# Inverse Problems for Dirac Operator with Boundary ```Cumhuriyet &Uuml;niversitesi Fen Fak&uuml;ltesi
Fen Bilimleri Dergisi (CFD), Cilt 38, No. 2 (2017)
ISSN: 1300-1949
Cumhuriyet University Faculty of Science
Science Journal (CSJ), Vol. 38, No. 2 (2017)
ISSN: 1300-1949
http://dx.doi.org/10.17776/cumuscij.308486
Inverse Problems for Dirac Operator with Boundary Conditions Involving
A Herglotz-Nevanlinna function
Yal&ccedil;ın G&Uuml;LD&Uuml;, A. Sinan &Ouml;ZKAN
Department of Mathematics, Faculty of Science, Cumhuriyet University 58140 Sivas, Turkey
Abstract. In this paper, we deal with the inverse problems for Dirac operator with rationally eigenvalue dependent boundary
condition and linearly eigenvalue dependent jump conditions. We prove that when Q (x ) is known on  1 ,1 then only
2 
one spectrum excluding a finite number of eigenvalues is sufficient to determine Q (x ) on the interval 0,1 and the other
coefficients of the problem. Moreover, it is shown that Q (x ) is uniquely determined by the classical spectral data, i.e.,
eigenvalues and normalising numbers.
Keywords. Inverse problem, impulsive differential equation, boundary value problem
Herglotz-Nevanlinna Fonksiyonu İ&ccedil;eren Sınır Koşullarına Sahip Dirac
Operat&ouml;r&uuml; i&ccedil;in Ters Problemler
&Ouml;zet. Bu makalede, sınır koşulları spektral parametreye rasyonel, s&uuml;reksizlik koşulları ise lineer şekilde bağlı olan
Dirac operat&ouml;r&uuml; i&ccedil;in ters problem ele alınmıştır. (&frac12;,1) aralığında Q(x) potansiyel fonksiyonu biliniyorken, tek
spektrumun sonlu sayıda &ouml;zdeğerlerin dışında (0,1) aralığında Q(x) potansiyel fonksiyonunu ve problemin diğer
katsayılarını tek olarak belirlediği ispatlanmaktadır. Ayrıca, Q(x) fonksiyonunun klasik spektral veriler, yani
&ouml;zdeğerler ve normalleştirici sayılar yardımıyla tek olarak belirlendiği g&ouml;sterilmektedir.
Anahtar kelimeler: Ters problem, s&uuml;reksiz Dirac operat&ouml;r, sınır-değer problem
1.
INTRODUCTION
We consider the boundary value problem L = L(Q, f ,  ,  ,  ) generated by the system of Dirac
differential equation
 1 1 
y := By '  Q( x) y = y, x  I :=  0,    ,1
 2 2 
(1)
 p( x) q( x) 
 0 1
y ( x)
, y ( x) =  1  subject to the boundary conditions
, Q ( x) = 
 q( x)  p( x) 
 1 0
 y2 ( x) 
with B = 
U ( y ) := y1 (0) = 0
(2)
V ( y ) := y 2 (1)  f ( ) y1 (1) = 0
(3)
_____________
* Corresponding author. Email address: [email protected]
http://dergi.cumhuriyet.edu.tr/cumuscij/index &copy;2016 Faculty of Science, Cumhuriyet University
GULDU, OZKAN
and the jump conditions
1
1

y1 (  0) = y1 (  0)

2
2
 1
1
1
 y 2 (  0) =  1 y 2 (  0)  (    ) y1 (  0).
2
2
 2
(4)
Here p(x) and q(x) are real valued functions in L2 (0,1) ;  is a spectral parameter;

,  and

are real numbers,  &gt; 0 ,  &gt; 0 ; f ( ) is a rational function of Herglotz–Nevanlinna type such that
N
fk
k =1   g k
f ( ) = a  b  
where a, b,
f k , gk
are real numbers, a &gt; 0,
f k &gt; 0, g1 &lt; g2 &lt; ... &lt; g N .
We shall note that if
f ( ) =  then the condition (3) will be as the condition y1 (1) = 0 .
Inverse problems for Dirac operator have been rather well studied and continued acceleratingly(see ,
, ,  and references therein)
When the spectrum of a differential operator and the potential on a half of the interval are known, one
can recover the differential operator on whole interval. Such problems are known as half-inverse
problems. The first study on the half-inverse problem for Sturm-Liouville operator was made by
Hochstadt and Lieberman in 1978 . They proved that if the potential function q  x  of Sturm-
1 
2 
Liouville equation is given over the interval  ,1 , then one spectrum is enough to determine q  x  on


the interval  0,
1
 . After that in 1984, Hald  showed that if the potential is known over half the
2
interval and if one boundary condition is given, then the eigenvalues uniquely determine the potential
and the other boundary condition. In 1994, T. N. Arutyunyan  proved that the potential Q (x ) is
uniquely determined by the eigenvalues and normalising coefficients. Later, in  and in , it is
provided some results in inverse spectral analysis with partial data on the potential.
Spectral problems involving eigenvalue dependent boundary conditions arise in various problems of
mathematics as well as in applications. Firstly, in 1973, Walter  was interested in an expansion
theorem for the this kind of eigenvalue problem. In 1977, Fulton  also examined the Sturm-Liouville
eigenvalue problem. Inverse problems for some classes of differential operators linearly eigenvalue
dependent are analysed in diverse papers (see [1, 2, 9, 17, 22]). More general boundary conditions are
observed in [4-6, 10, 12, 26, 29, 30, 34,35]. On the other hand, when f ( ) is a rational function of
Herglotz–Nevanlinna type, direct and inverse spectral problems for Sturm-Liouville operator were
investigated in [7, 8, 27].
Recently, also some new uniqueness results in inverse spectral analysis with partial information on the
potential for Sturm-Liouville and Dirac operator have been given (see , [28, 31-33]).
In the present paper, our research is connected to a Dirac equation with rationally eigenvalue dependent
boundary conditions and linearly eigenvalue dependent jump conditions. We prove that the potential in
205
Inverse Problems for Dirac Operator with Boundary
(0,1) and the remaining coefficients of the boundary value problem can be uniquely determined by the
1
potential in ( ,1) and one spectrum excluding finite eigenvalues. We also show that Q (x ) is uniquely
2
determined by the sequences of eigenvalues and normalising numbers.
2.
PRELIMINARIES
Firstly, we define an operator such that (1)-(4) can be regarded as an eigenvalue problem of it. See 
and  for a similar operator associated with the Sturm- Liouville problem.
Consider the space
H = L2 (0,1)  L2 (0,1)  CN 2
with the inner product defined by
Y Z
y ( x) z ( x)  y ( x) z ( x)dx   f
1
Y , Z :=
N
1
1
2
k
2
k =1
0
where
k

k
YN 1 Z N 1 YN  2 Z N  2

a

Y =  y1 ( x), y2 ( x), Y1, Y2 ,...YN 2  , Z = z1 ( x), z2 ( x), Z1 , Z 2 ,...Z N 2 .
(5)
Define the operator T
with the domain
D(T ) = {Y  H :
y1 (0) = 0, y1 (
y1 ( x)
and y 2 ( x ) are absolutely continuous in I , y  L2 (0,1),
1
1
 0)  y1 (  0) = 0,
2
2
YN 1 = ay1 (1),
1
YN  2 = y1 (  0)} such that
2
T Y  = Z = z1 ( x), z2 ( x), Z1, Z 2 ,...Z N 2 ,
 z ( x) 
where z ( x) =  1  = y ( x) ,
 z 2 ( x) 
Z i = g iYi  f i y1 (1), for i = 1, N
N
Z N 1 = y2 (1)  by1 (1)  Yk
k =1
and
1
1
1
Z N  2 = y2 (  0)   1 y1 (  0)  y1 (  0)
2
2
2
Theorem 1 Eigenvalues of the operator T coincide with eigenvalues of the problem L .
206
GULDU, OZKAN
Proof. The following equalities obtained from TY = Y are enough to see validity of this
theorem:
g1Y1  b1 y1 (1) = Y1
g 2Y2  b2 y1 (1) = Y2
...
g NY1  bN y1 (1) = YN
N
y2 (1)  by1 (1)  Yk = ay1 (1)
k =1
1
1
1
1
y2 (  0)   1 y2 (  0)  y1 (  0) = y1 (  0).
2
2
2
2
Let the function
 ( x,  ) = 1 ( x,  ) 2 ( x,  )
T
be the solution of (1) under the initial
conditions
0
 (0,  ) =  
 1
and under the jump conditions (4).  ( x,  ) satisfies the following integral equations:
For x &lt;
1
2
x
x
0
0
1 ( x,  ) = sin x   sin  ( x  t )  (t ,  )dt   cos  ( x  t )  (t ,  )dt
x
x
0
0
 2 ( x,  ) =  cos x   cos  ( x  t )  (t ,  )dt   sin  ( x  t )  (t ,  )dt
and for x &gt;
1
2
1 ( x,  ) =   sin x    sin  (1  x) 
 
1/2



sin  ( x  t )    sin  (1  x  t )   (t ,  )dt

cos  ( x  t )    cos  (1  x  t )   (t ,  )dt
0
 
1/2


0
207
(6)
Inverse Problems for Dirac Operator with Boundary
 sin  ( x  t )
x



(t ,  )  cos  ( x  t )  (t ,  ) dt
1/2
(    ) 

cos x  cos  (1  x)   [cos  ( x  t )  cos  (1  x  t )] (t ,  )dt
2 
0
1/2

1/2

  [sin  ( x  t )  sin  (1  x  t )]  (t ,  )dt 
0

2 ( x,  ) =   cos x    cos  (1  x)
 
1/2



cos  ( x  t )    cos  (1  x  t )   (t ,  )dt

sin  ( x  t )    sin  (1  x  t )   (t ,  )dt
0
 
1/2


0
 cos  ( x  t )
x



(t ,  )  sin  ( x  t )  (t ,  ) dt
1/2
(    ) 

sin x  sin  (1  x)   [sin  ( x  t )  sin  (1  x  t )] (t ,  )dt
2 
0
1/2

1/2

  [cos  ( x  t )  cos  (1  x  t )]  (t ,  )dt 
0

where
 (t,  ) = 1 (t,  ) p(t )  2 (t,  )q(t ),  (t,  ) = 1 (t,  )q(t )  2 (t,  ) p(t )
and
1
1
  =   .
2

Moreover, it is shown in  that,  ( x,  ) has a representation as follows:


x
1
2
 ( x,  ) =  0 ( x,  )  K ( x, t ) 0 (t ,  )dt , for x   0, 
0
where
 0 ( x,  ) = sin x  cos x  , K ( x, t ) = Kij ( x, t )i , j =1,2
T
Kij x,.  L2 (0,1) for each fixed x.
The following asymptotic relations can be proved easily;
208
(7)
GULDU, OZKAN

 sin x  o(exp x ),
1 ( x,  ) = 

 cos x  cos  (1  x)  o( exp x ),
 2

  cos x  o(exp x),
 2 ( x,  ) = 

 sin x  sin  (1  x)  o( exp x),
 2
1
2
1
x&gt;
2
x&lt;
1
2
1
x&gt;
2
x&lt;
where  = Im .
It is clear that f ( ) can be written as follows:
a ( )
b ( )
f ( ) =
where
a ( ) = a  b   g k    f k   g j  ,
N
N
N
k =1
k =1
j =1
j k
N
b( ) =   g k  .
k =1
Let the function
 ( x,  ) =  1 ( x,  )  2 ( x,  )
T
be the solution of (1) under the initial
conditions
 b ( ) 

  a ( ) 
 (1,  ) = 
(8)
and under the jump conditions (4).
Consider the function
 ( ) := W  ,   =  2 ( x,  ) 1 ( x,  )  1 ( x,  ) 2 ( x,  )
(9)
= a ( )1 (1,  )  b( ) 2 (1,  ) =  1 (0,  ).
It is obvious that  ( ) is entire function and the zeros, namely
the problem (1)-(4). Additionally, the equality
 n =  2 (0,  n ) =
n nZ of ( ) are eigenvalues of
 ( x, n ) =  n ( x,  )
b ( n )
.
1 (1,  n )
209
holds for all
x
and  , where
Inverse Problems for Dirac Operator with Boundary
3.
UNIQUENESS THEOREMS
3.1 According to the mixed given data
The first main result of this work is a generalized of Hochstadt and Lieberman theorem . We prove
1 
2 
that when Q (x) is known on  ,1 then only one spectrum excluding a finite number of eigenvalues
is sufficient to determine Q (x ) on the interval 0,1 and the coefficients  ,  and  . Together with
~
L , we consider the problem L
of the same form but with a different coefficients
 ~p ( x) q ( x)  ~ ~
~
~
Q ( x) = 
~ ,  ,  and  . It is assumed in what follows that if a certain symbol s
 q ( x)  p ( x) 
s with tilde denote the analogous object
denotes an object related to L , then the corresponding symbol ~
~
related to L . Let us denote by
Let
Z0
 ( x, n ),
the eigenfunction which corresponds to
n .
be any subset with N  1 elements of Z. Denote  := n nZ \ Z and consider the
0
following representation:
 ( ) = R ( )    n 
nZ 0



where R ( ) = C  1    , C is a constant which depends only on

n 
n  
requires minor modifications).
Theorem 2 If  = ~ ,
2 
~
 =
and  = ~ .
Before the proof of theorem, we need to prove the following lemma.
Lemma 1 i) The eigenvalues
d( )
 0 , i.e.,
d  = 
(the case (0) = 0
~
~
~
1 
f ( )  f ( ) and Q( x) = Q ( x) on  ,1 then Q( x) = Q ( x)
almost everywhere on 0,1 ,  = ~ and
ii)
n
n
are real numbers.
n nZ are simple zeros of ( ) .
n
210
GULDU, OZKAN
Proof. i) According to the Theorem 1, it is enough to show that eigenvalues of T are real. For
Y in D(T ), we calculate that
TYk Y k TYN 1Y N 1 TYN  2 Y N  2


fk
a

k =1
1
N
TY , Y = y ( x) y ( x)dx  
0
1


1
=  p( x) y1 ( x)  y2 ( x) dx   2 Re y2 ( x) y1 ( x)q( x)dx
2
2
0
0
1  1 
1  1 
 y1   0  y2   0   y1 0 y2 0  y1   0  y2   0 
2  2 
2  2 
1
N
g kYk  f k y1 (1)Y k
0
k =1
fk
 y1 1 y2 1   2 Re y2 ( x) y1 ( x)dx  
N


  y2 (1)  by1 (1)  Yk  y1 (1)
k =1 

1
1
 1
 1
   y2 (  0)   1 y2 (  0)  y1 (  0) y1 (  0).
2
2
 2
 2
By using the structure of T , we obtain
1


1
TY , Y =  p( x) y1 ( x)  y2 ( x) dx   2 Re y2 ( x) y1 ( x)q( x)dx
2
2
0
0
1
N
0
k =1
  2 Re y2 ( x) y1 ( x)dx  

N
gk
2
Yk  2 Re y1 (1)Y k
fk
k =1

2
1
 b y1 1   y1 (  0) .
2
2
We conclude that TY, Y are real for each Y in D(T ) . Thus, all eigenvalues of the operator T (or
problem L ) are real numbers.
ii) Assume
n  g k . If n = gk ,
Write the equation (1) for
the proof is similar.
 x, n  and  x,   respectively.
  2' ( x, n )  p x 1  x, n   qx  2 x, n  = n1 x, n ,
 '
 1 ( x, n )  q x 1 x, n   p x  2  x, n  = n 2  x, n 
211
Inverse Problems for Dirac Operator with Boundary
  2' ( x,  )  px  1 x,    q x  2  x,   =  1 x,  ,

'
 1 ( x,  )  qx  1 x,    p x  2  x,   =  2  x,  .
After some operations, we obtain

1 ( x,  n ) 2 ( x,  )   2 ( x,  n ) 1 ( x,  )


1
2
0

1
1
2




1
= (  n )  1 ( x,  )1 ( x, n )  2 ( x,  ) 2 ( x, n )dx.
0
Take into account of boundary and discontinuity conditions to get
 1 (0,  )  a( )1 (1, n )  b 2 (1, n )
 1 ( 1  0, n ) 2 ( 1  0,  )   2 ( 1  0, n ) 1 ( 1  0,  )
2
2
2
2
 1 ( 1  0, n ) 2 ( 1  0,  )   2 ( 1  0, n ) 1 ( 1  0,  )
2
2
2
2
1
= (  n )  1 ( x,  )1 ( x, n )  2 ( x,  ) 2 ( x, n )dx.
0
Using (4) and (9) we get
1
    b 1 (1, n ) f    f n    (  n )1 (  0, n ) 1 ( 1  0,  )
2
2
1
= (  n )  1 ( x,  )1 ( x, n )  2 ( x,  ) 2 ( x, n )dx.
0
If we divide both side of this equality by
(  n )
and take limit for
  n ,
1 2

1
 n  1 ( x, n )   22 ( x, n ) dx  12 (1, n ) f n   12 (  0, n ) =  (n ) .
2
0



Since  &gt; 0,  &gt; 0 and f n  &gt; 0 for all
Let us write the equation (1) for

n ,
~,
and 
B ( x,  )  Q( x) ( x,  ) =  ( x,  )
212
(n )  0.
GULDU, OZKAN
~
B~ ( x,  )  Q( x)~( x,  ) = ~( x,  )
Multiply these equalities by
~ T ( x,  ) and  T ( x,  ) respectively from left hand side and subtract then
we get
d
1 ( x,  )~2 ( x,  )  ~1 ( x,  ) 2 ( x,  )
dx
~
= Q( x)  Q ( x)  ( x,  )~ ( x,  ).

(10)

After integrating this equality on 0,1, taking into account the hypothesis
~
1 
Q( x) = Q ( x) on  ,1 , we find
2 

1 ( x,  )~2 ( x,  )  ~1 ( x,  ) 2 ( x,  ) 1/2
0  1/2 
1


1
2
=  p ( x )  ~
p ( x) ~ ( x,  ) T J ,  ( x,  ) dx
(11)
0
1 0 
 . Denote
 0 1
where .,. denotes the classical inner product of C2 and J := 
~ (1,  )   (1,  )
~ (1,  ).
H ( ) :=  2 (1,  )
1
1
2
(12)
One can calculate from (4) and (11) that
1
2
 2 (1,  )~1 (1,  )  1 (1,  )~2 (1,  ) =  p( x)  ~
p ( x) ~ ( x,  ) T J ,  ( x,  ) dx (13)
0
1
1
  1
 ~  1
 1  ~ 1 (  0,  )~2 (  0,  )  1   2 (  0,  )~1 (  0,  )
2
2
  2
  2



1
1
~
  ~    ~  ~ 1 (  0,  )~1 (  0,  ).
2
2
It is obtained from
~
f ( )  f ( ) that  2 (1, n )~1 (1, n )  1 (1, n )~2 (1, n ) = 0 . Therefore,
H (n ) = 0 for all n  .
Now, define F ( ) :=
H ( )
which is an entire function on  . From asymptotic relations of
R ( )
1
. Therefore, from Liouville’s Theorem,


 
1 ( x,  ) and  2 ( x,  ) , it is valid that F ( ) = O
213
Inverse Problems for Dirac Operator with Boundary
F ( )  0
and
1
2
 p ( x )  p ( x )   ( x ,  )
~
~
H ( )  0.
so
T
J ,  ( x,  ) dx =
0
One
can
calculate
that
1
2
1
 p( x)  ~p ( x)dx  o1, for   ,   R . Therefore
2 0
it can be written the following equality from (13),



1 ~
1  ~  
~
     ~  ~ 1  cos   o1    ~ sin   o1
2
2   

1
2
1
 p( x)  ~p ( x)dx = o1 for   ,   R.

20
We obtain from the last relation that,
1
2
~   
= = ~ = and so, 
 ~  ~
= ~,  = ~,
~
 = ,
 p( x)  p ( x)dx = 0. As a consequent,
~
0
1
2
 p ( x )  p ( x )   ( x,  )
~
~
T
J ,  ( x,  ) dx = 0 holds on the whole  -plane.
0
On the other hand, the following equality is valid,
x
x
0
0
~( x,  ) T J ,  ( x,  ) =  cos 2x  K 1 ( x, t ) cos 2tdt  K 2 ( x, t ) sin 2tdt
where
Ki ( x, t ),
x
i = 1,2, depend only on
and
t. It follows from (13) that
1
2
x
x


0 P( x)cos 2x  0 K1 ( x, t ) cos 2tdt  0 K 2 ( x, t ) sin 2tdtdx = 0
(14)
for all  , where P ( x) :=  p ( x)  ~
p ( x ). This can be rewritten as
1


2


0 cos 2  P( )   P( x) K1 ( x, t )dx dt


1
2
1
2
1
2
0

  sin 2t P ( x ) K 2 ( x, t ) dxdt = 0
214
(15)
GULDU, OZKAN
Therefore, we obtain from the completeness of the vector functions
cos 2
sin 2t  in
T
1
1
1
p ( x) for x  (0, ) . This completes the proof.
L2 (0, )  L2 (0, ) that P(x) = 0 , i.e. p( x) = ~
2
2
2
3.2 According to the classical spectral data
In , it is proven that the coefficients of the problem L are uniquely determined by the Weyl
function. We aim to prove uniqueness of the coefficients according to the eigenvalues and normalising
~
q~ ( x ) 
~ .
 q ( x)  p ( x) 
 p ( x)
~
~
numbers, namely spectral data. Consider the problem L with the coefficient Q ( x ) =  ~
For an element
Y =  y1 ( x), y2 ( x), Y1, Y2 ,...YN 2 
in H , the norm of Y
is defined by
2
Y := Y , Y . From (5), we get
1


N
Y =  y1 ( x)  y2 ( x) dx  
2
2
2
k =1
0
Let
n
Yk
fk
2

YN 1
a
2
Y
 N 2

2
be an eigenvalue of T (or the problem L ) and Y (n) eigenvector for
numbers  n := Y (n)
2
(16)
n . Then the
are called as normalizing numbers.
Lemma 2 The equality
 n = 12 ( x, n )   22 ( x, n )dx
1
0
 f ' (n )12 (1, n )  12 (d1  0, n )
is valid.
Proof. Let
n  g k .
Since
n = g k
is equivalent to
1 1, gk  = 0 ,
this case requires minor
modification in the following proof.
Using the structure of T and the inner product in (5), a direct calculation yields
1


Y (n) =  12 ( x, n )   22 ( x, n ) dx
2
0
N

k =1
Yk
fk
2

YN 1
a
2
 YN  2


2
215
Inverse Problems for Dirac Operator with Boundary
1


=  12 ( x, n )   22 ( x, n ) dx
0
N
fk
 a12 (1, n )
2
k =1 g k  n 
 12 (1, n )
1


=  12 ( x, n )   22 ( x, n ) dx
0
N


fk
1
 12 (1, n )a  
 12 (  0, n )
2
2
 k =1 g k  n  
1


1
=  12 ( x, n )   22 ( x, n ) dx  f ' (n )12 (1, n )  12 (  0, n )
2
0
Take into account Lemma 1 and Lemma 2 to get the following relation
' ( n ) =  n  n
(17)
The Weyl function is defined as follows;
m ( ) =
~ ( ) then
Theorem 3  If m( ) = m
 2 (0,  )
.
 ( )
(18)
~
L = L ; i.e. the Weyl function
m( ) determines
uniquely the problem L .
Theorem 4 If
 n ,  n  = ~n , ~n  then L = L~; i.e. the spectral data n , n  determine
uniquely the problem L .
Proof.
Since
~
n = n ,
~
( ) = ( ).
Therefore,
from
(17)
 n = ~n
so
 2 (0, n ) = ~2 (0, n ). Hence the function defined as
G ( ) :=
 2 (0,  )  ~2 (0,  )
 ( )
is an entire on  . Moreover, one can obtained that G( ) = o(1) for
  .
G ( )  0 and so
~
~ ( ) . Consequently, from Theorem 3, L = L
.
 2 (0,  )  ~2 (0,  ). From (18) we have m( ) = m
216
GULDU, OZKAN
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