Selçuk J. Appl. Math. Selçuk Journal of Vol. 5. No. 1. pp. 59-65, 2004 Applied Mathematics
On Spectrum of a Nonselfadjoint Singular Sturm-Liouville Problem with a Spectral Parameter in the Boundary Condition
Mevlüde Yak¬t Ongun
Süleyman Demirel University, Department of Mathematics, Cunur Campus, 32660, Isparta, Turkey;
myakit@fef.sdu.edu.tr
Received: March 10, 2004
Summary.In this paper we consider the nonselfadjoint singular Sturm-Liouville boundary value problems in limit-circle case with a spectral parameter in the boundary condition. The approach is based on the use of the maximal dissipa-tive operator.
Key words: Nonselfadjoint singular Sturm-Liouville problem, spectral para-meter, maximal dissipative operator spectrum, resolvent.
1991 Mathematics Subject Classi…cation: Primary 34B05, 34B40,34L05; Sec-ondary 47A10, 47B44
1. Introduction
The study of boundary value problems with spectral parameter in the boundary conditions is of great interest as a lot of problems of mathematical physics and mechanics [3; 4; 5; 11; 13] : It take place whenever the method separation of vari-ables is applied to solve the proper partial di¤erential equation with boundary conditions which have a directional derivative. A lot of studies have been de-voted to the boundary value problems with spectral parameters in the boundary conditions [4; 5; 6; 8; 13; 15]. In this paper we consider the following nonselfad-joint singular Sturm-Liouville boundary problem with a spectral parameter in the equation and the boundary condition:
(1) l(y) := 1
w(x)[ (py
0)0(x) + q(x)y(x)] = y(x); x 2 R
+:= [0; 1)
(3) 1W [y; v]1 2W [y; u]1=
0
1W [y; v]1
0
2W [y; u]1
where w; p and q are real-valued, Lebesgue measurable functions on R+; and
w;1
p; q 2 L
1
loc(R+) for almost all x 2 R+; w(x) > 0; is a complex spectral
parameter, 1; 2; 0 1; 0 22R:= ( 1; 1) and := 0 1 1 0 2 2 = 01 2 0 2 1> 0:
The nonselfadjoint (dissipative) singular boundary value problems with -independent boundary conditions have been investigated in [1; 2; 7] : In this pa-per, we construct the maximal dissipative operator Ah the spectrum of which
coincides with the spectrum of the boundary problem (1)-(3).
2. Preliminaries
We introduce the Hilbert space L2
w(R+) consisting of all complex-valued
func-tions y such thatR1
0 jy(x)j 2
w(x)dx < 1, with the inner product
(y; z) =
1
Z
0
y(x)z(x)w(x)dx:
Let us denote by L0the closure of the minimal symmetric operator generated
by (1) Let D0 be the domain of L0 and D the set of all functions y in L2w(R+)
whose elements y and py0 are locally absolutely continuous and `(y) 2 L2w(R+):
D is the domain of maximal operator L and L = L (see [9]) :
Assume that w; p; q be such that Weyl’s limit-circle case holds for the dif-ferential expression l, i.e., the symmetric operator L0 has defect index (2; 2) :
There are several su¢ cient conditions that guarantee Weyl’s limit-circle case (see [3]; [9] and [12]) : We give one such conditions that guarentee Weyl’s limit-circle case (see [3]) : Assume that q(x) has an absolutely continuous derivative q0(x) and q00(x) have constant sign in [x
0; 1) for su¢ ciently large x0; and that
q ! 1 and q0(x) = O (jqj ) as x ! +1; where 0 < x < 3
2: If the integral
R1
jq(x)j 12dx converges, then the operator L
0 has defect index (2; 2) :
Let us denote by u(x) and v(x) the solutions of `(y) = 0; (x 2 R+) satisfying
the initial conditions
u(0) = 1; (pu0) (0) = 0; v(0) = 0; (pv0) (0) = 1:
Clearly, u(x) and v(x) are linearly independent and their Wronskian is equal to one:
W [u; v]x:= (upv0 pu0v) j x
Since L0 has defect index (2; 2) ; we have u(x), v(x) 2 D: Green’s formula x Z 0 `(y)zdt x Z 0
y`(z)dt = W [y; z]x W [y; z]0
implies that for all functions y; z 2 D; the limit W [y; z]1:= limx!+1W [y; z]x
exists and is …nite where
(4) W [y; z]x= W [y; u]xW [z;v]x W [y; v]xW [z;u]x; x 2 R+:
3. Construction of Dissipative Operator in Adequate Hilbert Space Let us adopt the following notation:
R1(y) := 1W [y; v]1 2W [y; u]1
R0 1(y) := 0 1W [y; v]1 02W [y; u]1 B0 1(y) := y(0) B0 2(y) := (py0) (0) B1 1 (y) := W [y; v]1 B1 2 (y) := W [y; u]1
R0(y) := B20(y) hB10(y)
Then for arbitrary y; z 2 D; we have
(5) W [y; z]0= B10(y)B20(z) B20(y)B01(z);
(6) W [y; z]1= 1 hR1(y)R0
1(z) R01(y)R1(z)
i
R1(z) = R1(z); R0
1(z) = R01(z); B10(z) = B10(z); B02(z) = B02(z):
Let and ( 2 C) denote the solutions of (1) satisfying the conditions B0 1( ) = (0) = 1; B02( ) = p 0 (0) = h; B1 1 ( ) = 2 02; B21( ) = 1 01: Then by (5), we have 40( ) = W [ ; ]0= W [ ; ]0 = B0 1( )B20( ) + B20( )B10( ) = B0 2( ) + hB01( ) = R0( )
It follows from the equality (6) that
41( ) = W [ ; ]1= W [ ; ]1
= 1 R1( )R01( ) R01( )R1( ) = R1( ) R01( )
For the relation betwen the eigenvalues of the boundary value problem (1)-(3) and roots of 4( ) we have
Lemma 1. The eigenvalues of the boundary value problem (1)-(3) are exactly roots of 4( ), where 4( ) = 40( ) = 41( ) (see [10]).
We let G(x; ; ) = 8 < : (x) ( ) ( ) ; x (x) ( ) ( ) ; x:
It can be shown that G(x; :; ) satis…es the equation (1) and the boundary conditions (2)-(3); and G(x; ; ) is a Green function of the (1)-(3). Since the defect index of operator L0 is (2; 2), 2 L2(R+) and 2 L2(R+): We obtain
that G(x; ; ) is a Hilbert-Schmidt kernel and solution of the boundary value problem (1)-(3) can be expressed by
y(x; ) =
1
Z
0
G(x; ; )f ( )d := R y:
Thus, R is a Hilbert-Schmidt operator on space L2(R
+): Spectrum of the
boundary value problem (1)-(3) coincides with the roots ofM ( ) = 0: Since M ( ) is analytic and not identical zero and are linearly independent ; it means that functionM ( ) has at most a countable number of isolated zeros with …nite multiplicity and possible limit points at in…nity.
The Hilbert space H = L2(R
+) C is the set of vector-valued functions with
values in C2 equipped with the following inner product
(f; g)H= 1 Z 0 f1(x)g1(x)w(x)dx + 1 f2g2 where f = f1(x) f2 ; g = g1(x) g2 2 H: Put D(Ah) = f1(x) R01(f1) 2 H jf1(x) 2 D; R0(f1) = 0; f2= R 0 1(f1) :
We de…ne the operator Ah on D(Ah) by the equality
Ahf = el(f ) := `(f1)
R1(f1) :
Lemma 2. The operator Ah in Hilbert space H = L2(R+) C satis…es the
equation (7) (Ahf; g) (f; Ahg) = W [f1; g1]1 W [f1; g1]0 +1hR 1(f1)R01(g1) R01(f1)R1(g1) i :
For the proof of this lemma, see [10] :
De…nition 1. The linear operator T (with dense domain D(T )) acting on Hilbert space H is called dissipative if Im(T f; f ) 0 for all f 2 D(T ) and maximal dissipative if it does not have a proper dissipative extension.
Theorem 1The operator Ah is maximal dissipative in space H:
Proof. Let y 2 D(Ah) and D(Ah) is dense in H: Next by (7) and (5), we have
(Ahy; y) (y; Ahy) = W [y1; y1]1 W [y1; y1]0 (8) +1 hR1(y1)R01(y1) R01(y1)R1(y1) i = W [y1; y1]0 = B10(y1)B20(y1) + B20(y1)B10(y1) = hB10(y1)B10(y1) + hB10(y1)B01(y1) = (h h) B10(y1) 2 : (9) It follows from Im (Ahy; y) = Im h B10(y1) 2 0; that Ah is a dissipative
operator in H:In order to prove that Ahis a maximal dissipative operator in H
it su¢ ces to check that
(10) (Ah I) D(Ah) = H; Im < 0
To prove (10), let y 2 H; Im < 0 and put
^ = D e Gx; ; y E R0 1( eGx; ; y) ! ; where e Gx; = G(x; :; ) R0 1(G(x; :; )) = G(x; :; )(x) ( ) ! ; G(x; ; ) = 8 < : (x): ( ) ( ) ; x (x): ( ) ( ) ; x:
The function x ! (G(x; :; ); y1) satis…es the equation l(y) y = y1 (x 2 R+)
and boundary conditions (2) and (3). We arrive at^2 D(Ah) for all y 2 H and
for Im < 0. For each y 2 H and for Im < 0 , (Ah I) ^
= y: Consequently, in case of Im < 0; we have (Ah I) D(Ah) = H: Theorem 1 is proved.
Theorem 2 The operator Ah has not any real eigenvalue; therefore, all its
eigenvalues lie in the open upper half line for Im > 0:
Proof. Suppose that the operator Ah has a real eigenvalue 0: The function
Im 0ky0k2 = 0; then we get from (8) that y0(0) = 0: This and the boundary
condition (2) also imply the equality (py0
0) (0) = 0: Thus,
(11) y0(0) = (py00) (0) = 0:
Let y = (x; 0) and z = '(x; 0): Then using (5) we get
W [ ; ']0= B10( )B20(') B20( )B10(')
The right-side is equal to 0 in view of (11) while the left-side, being the value of the Wronskian of the solutions (x; 0) and '(x; 0) of (1) with = 0, is
equal to 1. This contradiction completes the proof of theorem.
De…nition 2. The system of functions y0; y1;:::;yn is called a chain of
eigen-functions and associated eigen-functions of the boundary problem (1) and (3) corre-sponding to the eigenvalue 0; if the conditions
l(y0) = 0y0 R1(y0) 0R01(y0) = 0 R0(y0) = 0 l(ys) 0ys ys 1= 0 R1(ys) 0R10 (ys) R01(ys 1) = 0 R0(ys) = 0; s = 1; 2; :::; n are realized.
From Theorem 1 and De…nition 2 we have
Corollary.Including their multiplicity, the eigenvalues of the boundary value problem (1) and (3) and the eigenvalues of the dissipative operator Ahcoincide.
Each chain of eigenfunctions and associated functions of the boundary value problem (1)-(3) correspond to the chain of eigenvectors and associated vectors y0; y1;:::;yn of the operator Ahcorresponding to the same eigenvalue 0:
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