C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat. Volum e 67, N umb er 2, Pages 156–164 (2018) D O I: 10.1501/C om mua1_ 0000000870 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
SELFADJOINT SINGULAR DIFFERENTIAL OPERATORS FOR FIRST ORDER
PEMBE IPEK AND ZAMEDDIN I. ISMAILOV
Abstract. The parametrization of all selfadjoint extensions of the minimal operator generated by …rst order linear symmetric singular di¤erential-operator expression in the Hilbert space of vector-functions de…ned at the right semi-axis has been given. To this end we use the Calkin-Gorbachuk method. Finally, the structure of spectrum set of such extensions is researched.
1. Introduction
It is known that fundamental question on the parametrization of selfadjoint ex-tensions of the linear closed densely de…ned with equal de…ciency indices symmetric operators in a Hilbert space has been investigated by J. von Neumann [11] and M. H. Stone [10] …rstly. Applications of these results to any scaler linear even or-der symmetric di¤erential operators and representation of all selfadjoint extensions in terms of boundary conditions have been investigated by I. M. Glazman-M. G. Krein- M. A. Naimark (see [5,8]). In mathematical literature there is co-called Calkin-Gorbachuk method (see [6,9]).
The motivation of this paper originates from the interesting researches of W. N. Everitt, L. Markus, A. Zettl, J. Sun, D. O’Regan, R. Agarwal [2,3,4,12] in scaler cases. Throughout this paper A. Zettl’s and J. Suns’s view about these topics is to be taken into consideration [12]. A selfadjoint ordinary di¤erential operator in a Hilbert space is generated by two things:
(1) a symmetric ( formally selfadjoint) di¤erential expression;
(2) a boundary condition which consists selfadjoint di¤erential operators.
And also the geometrical place in plane of the spectrum of given selfadjoint di¤er-ential operator is one of the important questions of this theory.
In this work in Section 3 the representation of all selfadjoint extensions of the symmetric singular di¤erential operator, generated by …rst order symmetric
Received by the editors: April 12, 2017, Accepted: June 28, 2017. 2010 Mathematics Subject Classi…cation. 47A10, 47B25.
Key words and phrases. Symmetric and selfadjoint di¤erential operators, de…ciency indices, spectrum.
c 2 0 1 8 A n ka ra U n ive rsity. C o m m u n ic a tio n s Fa c u lty o f S c ie n c e s U n ive rs ity o f A n ka ra -S e rie s A 1 M a t h e m a tic s a n d S t a tis tic s . C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra -S é rie s A 1 M a t h e m a tic s a n d S t a tis t ic s .
di¤erential-operator expression (for the de…nition see [4]) in the Hilbert spaces of vector-functions de…ned at the semi-axis in terms of boundary conditions are de-scribed. In Section 4 the structure of spectrum of these selfadjoint extensions is investigated.
2. Statement of the Problem
Let us H is a separable Hilbert space and a 2 R. In the Hilbert space L2(H; (a; 1)) consider the following di¤erential-operator expression in a form (for scaler case see [4]) l(u) = i u0+1 2i 0u + Au; where: (1) : (a; 1) ! (0; 1); (2) 2 ACloc(a; 1); (3) R1 a ds (s) < 1; (4) A = A : D(A) H ! H:
By standard way the minimal operator L0corresponding to di¤erential-operator
expression l(:) in L2(H; (a; 1)) can be de…ned (see [7]). The operator L = (L 0)
is called the maximal operator corresponding to l(:) in L2(H; (a; 1)) (see [7]).
It is clear that
D(L) = fu 2 L2(H; (a; 1)) : l(u) 2 L2(H; (a; 1)g; D(L0) = fu 2 D(L) : (p u)(a) = (p u)(1) = 0g:
In this case the operator L0is symmetric and is not maximal in L2(H; (a; 1)).
In this paper, …rstly the represention of all selfadjoint extensions of the mini-mal operator L0 will be described. Secondly, structure of the spectrum of these
extensions shall be researched.
In special case when H = C the similar questions was investigated in [4] using the Glazman-Krein-Naimark method.
In left and right semi-in…nitive intervals case the similar problems have been surveyed in [1].
3. Description of Selfadjoint Extensions
In this section, the general representation of selfadjoint extensions of the minimal operator L0 will be investigated by using the Calkin-Gorbachuk method.
Firstly, let us prove the following proposition.
Lemma 1. The de…ciency indices of the operator L0 is in form (m(L0); n(L0)) =
Proof. For the simplicity of calculations it will be taken A = 0: It is clear that the general solutions of following di¤erential equations
i (t)u0 (t) +1 2i
0(t)u (t) iu (t) = 0;
in the L2(H; (a; 1)) are in forms
u (t) = exp 0 @ t Z c 2 0(s) 2 (s) ds 1 A f; f 2 H; t > a; c > a:
From these representations, we have
ku+k2L2(H;(a;1)) = 1 Z a ku+(t)k2Hdt = 1 Z a exp 0 @ t Z c 2 + 0(s) (s) ds 1 A dtkfk2 H = 1 Z a (c) (t)exp 0 @ t Z c 2 (s)ds 1 A dtkfk2 H = (c) 2 1 Z a exp 0 @ t Z c 2 (s)ds 1 A d 0 @ t Z c 2 (s)ds 1 A kfk2 H = (c) 2 2 4exp 0 @ a Z c 2 (s)ds 1 A exp 0 @ 1 Z c 2 (s)ds 1 A 3 5 kfk2 H < 1:
On the other hand it is clear that for any f 2 H the solution ku k2L2(H;(a;1)) = 1 Z a ku (t)k2Hdt = 1 Z a exp 0 @ t Z c 2 0(s) (s) ds 1 A dtkfk2 H = 1 Z a (c) (t)exp 0 @ t Z c 2 (s)ds 1 A dtkfk2 H = (c) 2 1 Z a exp 0 @ t Z c 2 (s)ds 1 A d 0 @ t Z c 2 (s)ds 1 A kfk2 H = (c) 2 2 4exp 0 @ 1 Z c 2 (s)ds 1 A exp 0 @ a Z c 2 (s)ds 1 A 3 5 kfk2 H< 1:
It follows from that n(L0) = dim ker(L iE) = dimH. This completes the
proof of theorem consequently, the minimal operator L0has at least one selfadjoint
extensions (see [6]).
De…nition 1. Let H be any Hilbert space and S : D(S) H ! H be a closed densely de…ned symmetric operator in the Hilbert space H having equal …nite or in…nite de…ciency indices. A triplet (H; 1; 2); where H is a Hilbert space, 1and
2are linear mappings from D(S ) into H, is called a space of boundary values for
the operator S if for any f; g 2 D(S )
(S f; g)H (f; S g)H= ( 1(f ); 2(g))H ( 2(f ); 1(g))H
while for any F1; F22 H; there exists an element f 2 D(S ) such that 1(f ) = F1
and 2(f ) = F2.
Lemma 2. The triplet (H; 1; 2);
1: D(L) ! H; 1(u) = 1 p 2(( p u)(1) (p u)(a)); 2: D(L) ! H; 2(u) = 1 ip2(( p u)(1) + (p u)(a)); u 2 D(L) is a space of boundary values of the minimal operator L0 in L2(H; (a; 1)):
Proof. In this case the direct calculations show for arbitrary u; v 2 D(L) that (Lu; v)L2(H;(a;1)) (u; Lv)L2(H;(a;1)) = (i u0+
1 2i 0u + Au; v) L2(H;(a;1)) (u; i v0+1 2i 0v + Av) L2(H;(a;1)) = (i u0; v)L2(H;(a;1))+ 1 2(i 0u; v) L2(H;(a;1)) (u; i v0)L2(H;(a;1)) (u;
1 2i 0v) L2(H;(a;1)) = ih( u0; v)L2(H;(a;1))+ ( 0u; v)L2(H;(a;1)) +( u; v0)L2(H;(a;1)) i = ih(( u)0; v)L2(H;(a;1))+ ( u; v0)L2(H;(a;1)) i = i (( u; v))0L2(H;(a;1)) = i ((p u;p v))0L2(H;(a;1)) = i ((p u)(1); (p v)(1))H ((p )u(a); (p )v(a))H = ( 1(u); 2(v))H ( 2(u); 1(v))H:
Now for any given elements f; g 2 H; let us …nd the function u 2 D(L) satisfying
1(u) =
1 p
2(( p
u)(1) (p u)(a)) = f and 2(u) = 1 ip2((
p
u)(1) + (p u)(a)) = g: From this
(p u)(1) = (ig + f)=p2 and (p u)(a) = (ig f )=p2 is obtained.
If we choose the function u in following form u(t) = p1
(t)(1 e
a t)(ig + f )=p2 +p1
(t)e
a t(ig f )=p2;
u 2 D(L); 1(u) = f and 2(u) = g:
Finally, using the method given in [6], we can introduce the following result. Theorem 1. If eL is a selfadjoint extension of the minimal operator L0in L2(H; (a; 1))
, then it is generated by the di¤ erential-operator expression l(:) and boundary
con-dition
(p u)(1) = W (p u)(a);
where W : H ! H is a unitary operator. Moreover, the unitary operator W in H is determined uniquely by the extension eL, i.e. eL = LW and vice versa.
Proof. It is known from [6] or [9] that all selfadjoint extensions of the minimal operator L0are described by di¤erential-operator expression l(:) and the boundary
condition
where V : H ! H is a unitary operator. So from Lemma 2, we have
(V E) ((p u)(1) (p u)(a)) + (V + E) ((p u)(1) + (p u)(a)) = 0: Hence, we obtain
(p u)(a) = V (p u)(1): Choosing W = V 1 in last boundary condition, we have
(p u)(1) = W (p u)(a):
4. The Spectrum of the Selfadjoint Extensions
In this section the structure of the spectrum of the selfadjoint extensions LW of
the minimal operator L0 in L2(H; (a; 1)) will be investigated.
First of all let us prove the following result.
Theorem 2. The spectrum of any selfadjoint extension LW is in form
(LW) = 8 < : 2 C : = 0 @ 1 Z a ds (s) 1 A 1 (2n arg ); n 2 Z; 2 0 @W exp 0 @ iA 1 Z a ds (s) 1 A 1 A 9 = ;: Proof. Consider the following problem to spectrum of the extension LW
l(u) = u + f; u; f 2 L2(H; (a; 1)); 2 R; (p u)(1) = W (p u)(a);
that is,
i (t)u0(t) +1 2i
0(t)u(t) + Au(t) = u(t) + f (t); t > a;
(p u)(1) = W (p u)(a):
The general solution of the last di¤erential equation is in the following form u(t; ) = s (c) (t)exp 0 @i(A E) t Z c ds (s) 1 A f + pi (t) 1 Z t exp 0 @i(A E) t Z s d ( ) 1 Apf (s) (s)ds; f 2 H; t > a; c > a: In this case k s (c) (t)exp 0 @i(A E) t Z c ds (s) 1 A f k2 L2(H;(a;1))= (c) 1 Z a dt (t)kf k 2 H < 1
and kpi (t) 1 Z t exp 0 @i(A E) t Z s d ( ) 1 Apf (s) (s)dsk 2 L2(H;(a;1)) = 1 Z a 1 (t)k 1 Z t exp 0 @i(A E) t Z s d ( ) 1 Apf (s) (s)dsk 2 Hdt 1 Z a 1 (t) 2 4 1 Z t k exp 0 @i(A E) t Z s d ( ) 1 A kHkf(s)kp H (s) ds 3 5 2 dt 1 Z a 1 (t) 0 @ 1 Z t ds (s) 1 A 0 @ 1 Z t kf(s)k2Hds 1 A dt 1 Z a 1 (t) 0 @ 1 Z a ds (s) 1 A 0 @ 1 Z a kf(s)k2Hds 1 A dt = 1 Z a dt (t) 1 Z a ds (s)kf(s)k 2 L2(H;(a;1))ds = 0 @ 1 Z a dt (t) 1 A 2 kfk2L2(H;(a;1))< 1:
Hence for u( :; ) 2 L2(H; (a; 1)) for 2 R. From this and boundary condition,
we have 0 @exp 0 @ i 1 Z a ds (s) 1 A W exp 0 @ iA 1 Z a ds (s) 1 A 1 A exp 0 @iA 1 Z c ds (s) 1 A exp 0 @ i a Z c ds (s) 1 A f = pi (c)W 1 Z a exp 0 @i(A ) a Z s d ( ) 1 Apf (s) (s)ds
In order to get 2 (LW); the necessary and su¢ cient condition is
exp 0 @ i 1 Z a ds (s) 1 A = 2 0 @W exp 0 @ iA 1 Z a ds (s) 1 A 1 A Consequently, 1 Z a ds (s)= 2n arg ; n 2 Z;
that is, = 0 @ 1 Z a ds (s) 1 A 1 (2n arg ); n 2 Z: This completes proof of theorem.
Example. All selfadjoint extensions L' of the minimal operator L0 generated by
di¤erential expression l(u) = it2@u(t; x) @t + itu(t; x) + Au; A : D(A) L2(0; 1) ! L2(0; 1); where Av(t) = @ 2v(t) @t2 ; D(A) = u 2 W22(0; 1) : v(0) = v(1); v0(0) = v0(1) ;
in the Hilbert space L2((1; 1) (0; 1)) in terms of boundary conditions are described by following form
(tu(t; x))(1) = ei'(tu(t; x))(1); ' 2 [0; 2 ); x 2 (0; 1): Moreover, the spectrum of such extension is
(L') = f 2 C : = 2n + (' ); n 2 Z; 2 (A)g :
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Current address : Pembe IPEK: Institute of Natural Sciences, Karadeniz Technical University, 61080, Trabzon, Turkey
E-mail address : ipekpembe@gmail.com
ORCID Address: http://orcid.org/0000-0002-6111-1121
Current address : Zameddin I. ISMAILOV: Institute of Natural Sciences, Karadeniz Technical University, 61080, Trabzon, Turkey
E-mail address : zameddin.ismailov@gmail.com
