Gen. Math. Notes, Vol. 31, No. 1, November 2015, pp. 72-84
ISSN 2219-7184; Copyright © ICSRS Publication, 2015
www.i-csrs.org
Available free online at http://www.geman.in
Eigenvalue Problem for Discontinuous Dirac
System with Eigenparameter in a
Transmission Condition
Fatma Hıra1 and Nihat Altınışık2
1
Department of Mathematics, Faculty of Science and Arts
Hitit University, 19030, Çorum, Turkey
E-mail: fatmahira@yahoo.com.tr
2
Department of Mathematics, Faculty of Science and Arts
Ondokuz Mayıs University, 55139, Samsun, Turkey
E-mail: anihat@omu.edu.tr
(Received: 14-9-15 / Accepted: 19-10-15)
Abstract
We deal with an eigenvalue problem for discontinuous Dirac system which
includes an eigenvalue parameter in a transmission condition. We investigate
asymptotic behavior of eigenvalues and eigenfunctions of this Dirac system.
Keywords: Asymptotic
eigenvalue problem.
1
approximation,
discontinuous
Dirac
system,
Introduction
The basic and comprehensive results about Dirac operators were given in [12].
The oscillation property and asymptotic formulas for the eigenvalues were given
in [10] and the computation of the eigenvalues of Dirac systems using a
regularized sampling method was investigated in [5]. They are examples for
Eigenvalue Problem for Discontinuous Dirac…
73
continuous Dirac systems (see also [4, 17]). Direct and inverse problems for Dirac
operators with discontinuities inside an interval were investigated in [3], the
computation of the eigenvalues of discontinuous Dirac system using Hermite
interpolation technique was given in [15]. In these discontinuous works, neither
the boundary conditions nor the discontinuity condition depend on the spectral
parameter. Inverse spectral problems and uniqueness theorems of the inverse
problems for the same Dirac operator with eigenvalue dependent boundary and
jump conditions were studied in [11, 18], respectively. Also, in [8] the authors
investigated asymptotic behavior of eigenvalues and constructed Green's function
for a Dirac system which has two points of discontinuity and contains an
eigenparameter in a boundary and two transmission conditions.
We consider the following discontinuous Dirac system
u2' ( x ) − r1 ( x ) u1 ( x ) = λ u1 ( x ) 
, x ∈ I,
u1' ( x ) + r2 ( x ) u2 ( x ) = −λ u2 ( x ) 
(1)
B1 ( u ) := u1 ( −1) sin α + u2 ( −1) cos α = 0,
(2)
B2 ( u ) := u1 (1) sin β + u2 (1) cos β = 0,
(3)
and transmission conditions
T1 ( u ) := u1 ( 0 − ) − u1 ( 0+ ) = 0,
(4)
T2 ( u ) := u2 ( 0− ) − u2 ( 0 + ) + λδ u1 ( 0− ) = 0,
(5)
 u1 ( x ) 
where I := [ −1, 0 ) ∪ ( 0,1] ; λ complex spectral parameter; u ( x ) = 
; the
 u ( x ) 
 2

real valued function r1 (.) and r2 (.) are continuous in [ −1, 0 ) and ( 0,1] and have
finite limits r1 ( 0± ) := lim± r1 ( x ) ,
α , β ∈ [ 0, π ) .
x →0
r2 ( 0± ) := lim± r2 ( x ) ; δ ∈ ℝ and δ > 0 ;
x →0
Defined by equations (4) and (5) are called transmission (or jump; discontinuity)
conditions. Boundary-value problems with transmission conditions inside the
interval often appear in mathematical physics, mechanics, electronics, geophysics
and other branches of natural sciences ( see [13, 16]).
In this paper, we investigate asymptotic behavior of eigenvalues and vectorvalued eigenfunctions of a discontinuous Dirac system which contains an
eigenparameter in a transmission condition with the same eigenparameter in the
74
Fatma Hıra et al.
equations. For studying these asymptotic behaviors, we derive the integral
equations similarly to the discontinuous Sturm Liouville problem (see [1, 2, 6])
and solve these integral equations by the method of successive approximations.
Finally, we obtain asymptotic formulas for eigenvalues depending on whether
Q1 ( 0 ) and Q2 (1) are zero or not, where
0
1
1
1
Q1 ( 0 ) = ∫ ( r1 ( y ) + r2 ( y ) ) dy, Q2 (1) = ∫ ( r1 ( y ) + r2 ( y ) ) dy.
2 −1
20
2
Spectral Properties
Let L := L2 ( −1,0 ) ⊕ L2 ( 0 ,1) , then we define the Hilbert space H := L ⊕ ℂ with
an inner product
u ( .) ,v ( .)
1
H
:=
1
τ
∫ u ( x ) v ( x ) dx + δ hk ,
(6)
−1
 u( x)
 v ( x) 
where τ denotes the matrix transpose, u ( x ) = 
 , v ( x) = 
∈ H ;
 h 
 k 
 u1 ( x ) 
 v1 ( x ) 
u ( x) = 
 , v ( x) = 
 ∈ L, ui ( .) ,vi ( .) ∈ L2 ( −1,1) ( i = 1, 2 ) ; h,k ∈ ℂ .
 u ( x) 
 v ( x) 
 2

 2

Equation (1) can be written as
ℓ ( u ) := Bu' ( x ) − Ω ( x ) u ( x ) = λ u ( x ) ,
(7)
 u1 ( x ) 
 r1 ( x )
0 
 0 1
where B = 
.
 , u ( x ) = 
 , Ω ( x) =  0

r2 ( x ) 
 −1 0 

 u2 ( x ) 
For functions u ( x ) , which is defined on
I
and has finite limit
u ( 0± ) := lim± u ( x ) , by u(1) ( x ) and u( 2 ) ( x ) we denote the functions
x →0
u ( x ) , x ∈ [ −1, 0 ) ,

u(1) ( x ) = 
−
x = 0,
u 0 ,
( )
u ( x ) , x ∈ ( 0 ,1] ,

u( 2 ) ( x ) = 
+
u 0 , x = 0 ,
( )
which are defined on I1 := [ −1,0] and I 2 := [ 0,1] , respectively.
Lemma 2.1 The eigenvalues of the problem (1)-(5) are real.
Eigenvalue Problem for Discontinuous Dirac…
75
Proof: Assume the contrary that λ0 is a nonreal eigenvalue of the problem (1)-(5).
 u1 ( x ) 
Let u ( x ) = 
 be a corresponding (non-trivial) eigenfunction. By equation
 u ( x) 
 2

(1), for x ∈ I , we have
{
} (
d
u1 ( x ) u2 ( x ) − u1 ( x ) u2 ( x ) = λ 0 − λ0
dx
){ u1 ( x ) 2 + u2 ( x ) 2 } .
Integrating the above equation through [ −1,0 ) and ( 0,1] , we obtain, respectively,
(
(
)
0

2
2
λ 0 − λ0  ∫ u1 ( x ) + u2 ( x ) dx  = u1 0− u2 0− − u1 0− u2 0−


 −1

)
( ) ( ) ( ) ( )
(
(8)
)
− u1 ( −1) u2 ( −1) − u1 ( −1) u2 ( −1) ,
(
1
)
(
λ 0 − λ0  ∫ u1 ( x ) + u2 ( x )
0
2
2
)

dx  = u1 (1) u2 (1) − u1 (1) u2 (1)


( ( ) ( ) ( ) ( )) .
(9)
− u1 0+ u2 0+ − u1 0+ u2 0+
Then from the boundary conditions (2), (3) and the transmission conditions (4),
(5), we have, respectively,
u1 ( −1) u2 ( −1) − u1 ( −1) u2 ( −1) = 0 , u1 (1) u2 (1) − u1 (1) u2 (1) = 0 ,
and
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (
) ( ) ( )
u1 0+ u2 0+ − u1 0+ u2 0+ = u1 0− u2 0− − u1 0− u2 0− + δ λ0 − λ0 u1 0− u1 0−
Since λ0 ≠ λ 0 and δ > 0 , it follows that
∫(
1
−1
u1 ( x ) + u2 ( x )
2
2
)
( )
dx + δ u1 0−
2
=0.
(10)
Then ui ( x ) = 0 ( i = 1, 2 ) and this is a contradiction. Consequently, λ0 must be
real.
□
 u1 ( x, λ ) 
 v1 ( x, µ ) 
Lemma 2.2 Two eigenfunctions u ( x, λ ) = 
 and v ( x, µ ) = 

 u ( x, λ ) 
 v ( x, µ ) 
 2

 2

corresponding to different eigenvalues λ and µ , are orthogonal, i.e.,
76
1
∫u
Fatma Hıra et al.
τ
( x, λ ) v ( x, µ )dx +
−1
1
δ
hk = 0
Now we will construct a special fundamental system of solutions of equation (1).
By virtue of Theorem 1.1 in [12], we shall define two solutions of equation (1)
 φ1 ( ., λ ) 
 χ1 ( ., λ ) 
 , χ ( x, λ ) = 

 φ ( ., λ ) 
 χ ( ., λ ) 
 2

 2

φ ( x, λ ) = 
where
φ11 ( x, λ ) , x ∈ [ −1, 0 ) ,
φ21 ( x, λ ) , x ∈ [ −1, 0 ) ,
φ1 ( x,λ ) = 
φ2 ( x,λ ) = 
φ12 ( x, λ ) , x ∈ ( 0 ,1] ,
φ22 ( x, λ ) , x ∈ ( 0 ,1] ,
 χ11 ( x, λ ) , x ∈ [ −1, 0 ) ,
 χ 21 ( x, λ ) , x ∈ [ −1, 0 ) ,
χ 2 ( x, λ ) = 
 χ12 ( x, λ ) , x ∈ ( 0 ,1] ,
 χ 22 ( x, λ ) , x ∈ ( 0 ,1] ,
χ1 ( x,λ ) = 
(11)
(12)
 φ11 ( x, λ ) 
as follows: Let u = 
 be the solution of equation (1) on [ −1,0] , which
 φ ( x, λ ) 
21


satisfies the initial condition (2);
 u1 ( −1)   cos α 
.

=
 u ( −1)   − sin α 
 2

(13)
 φ12 ( x, λ ) 
u=
 of equation
 φ ( x, λ ) 
 22

(1) on [ 0,1] by means of the solution φ1 ( x,λ ) and by the transmission conditions
(4) and (5);
After defining this solution, we define the solution
(
)
)

φ11 0− ,λ
 u1 ( 0 )  

=
 u (0)  
−
−
 2
  φ21 0 ,λ + λδφ11 0 ,λ

(
(
)

.



(14)
Hence φ ( x, λ ) satisfies equation (1) on I , the boundary condition (2) and the
transmission conditions (4) and (5).
Eigenvalue Problem for Discontinuous Dirac…
77
 χ12 ( x, λ ) 
Similarly, first we define the solution u = 
 of equation (1) on [ 0,1] by
 χ ( x, λ ) 
 22

the initial condition (3);
 u1 (1)   cos β 
.

=
 u (1)   − sin β 
 2 
(15)
 χ11 ( x, λ ) 
After defining this solution, we define the solution u = 
 of equation
 χ ( x, λ ) 
 21

(1) on [ −1,0] by the transmission conditions (4) and (5);
(
)
)

χ12 0+ ,λ
 u1 ( 0 )  

=
 u (0)  
+
+
 2
  χ 22 0 ,λ − λδχ12 0 ,λ

(
(
)

.



(16)
Hence χ ( x,λ ) satisfies equation (1) on I , the boundary condition (3) and the
transmission conditions (4) and (5).
Let W (φ , χ )( x, λ ) denote the Wronskian of φ (., λ ) and χ (., λ ) . Since the
Wronskian W (φ , χ )( x, λ ) is independent on variable x ∈ I i ( i = 1, 2 ) and φ (., λ )
and χ (., λ ) are entire functions of parameter λ for each x ∈ I i ( i = 1, 2 ) , then the
function
ωi ( λ ) := W (φ , χ )( x, λ ) = φ1i ( x, λ ) χ 2i ( x, λ ) − φ2i ( x, λ ) χ1i ( x, λ ) ,
(17)
are entire function of parameter λ . After a short calculation, we see that
ω1 ( λ ) = ω2 ( λ ) for each λ ∈ ℂ . The zeros of the functions ω1 ( λ ) and ω2 ( λ )
coincide. Then, we may take into consideration the characteristic function ω ( λ )
as
ω ( λ ) := ω1 ( λ ) = ω2 ( λ ) .
(18)
Lemma 2.3 All eigenvalues of the problem (1)-(5) are just zeros of the function
ω ( λ ) . Moreover, every zero of ω ( λ ) has multiplicity one.
Proof: Since the function φ1 ( .,λ ) and φ2 ( .,λ ) satisfy the boundary condition (2)
and the transmission conditions (4) and (5) to find the eigenvalues of the problem
78
Fatma Hıra et al.
(1)-(5), we have to insert the functions φ1 ( .,λ ) and φ2 ( .,λ ) in the boundary
condition (3) and find the roots of this equation.
By equation (1) we obtain for λ , µ ∈ ℂ , λ ≠ µ ,
d
{φ1 ( x, λ ) φ2 ( x, µ ) − φ1 ( x, µ ) φ2 ( x, λ )}= (µ − λ ) {φ1 ( x, λ ) φ1 ( x, µ ) + φ2 ( x, λ ) φ2 ( x, µ )}
dx
Integrating the above equation through [ −1, 0 ) and ( 0,1] , and taking into account
the initial conditions (13), (14) and (16), we obtain
φ12 (1, λ ) φ22 (1, µ ) − φ12 (1, µ ) φ22 (1, λ ) = ( µ − λ ) {δφ11 ( 0, λ ) φ11 ( 0, µ )
0
+ ∫ (φ11 ( x, λ ) φ11 ( x, µ ) + φ21 ( x, λ ) φ21 ( x, µ ) ) dx
(19)
−1
1

+ ∫ (φ12 ( x, λ ) φ12 ( x, µ ) + φ22 ( x, λ ) φ22 ( x, µ ) ) dx  .
0

Dividing both sides of equality (19) by ( µ − λ ) and by letting µ → λ , we arrive
to the relation
0
∂φ12 (1, λ )
∂φ22 (1, λ )
2
2
2
− φ12 (1, λ )
= δ φ11 ( 0, λ ) + ∫ φ11 ( x, λ ) + φ21 ( x, λ ) dx
φ22 (1, λ )
∂λ
∂λ
−1
)
(
1
(
+ ∫ φ12 ( x, λ ) + φ22 ( x, λ )
0
2
2
) dx.
(20)
We show that equation
ω ( λ ) = φ12 (1, λ ) sin β + φ22 (1, λ ) cos β = 0 ,
(21)
has only simple roots. Assume the converse, that is, equation (21) has a double
root λ0 . Then the following two equations hold
φ12 (1, λ0 ) sin β + φ22 (1, λ0 ) cos β = 0 ,
(22)
∂φ12 (1, λ0 )
∂φ (1, λ0 )
sin β + 22
cos β = 0 .
∂λ
∂λ
(23)
The equation (22) and (23) imply that
Eigenvalue Problem for Discontinuous Dirac…
φ22 (1, λ0 )
∂φ12 (1, λ0 )
∂λ
− φ12 (1, λ0 )
∂φ22 (1, λ0 )
∂λ
79
= 0.
(24)
Combining (24) and (20), with λ = λ0 , we obtain
∫ ( φ ( x, λ )
1
1
2
0
+ φ2 ( x, λ0 )
−1
2
) dx +δ φ
11
( 0, λ0 )
2
=0.
(25)
It follows that φ1 ( x,λ0 ) = φ2 ( x,λ0 ) = 0 , which is imposible. This proves the
lemma.
□
3
Asymptotic Approximate Formulas
Now we derive asymptotic formulas for the eigenvalues and for the vector-valued
eigenfunctions of the problem (1)-(5). Similarly to the Sturm-Liouville problem, it
follows that there exist infinitely many eigenvalues. We essentially used the
integral equations for the fundamental system of solutions of the Sturm-Liouville
equation (see [1, 2, 6, 7, 9, 12, 14]). In the case of the Dirac system, the explicit
form of the vector-valued function φ ( .,λ ) from equalities (13) and (14), we
obtain the following integral equations:
φ11 ( x,λ ) = cos ( λ ( x + 1) − α ) −
x
∫ sin ( λ ( x − y ) ) r1 ( y )φ11 ( y,λ ) dy
−1
(26)
x
−
∫ cos ( λ ( x − y ) ) r2 ( y )φ21 ( y,λ ) dy,
−1
φ21 ( x,λ ) = sin ( λ ( x + 1) − α ) +
x
∫ cos ( λ ( x − y ) ) r1 ( y )φ11 ( y,λ ) dy
−1
(27)
x
−
∫ sin ( λ ( x − y ) ) r2 ( y )φ21 ( y,λ ) dy,
−1
(
)
( (
)
(
))
φ12 ( x, λ ) = φ11 0− ,λ cos ( λ x ) − φ21 0− ,λ + λδφ11 0− ,λ sin ( λ x )
x
x
0
0
− ∫ sin ( λ ( x − y ) ) r1 ( y )φ12 ( y, λ ) dy − ∫ cos ( λ ( x − y ) ) r2 ( y )φ22 ( y, λ ) dy,
(28)
80
Fatma Hıra et al.
(
( (
)
)
(
))
φ22 ( x,λ ) = φ11 0− ,λ sin ( λ x ) + φ21 0− , λ + λδφ11 0− ,λ cos ( λ x )
x
x
0
0
+ ∫ cos ( λ ( x − y ) ) r1 ( y )φ12 ( y, λ ) dy − ∫ sin ( λ ( x − y ) ) r2 ( y )φ22 ( y, λ ) dy,
(29)
for the components of the vector-valued function φ ( .,λ ) .
Lemma 3.1 For λ → ∞ ,
the solution φ (., λ ) has the following asymptotic
representation uniformly with respect to x, x ∈ I ,
1
φ11 ( x,λ ) = cos ( λ ( x + 1) − α ) − Q1 ( x ) sin ( λ ( x + 1) − α ) + O   ,
λ 
1
φ21 ( x,λ ) = sin ( λ ( x + 1) − α ) + Q1 ( x ) cos ( λ ( x + 1) − α ) + O   ,
λ
(30)
(31)
φ12 ( x,λ ) = λδ ( Q1 ( 0 ) sin ( λ − α ) − cos ( λ − α ) ) ( sin ( λ x ) + Q2 ( x ) cos ( λ x ) )
+ (1 − Q1 ( 0 ) Q2 ( x ) ) cos ( λ ( x + 1) − α ) − ( Q1 ( 0 ) + Q2 ( x ) ) sin ( λ ( x + 1) − α )
1
− δ R1 ( x ) ( Q1 ( 0) sin ( λ − α ) − cos ( λ − α ) ) sin ( λ x ) + O   ,
λ
φ22 ( x,λ ) = λδ ( Q1 ( 0 ) sin ( λ − α ) − cos ( λ − α ) ) ( Q2 ( x ) sin ( λ x ) − cos ( λ x ) )
(32)
+ (1 − Q1 ( 0 ) Q2 ( x ) ) sin ( λ ( x + 1) − α ) + ( Q1 ( 0 ) + Q2 ( x ) ) cos ( λ ( x + 1) − α )
1
− δ R2 ( x ) ( Q1 ( 0 ) sin ( λ − α ) − cos ( λ − α ) ) cos ( λ x ) + O   ,
λ
(33)
where
x
1
Q1 ( x ) = ∫ ( r1 ( y ) + r2 ( y ) ) dy,
2 −1
x
1
Q2 ( x ) = ∫ ( r1 ( y ) + r2 ( y ) ) dy,
20
1
( r1 ( x ) − r2 ( x ) + r1 ( 0 ) − r2 ( 0 ) ) ,
4
1
R2 ( x ) = ( r1 ( x ) − r2 ( x ) − r1 ( 0 ) + r2 ( 0 ) ) .
4
(34)
R1 ( x ) =
(35)
Proof: These asymptotic formulas are obtained by solving the systems (26)-(29)
by the method of successive approximations.
□
Eigenvalue Problem for Discontinuous Dirac…
81
The characteristic function of the problem (1)-(5) is defined by
ω ( λ ) = φ12 (1, λ ) sin β + φ22 (1, λ ) cos β .
(36)
The function ω ( λ ) is an analytic functions in λ and its zeros are precisely
eigenvalues of the problem (1)-(5).
Theorem 3.1 The characteristic function ω ( λ ) has the following asymptotic
representation:
ω ( λ ) = λδ ( Q1 ( 0 ) sin ( λ − α ) − cos ( λ − α ) ) ( Q2 (1) sin ( λ + β ) − cos ( λ + β ) )
+ (1 − Q1 ( 0 ) Q2 (1) ) sin ( 2λ − α + β ) + ( Q1 ( 0 ) + Q2 (1) ) cos ( 2λ − α + β )
 r ( 0 ) − r1 ( 0 )
− δ ( Q1 ( 0 ) sin ( λ − α ) − cos ( λ − α ) )  2
cos ( λ + β )
4

+
r1 (1) − r2 (1)
4
(37)

1
cos ( λ − β )  + O   .
λ

Proof: By substituting the obtained asymptotic formulas for φ12 ( ., λ ) and
φ22 (., λ ) (equations (32) and (33)) in the definitions of (36), then (37) can be
□
obtained.
Theorem 3.2 The eigenvalues λn of the problem (1)-(5) have the following
asymptotic representation:
Case 1. If Q1 ( 0 ) ≠ 0, Q2 (1) ≠ 0, then


1
1
 
1
− β + arccot ( Q2 (1) ) + O   ,
n
λɶ =  n +  π + α + arccot ( Q1 ( 0 ) ) + O   ,
2
n

1

λɶɶ =  n +  π
2

(38)
Case 2. If Q1 ( 0 ) = 0, Q2 (1) ≠ 0, then


1
1
 
λɶ =  n +  π + α + O   ,
2
n

1

1
λɶɶ =  n +  π − β + arccot ( Q2 (1) ) + O   ,
2

n
(39)
82
Fatma Hıra et al.
Case 3. If Q1 ( 0 ) ≠ 0, Q2 (1) = 0, then


1
1
 
λɶ =  n +  π + α + arccot ( Q1 ( 0 ) ) + O   ,
2
n

(40)
1

1
λɶɶ =  n +  π − β + O   ,
2

n
Case 4. If Q1 ( 0 ) = 0, Q2 (1) = 0, then


1
1
 
1
− β + O ,
n
λɶ =  n +  π + α + O   ,
2
n

1

λɶɶ =  n +  π
2

where Q1 ( 0 ) =
0
(41)
1
1
1
r1 ( y ) + r2 ( y ) ) dy, Q2 (1) = ∫ ( r1 ( y ) + r2 ( y ) ) dy.
(
∫
2 −1
20
Proof: The asymptotic formula (37) can be written in the form
ω ( λ ) = λδ ( Q1 ( 0 ) sin ( λ − α ) − cos ( λ − α ) ) ( Q2 (1) sin ( λ + β ) − cos ( λ + β ) ) + O (1) .
(42)
And from equation (42) denoting
w1 ( λ ) = λδ ( Q1 ( 0 ) sin ( λ − α ) − cos ( λ − α ) ) ( Q2 (1) sin ( λ + β ) − cos ( λ + β ) )
and
w2 ( λ ) = O (1) .
According to [1-4], see also [11, 18], the proof is completed by using well-known
Rouche's theorem.
□
By putting asymtotic formulas (38)-(41) for eigenvalues λn in the asymptotic
formulas (30)-(33) for the solutions φ (., λ ) , the asymptotic formulae of the eigenvector-functions φ (., λn ) can be obtained.
Eigenvalue Problem for Discontinuous Dirac…
83
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