SEPARATION PROBLEM FOR STURM-LIOUVILLE EQUATION WITH OPERATOR COEFFICIENT

Let H be a separable Hilbert Space. Denote by H1 = L2(a, b; H) the set of function defned on the interval a < x < b (−∞ ≤ a < x < b ≤ ∞) whose values belong to H strongly measurable [12] and satisfying the condition ∫ b a ||f(x)||Hdx < ∞ If the inner product of function f(x) and g(x) belonging to H1 is defined by (f, g)1 = ∫ b a (f(x), g(x))Hdx then H1 forms a separable Hilbert space . We study separation problem for the operator formed by −y′′ + Q(x)y SturmLiouville differential expression in L2(−∞,∞;H) space has been proved where Q(x) is an operator which transforms at H in value of x,self-adjoint,lower bounded and its inverse is complete continous.

If the inner product of function f (x) and g(x) belonging to H 1 is defined by then H 1 forms a separable Hilbert space .We study separation problem for the operator formed by −y + Q(x)y Sturm-Liouville differential expression in L 2 (−∞, ∞; H) space has been proved where Q(x) is an operator which transforms at H in value of x,self-adjoint,lower bounded and its inverse is complete continous.

Introduction
Let us consider (1) −y + Q(x)y differential expression in H 1 = L 2 (−∞, ∞; H).Let us assume that Q(x), satisfies the following conditions as a self-adjoint operator making transformation at H in each value obtained (−∞, ∞) interval of x 1) Q(x) operator family have the same definition set indepent of x (−∞ < x < ∞) Let us show this set by D.
2) Let Q(x) ≥ I and for ∀f ∈ D Q(x)f be a strong countinous function in (−∞, ∞) and Let us form L 0 operator by ( 2) expression.Let the definition set D(L 0 ) of L 0 be by the functions y(x) satisfying the following conditions: 1) Let y(x) having compact support in (−∞, ∞),Q(x)y(x),y (x) be continious. 2) We have formed Since H which is the definition set of Q(x) is dense almost everywhere and since In this work we study the seperability of the operator L. According to the definition of seperability when y(x) is any function belonging to D(L) we will show that y (x) and Q(x)y(x) functions also belong to L 2 (−∞, ∞; H) space.
Let us show the resolvent of L operator in reguler λ value (λ is complex number) by R λ = (L − λI) −1 .According to the definition of Resolvent, R λ operator is bounded operator in H 1 space.
Proof: Let y(x) be an arbitrary element belonging to Many books, by B. M. Levitan and I. S. Sargsyan [8], E. C. Titcmarch [11], M. Otelbayev [10], and papers by T.C.Fulton and S.A.Pruess [6] belonging to singular Sturm-Liouville problem have been written.English mathematicians W. N. Everitt and M. A. Giertz [3], [4], [5] have proved that they introduced separation definition for operator L consisting of expression (1) being real valued function Q(x) with series papers and studies, and separation theorem of Q(x) for operator L in various conditions.For the separability problem,there are works by M. Bayramoglu and A. Abudov [1], K. Boymatov [2], A. Izmaylov and M. Otelbayev [13], M. Otelbayev [14] and many mathematicians; it has taken big place in the book [9] by the K. Minbayev and M. Otelbayev and given many references in the book.
Localization method for seperability of L operator.This method was firstly used by R.Ismagilov [7], and were developed by M.Otelbayev [10].We use that developed method.

Z. Oer
boundary conditions in space L 2 (j − 1, j + 1; H).Each L j operator is a positive defined self-adjoint operator.Let w(x) be a function satisfying the following properties and differentiable defined in (−∞, ∞) Let Let us show Since L + λI and L j + λI operators coincide in the interval (j − 1.5, j + 1.5) and ϕ j (x) has compact support in the interval (j − 1.5, j + 1.5), we can write (6) If we consider ϕ j ψ j = ψ j and condition (4), the expression (6) can be written as If we apply (L + λI) −1 operator to both side of ( 7) we can write ( 8) Let us evalute the norm of B λ operator transforming in In (10) equality it is considered that support of ϕ j and ϕ j+k (k ≥ 2) functions are not intersected.Let us evaluate the terms of the first sum on the right side of (10) (11) Thus, Second sum is Let us prove the following lemma.
Lemma 2: Proof: Let us consider y(j − 1) = y(j + 1) = 0 by multiply with y(x) both sides of equation −y If we consider in this inequality that Q(x) = Q * (x) ≥ I and use the Schwartz inequality, we obtain Here L j 2 = L 2 (j − 1, j + 1).If we consider j+1 j−1 is obtain from the last inequality.Since f is the arbitary element of Thus, the terms belonging to the second sum in ( 10) are ( 12) If we consider ( 10), ( 11), ( 12) inequalities we find Thus, B λ ≤ 2 c 1 λ where in big pozitive values of λ, it is found that B λ is as small as desired.Therefore, we can write and Lemma is proved.Let us show that Q(x)(L j + λI) −1 is finite.When |x − y| ≤ 2 let us assume that Let us consider the following boundary value problem Let us write this problem as ( 14) Then v = A j y .If we consider these, we can write (14) as Let us evaluate the norm of operator (Q(x) − Q(j))A −1 j .For this let us write the expression as If we use the condition (13) we can write Now let us prove the following Lemma.

Lemma 5:
The set of Q(j)A −1 j operator is smooth bounded with 1.
Q(j)A −1 j ≤ 1 Proof: We will prove the Lemma by using opening formula according to eigenvector of Q(j) operator.
Let us find where Q −1 (j) complete continous and therefore its spectrum is pure-disjoint.

These form a base in H.
( If we put these equations in (17) we find Thus, (17),(18) problems are transformed The eigenvalues of these problems are The corresponding normalized eigenvectors are the following: if j is odd; if j even and k is odd; if k is even; Let us consider (23) in (19) Since {ϕ m,j (x)} ∞ m=0 functions form a complete ortonormal system in L 2 (j − 1, j + 1) ,and the element {g k (j)} ∞ j=1 form a complete ortonormal system in H {ϕ m,j (x)g k (j)} ∞ m=0,k=1 forms a complete ortonormal system in Hilbert space.
Since the system {ϕ m,j (x)g k (j)} ∞ m=0,k=1 forms a complete ortonormal system in L 2,j = L 2 (j − 1, j + 1; H) if we use Parseval equation we can write from (24) If we consider here condition Therefore, inequality (26) becomes Here, if we consider y(x) = L −1 j f , the last inequality takes the form of If we consider, then we find Thus, we have proved the theorem of separability.
Theorem: When the conditions 1) -3) are satisfied, L operator is separable in H 1 .