Electronic Journal of Differential Equations, Vol. 2017 (2017), No. 143, pp. 1{8. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu FIRST-ORDER SELFADJOINT SINGULAR DIFFERENTIAL OPERATORS IN A HILBERT SPACE OF VECTOR FUNCTIONS PEMBE IPEK, BULENT¨ YILMAZ, ZAMEDDIN I. ISMAILOV Communicated by Ludmila Pulkina Abstract. In this article, we give a representation of all selfadjoint extensions of the minimal operator generated by first-order linear symmetric multipoint singular differential expression, with operator coefficient in the direct sum of Hilbert spaces of vector-functions defined at the semi-infinite intervals. To this end we use the Calkin-Gorbachuk method. Finally, the geometry of spectrum set of such extensions is researched. 1. Introduction In the first years of the previous century, von Neumann [11] and Stone [10] investigated the theory of selfadjoint extensions of linear densely defined closed symmetric operators in a Hilbert spaces. Applications to scalar linear even or- der symmetric differential operators and description of all selfadjoint extensions in terms of boundary conditions due to Glazman in his seminal work [5] and in the book of Naimark [8]. In this sense the famous Glazman-Krein-Naimark (or Everitt- Krein-Glazman-Naimark) Theorem in the mathematical literature it is to be noted. In the mathematical literature there is another method co-called Calkin-Gorbachuk method. (see [6, 9]). Our motivation for this article originates from the interesting researches of Everitt, Markus, Zettl, Sun, O'Regan, Agarwal [2, 3, 4, 12] in scalar cases. Through- out this paper we consider Zettl and Suns's view about these topics [12]. A selfad- joint ordinary differential operator in Hilbert space is generated by two things: (1) a symmetric (formally selfadjoint) differential expression; (2) a boundary condition which determined selfadjoint differential operators; And also for a given selfadjoint differential operator, a basic question is: What is its spectrum? In this work in Section 3 the representation of all selfadjoint extensions of a multipoint symmetric quasi-differential operator, generated by first-order symmet- ric differential-operator expression (for the definition see [4]) in the direct sum of Hilbert spaces of vector-functions defined at the semi-infinite intervals in terms of 2010 Mathematics Subject Classification. 47A10, 47B25. Key words and phrases. Multipoint singular differential expression; deficiency indeces; symmetric and selfadjoint differential operator; spectrum. c 2017 Texas State University. Submitted March 17, 2018. Published June 17, 2017. 1 2 P. IPEK, B. YILMAZ, Z. I. ISMAILOV EJDE-2017/143 boundary conditions are described. In sec. 4 the structure of spectrum of these selfadjoint extensions is investigated. 2. Statement of the problem 2 2 In the direct sum H = L (H; (−∞; a1))⊕L (H; (a2; 1)), H is a separable Hilbert space, and a1; a2 2 R will be considered for the multipoint differential-operator expression in the form l(u) = (l1(u1); l2(u2)); 1 l (u ) = iρ u0 + iρ0 u + A u ; k = 1; 2; k k k k 2 k k k k where (1) ρ1 :(−∞; a1) ! (0; 1), ρ2 :(a2; 1) ! (0; 1); (2) ρ1 2 ACloc(−∞; a1) and ρ2 2 ACloc(a2; 1); (3) R a1 ds = 1, R 1 ds = 1; −∞ ρ1(s) a2 ρ2(s) ∗ (4) Ak = Ak : D(Ak) ⊂ H ! H, k = 1; 2. 1 2 The minimal operators L0 and L0 corresponding to differential-operator expres- 2 2 sions l1 and l2 in L (H; (−∞; a1)) and L (H; (a2; 1)), respectively, can be defined 1 ∗ 2 ∗ by a standard processes, see[7]. The operators L1 = (L0) and L2 = (L0) are max- 2 2 imal operators corresponding to l1 and l2 in L (H; (−∞; a1)) and L (H; (a2; 1)), respectively. In this case the operators 1 2 1 2 L0 = L0 ⊕ L0 and L = L ⊕ L will be indicating the minimal and maximal operators corresponding to differential expression on H, respectively. It is clear that 1 2 2 D(L ) = fu1 2 L (H; (−∞; a1)) : l1(u1) 2 L (H; (−∞; a1)g; 1 1 p D(L0) = fu1 2 D(L ):( ρ1u1)(a1) = 0g and 2 2 2 D(L ) = fu2 2 L (H; (a2; 1)) : l2(u2) 2 L (H; (a2; 1)g; 2 2 p D(L0) = fu2 2 D(L ):( ρ2u2)(a2) = 0g: 3. Description of Selfadjoint Extensions In this section using the Calkin-Gorbachuk method will be investigated the gen- eral representation of selfadjoint extensions of minimal operator L0. Firstly we prove the following result. 1 2 Lemma 3.1. The deficiency indices of the operators L0 and L0 are of the form 1 1 2 2 (m(L0); n(L0)) = (0; dim H); (m(L0); n(L0)) = (dim H; 0): Proof. Now for simplicity we assume that A1 = A2 = 0. It is clear that the general solutions of differential equations 1 iρ (t)u0 (t) + iρ0 (t)u (t) ± iu (t) = 0; t < a ; 1 1± 2 1 1± 1± 1 1 iρ (t)u0 (t) + iρ0 (t)u (t) ± iu (t) = 0; t > a 2 2± 2 2 2± 2± 2 EJDE-2017/143 SELFADJOINT SINGULAR DIFFERENTIAL OPERATORS 3 2 2 in L (H; (−∞; a1)) and L (H; (a2; +1)) are in the form Z c1 0 2 ± ρ1(s) u1±(t) = exp ± ds f1; f1 2 H; t < a1; c1 < a1 t 2ρ1(s) and Z t 0 2 ± ρ2(s) u2±(t) = exp ∓ ds f2; f2 2 H; t > a2; c2 > a2 c2 2ρ2(s) respectively. From these representations we have 2 ku k 2 1+ L (H;(−∞;a1)) Z a1 2 = ku1+(t)kH dt −∞ Z a1 Z c1 0 2 + ρ1(s) 2 = exp ds dtkf1kH −∞ t ρ1(s) Z a1 Z c1 ρ1(c1) 2 2 = exp ds dtkf1kH −∞ ρ1(t) t ρ1(s) Z a1 Z c1 Z c1 ρ1(c1) 2 2 2 = exp ds d − ds kf1kH 2 −∞ t ρ1(s) t ρ1(s) Z c1 Z c1 ρ1(c1)h 2 2 i 2 = − exp ds − exp ds kf1kH = 1: 2 a1 ρ1(s) −∞ ρ1(s) Consequently, 1 dim ker(L0 + iE) = 0 On the other hand it is clear that 2 ku k 2 1− L (H;(−∞;a1)) Z a1 2 = ku1−(t)kH dt −∞ Z a1 Z c1 0 2 + ρ1(s) 2 = exp − ds dtkf1kH −∞ t ρ1(s) Z a1 Z c1 ρ1(c1) 2 2 = exp − ds dtkf1kH −∞ ρ1(t) t ρ1(s) Z a1 Z c1 Z c1 ρ1(c1) 2 2 2 = exp − ds d − ds kf1kH 2 −∞ t ρ1(s) t ρ1(s) Z c1 Z c1 ρ1(c1)h 2 2 i 2 = exp − ds − exp − ds kf1kH 2 a1 ρ1(s) −∞ ρ1(s) Z c1 ρ1(c1) 2 2 = exp − ds kf1kH < 1: 2 a1 ρ1(s) Therefore, Z t 0 2 − ρ1(s) 2 u1−(t) = exp ds f1 2 L (H; (−∞; a1)): a1 2ρ1(s) Hence 1 dim ker(L0 − iE) = dim H In a similar way it can be shown that 2 2 2 2 m(L0) = dim ker(L0 + iE) = dim H and n(L0) = dim ker(L0 − iE) = 0 4 P. IPEK, B. YILMAZ, Z. I. ISMAILOV EJDE-2017/143 This completes the proof. Consequently, the minimal operator L0 has selfadjoint extensions; see [6]. To describe these extensions we need to obtain the space of boundary values. Definition 3.2 ([6]). Let H be any Hilbert space and S : D(S) ⊂ H ! H be a closed densely defined symmetric operator in the Hilbert space H having equal finite or infinite deficiency indices. A triplet (B; γ1; γ2), where B is a Hilbert space, ∗ γ1 and γ2 are linear mappings from D(S ) into B, 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 = (γ1(f); γ2(g))B − (γ2(f); γ1(g))B ∗ while for any F1;F2 2 B, there exists an element f 2 D(S ) such that γ1(f) = F1 and γ2(f) = F2. Lemma 3.3. Let 1 p p γ1 : D(L) ! H; γ1(u) = p ( ρ1u1)(a1) + ( ρ2u2)(a2) ; i 2 1 p p γ2 : D(L) ! H; γ2(u) = p (( ρ1u1)(a1) − ( ρ2u2)(a2)) ; 2 where u = (u1; u2) 2 D(L). Then the triplet (H; γ1; γ2) is a space of boundary values of the minimal operator L0 in H. Proof. For any u = (u1; u2), v = (v1; v2) 2 D(L) (Lu; v)H − (u; Lv)H 2 2 2 = (L1u1; v1)L (H;(−∞;a1)) + (L2u2; v2)L (H;(a2;1)) − (u1;L1v1)L (H;(−∞;a1)) 2 − (u2;L2v2)L (H;(a2;1)) 0 i 0 = (iρ u + ρ u + A u ; v ) 2 1 1 2 1 1 1 1 1 L (H;(−∞;a1)) 0 i 0 − (u ; iρ v + ρ v + A v ) 2 1 1 1 2 1 1 1 1 L (H;(−∞;a1)) 0 i 0 + (iρ u + ρ u + A u ; v ) 2 2 2 2 2 2 2 2 2 L (H;(a2;1)) 0 i 0 − (u ; iρ v + ρ v + A v ) 2 2 2 2 2 2 2 2 2 L (H;(a2;1)) 0 0 2 2 = i (ρ1u1; v1)L (H;(−∞;a1)) + (u1; ρ1v1)L (H;(−∞;a1)) i 0 0 + (ρ u ; v ) 2 + (u ; ρ v ) 2 2 1 1 1 L (H;(−∞;a1)) 1 1 1 L (H;(−∞;a1)) 0 0 2 2 + i (ρ2u2; v2)L (H;(a2;1)) + (u2; ρ2v2)L (H;(a2;1)) i 0 0 + (ρ u ; v ) 2 + (u ; ρ v ) 2 2 2 2 2 L (H;(a2;1)) 2 2 2 L (H;(a2;1)) 0 0 0 2 2 2 = i (ρ1u1; v1)L (H;(−∞;a1)) + (u1; ρ1v1)L (H;(−∞;a1)) + i(ρ1u1; v1)L (H;(−∞;a1)) 0 0 0 2 2 + i (ρ2u2; v2)L (H;(a2;1)) + (u2; ρ1v2)L (H;(a2;1)) + i(ρ2u2; v2)L(H;(a2;1)) 0 0 2 2 = i (ρ1u1 + ρ1u1; v1)L (H;(−∞;a1)) + (ρ1u1; v1)L (H;(−∞;a1)) 0 0 2 2 + i (ρ2u2 + ρ2u2; v2)L (H;(a2;1)) + (ρ2u2; v2)L (H;(a2;1)) 0 0 = i[((ρ u ); v ) ] 2 + i [((ρ u ); v ) ] 2 1 1 1 L (H;(−∞;a1)) 2 2 2 L (H;(a2;1)) EJDE-2017/143 SELFADJOINT SINGULAR DIFFERENTIAL OPERATORS 5 p p p p = i (( ρ2u2)(a2); ( ρ2v2)(a2))H − (( ρ1u1)(a1); ( ρ1v1)(a1))H = (γ1(u); γ2(v))H − (γ2(u); γ1(v))H : Now for any element f1; f2 2 H let us find the function u = (u1; u2) 2 D(L) such that 1 p p γ1(u) = p (( ρ1u1)(a1) + ( ρ2u2)(a2)) = f1; i 2 1 p p γ2(u) = p (( ρ1u1)(a1) − ( ρ2u2)(a2)) = f2 2 From here the following two expressions are obtained p p p p ( ρ1u1)(a1) = (if1 + f2)= 2; ( ρ2u2)(a2) = (if1 − f2)= 2: If we choose the functions u1(·); u2(·) in the following forms 1 p t−a1 u1(t) = p e (if1 + f2)= 2; t < a1; ρ1(t) 1 p a2−t u2(t) = p e (if1 − f2)= 2; t > a2; ρ2(t) p p then it is clear that u = ( ρ1u1; ρ2u2) 2 D(L) and γ1(u) = f1, γ2(u) = f2.