
where Aj( tx, ) 0 for t x. From (12) we get that if the function Aj( tx, ) )
0 ) , (
(Aj x t for t x satisfies the integral equation
2 2 2 1 ) ( ) ) ( ( 2 1 ) , ( 2 2 x t R t x p i ds s R c s q t x A j t x j j j x s t x s t j x t x j j t x j j d p s R s ds q s c ds A s d c I ic ( ( ) ) ( , ) 2 1 ) ( ) ( ) ( 0 2 1
x t d s A ds s p d c I ic x s t x s t j x x t j j ( ) ( , ) , ) ( 0 1 (13) then the function fj(x, ) expressed by (5) is the Jost solution of the equation (1) and conversely, if the function (5) is the Jost solution of the equation (1) then the kernel
) , ( tx
Aj satisfies the integral equation (13). Let us set t x 2v,t x 2u and
). , ( ) , (u v A u v u v
Hj j Then the integral equation (13) takes the following form:
u j j j j p u R u i ds s R c s q v u H ( ) ( ) 2 1 ) ( ) ) ( ( 2 1 ) , ( 2 v v j j u u j j j d p s R s ds d q c H d c I ic 0 0 2 1 ) , ( ) ) ( ( ) ( ) ( ) 2 ( u v j j v d i p u H u d H v p i 0 ) , ( ) ( ) , ( ) ( v v j u j j d d p H d c I ic 0 0 1 ) , ( ) ( ) 2 ( 2 (14)
To solve the integral equation (14) we apply the method of successive approximations. Let us define 1 (0) 2 0 (2 ) 1 1 ( , ) ( ( ) ) ( ) ( ) ( ) ( ) ( ) , 2 2 v j j j j j j j u u I c H u v q s c R s ds p u R u ic d p s R s ds i ,... 2 , 1 , ) , ( ) ( ) 2 ( 2 ) , ( ) ( )) , ( ) ( ) , ( ) ) ( ( ) , ( 0 ) 1 ( 0 1 0 ) 1 ( 0 ) 1 ( 2 ) ( m d H p d d c I ic d u H u p i d v H v p i d H c q d v u H v m j u v j j v j u m j v m j j u m j
It is easy to obtain the estimates
! ) ( ) ( 2 1 ) , ( ), ( 2 1 ) , ( ( ) ) 0 ( m v u u v u H u v u Hj j j m j j (15)
From (15) we have that the series 0 ) ( ) , ( m m j u v
H converges absolutely and uniformly, the sum Hj( vu, ) of the series is a unique solution of the integral equation (14) and
)) ( exp( ) ( 2 1 ) , (u v u u v Hj j j (16) Since u x t,v t x and Hj(u,v) Aj(x,t) the estimation (7) immediately is
obtained from (16). The equality (8) is obtained from (13). The proof is completed. (8) implies that , )] ( ) ( [ ) ( 2 1 ) , ( 2 2 i (x) x j j j e dt t p c t q x ip x x A (17) where x
j(x) p(t)dt. Setting Bj(x,t) ReAj(x,t),Kj(x,t) ImAj(x,t) and taking into account the expression (6) we have
x j j j j j(x) 2 [B (t,t)sin (t) K (t,t)cos (t)]dt, (18) )] ( cos ) , ( ) ( sin ) , ( [ 2 ) ( ) ( 2 2 x x x K x x x B dx d x p c x q j j j j j (19)
From (5) it is also obtained that uniformly for j the following asymptotic formulas are hold:
x o e x f ikj j( , ) [1 (1)], ) ( x o e ik e x f j ikj j i j x )], 1 ( ) ( [ ) , ( ' ( ) ( ) (20)
where "+" corresponds to j 1 and "-" corresponds to j 2.
Since p(x) and q(x) are real functions then for all j, cj the equation (1) has solutions fj(x, ) and fj(x, ). Let
) ( ' ) ( ) ( ) ( ' ] , [y1 y2 y1 x y2 x y1 x y2 x
W is the Wronskian of the functions y1(x) and
). ( 2 x
y From the asymptotic formulas (20) we have . 2 ) 1 ( ) , ( ), , ( j j 1 j j x f x ik f W (21) Therefore, for j , cj the functions fj(x, ) and fj(x, ). are linearly independent solutions of the equation (1) and the following equalities take place:
, , ), , ( ) ( ) , ( ) ( ) , ( 1 1 1 1 1 1 2 x a f x b f x c f (22) , , ), , ( ) ( ) , ( ) ( ) , (x a f x b f x c f (23)
Taking into account (21) we have , , ], , [ 2 1 ) ( 1 2 j j j j W f f c ik a (24) , , ], , [ 2 1 ) ( 2 1 1 1 1 1 W f f c ik b 2 2 1 2 2 2 [ , ], , 2 1 ) ( W f f c ik b (25) From (24) and (25) we obtain
, , ) ( ) ( ) ( 1 ) ( ) ( ) ( 2 2 2 1 2 2 2 1 2 b c k k a k k . , ) ( ) ( ) ( 1 ) ( ) ( ) ( 1 2 1 2 1 2 1 2 1 b c k k a k k (26) We also obtain that
1 2 2 1 1 2 2 1 1( )a( ) k ( )a ( ),k ( )b( ) k ( )b ( ), c k (27)
Moreover, since fj(x, ) can be analytically continued into the upper sheet j then from (24), (25) we conclude that the function aj( ) also can be analytically continued into j : j j j W f f ik a [ , ], 2 1 ) ( 1 2 (28) It is clear that for which is not real or for ( c2,c2) the equation (1) has
unique solution f2(x, )(f1(x, )) belonging to L2( ,0)(L2(0, )). Therefore, the equation (1) has the solution belonging to L2( , ) if and only if is a root of the equation aj( ) 0. Hence, any zero of aj( ) will be called an eigenvalue of the operator corresponding to the equation (1). From (26) we have that the function aj( )
has not any zero on j and therefore the zeros of the function aj( ) are placed in the
interval ( c2,c2) or in the sheet j .
Proof: We know that the zeros are placed in ( c2,c2) j . Suppose that aj( ) has
an infinite number of zeros k(j) (k 1,2,...) Since aj( k( j)) 0 from (28) we obtain that the functions f1(x, k(j)) and f2(x, k(j)) are linearly dependent. Let us denote
,...) 3 , 2 , 1 ( ) , ( ) ( 1 ( ) ) ( k x f x
yk j k j and let L0(j) to be the minimal closed operator generated on L2( , ) by differential expression 2 ( ) 2.
2 j c x q dx d It is clear that the operator L0(j) is selfadjoint. Since
) ( ) ( ) ( 2 ) ( ( ) ( ) ( ) 2 12 ( ) ) ( ) ( 0 y x p x y x c y x L j k j k j k j k j k j then , , , , , , ) ( ) ( ) ( ) ( 2 1 ) ( ) ( ) ( 0 2 ) ( ) ( ) ( ) ( ) ( j k j k j k j k j k j k j j k j k j k j k j k y y y y c y y L y py y py (29) where .,. is the inner product in L2( , ). Taking into account
j j
k ( c2,c2) )
(
from (29) it is obtained that
,... 2 , 1 , 0 , ( ) ) ( ) ( 0 y y k L j k j k j (30) Since yk j x eikj k j x o x )], 1 ( 1 [ ) ( ( ) ) ( ()
and the numbers k( j) are distinct from (30) we obtain that the selfadjoint operator L0(j) has an infinite number of negative eigenvalues. Therefore, our assumption is not true and aj( ) may have only a finite
number of zeros.
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