Abstract

For the expression of all boundedly solvable extensions of the minimal operator generated by linear singular differential-operator expression for first order it has been applied Operator Theory Methods. Lastly, geometry of spectrum of these extensions is investigated.

1. Introduction

It is known that many solvability problems arising in life sciences can be expressed as boundary value problems for linear functional equations in corresponding functional spaces.

The solvability of the considered problems may be seen as boundedly solvability of linear differential operators in corresponding functional Banach spaces. Note that the theory of boundedly solvable extensions of a linear densely defined closed operator in Hilbert spaces was presented in the important works of Vishik in ^{1, 2}.

Let us recall that an operator on any Hilbert space is called boundedly solvable, if S is one-to-one and onto, and

The main aim of this work is to describe of all boundedly solvable extensions of the minimal operator generated by first-order linear quasi differential-operator expression in the Hilbert space of vector-functions at right semi-axis in terms of boundary conditions. Lastly, the structure of spectrum of these extensions will be investigated.

2. Statement of the Problem

Let H be a separable Hilbert space and In the weighted Hilbert space of H-valued vector-functions defined at the interval consider the following linear quasi-differential expression with operator coefficient for first order in a form

where for operator-function is satisfied

By the standard way the minimal and maximal operators corresponding to differential expression in can be defined (see ³).

In this case and (see sec.3).

In this work, firstly all boundedly solvable extensions of the minimal operator generated by first order linear singular differential-operator expression in the weighted Hilbert space of vector-functions at right semi-axis in terms of boundary conditions. Later on, the structure of spectrum of these type extensions will be investigated.

3. Description of Boundedly Solvable Extensions

In this section using the Vishik's methods all boundedly solvable extensions of the minimal operator in weighted Hilbert spaces

Before of all note that using the knowing standard way the minimal and the maximal operators generated by differential expression

in Hilbert space can be defined (see ³).

Later on, by will be defined the family of evolution operators corresponding to the homogeneous differential-operator equation

with boundary condition

The operator is linear continuous and boundedly solvable in And also for any

(for detail analysis see ⁴).

If introduce the following operator

then it is easily to check that

Therefore

Hence it is clear that if is some extension of the minimal operator that is, then

Now we prove the following assertion.

Theorem 3.1. and

Proof. Consider the following boundary value problem in

Then the general solution of above differential equation is in form

From this and boundary conditions we have

Consequently,

On the other hand it is clear that the general solution of following differential equation in

in form

This means that

So

Theorem 3.2. Each solvable extension of the minimal operator in is generated by the differential-operator expression with boundary condition

where is a identity operator in H. The operator B is determined uniquely by the extension i.e

On the contrary, the restriction of the maximal operator L to the manifold of vector-functions satisfy the above boundary condition for some bounded operator is a boundedly solvable extension of the minimal operator in

Proof. Firstly, all boundedly solvable extensions of the minimal operator in in terms of boundary conditions will be described.

Consider the following so-called Cauchy extension

of the minimal operator It is clear that is a boundedly solvable extension of minimal operator and

Indeed, for any we have

Now assumed that is a solvable extension of the minimal operator in In this case it is known that the domain of can be written as a direct sum

where (see ^{1, 2}).

It is easily to see that

Therefore each function can be written in following form

And from this we have

Hence

From these relations it is obtained that

Then the last equality can be written in form

where

On the other hand note that the uniquenses of the operator is clear from ^{1, 2}. Therefore, This completes of necessary part of assertion.

On the contrary, if is a operator generated by and boundary condition

then is boundedly invertible and

Consequently, assertion of theorem for the boundedly solvable extension of the minimal operator is true.

The extension of the minimal operator is boundedly solvable in if and only if the operator is a boundedly solvable extension of the minimal operator in Then if and only if

Since for some then we have

This completes the proof of theorem.

4. Spectrum of Boundedly Solvable Extensions

In this section the structure of spectrum of boundedly solvable extensions of the minimal operator in will be investigated.

Firstly, prove the following result.

Theorem 4.1. If is a boundedly solvable extension of the minimal operator and corresponding boundedly solvable extension of the minimal operator then it is true

Proof. Consider the following problem to spectrum for any boundedly solvable extension in that is

From this it is obtained that

Then we have

Therefore, the validity of the theorem is clear.

Now prove the main theorem on the structure of spectrum.

Theorem 4.2. The spectrum of the boundedly solvable extension of the minimal operator in has the form

Proof. By Theorem 4.1. for this it is sufficiently the investigate the spectrum of the corresponding boundedly solvable extension of the minimal operator in

Now consider the following problem to spectrum for the extension that is,

Then

with boundary condition

It is clear that a general solution of the above differential equation has the form

From this and boundary condition it is obtained that

In case when from the last relation it is established that

Consequently, in this case the resolvent operator of is in form

Now assumed that Then from the mentioned above equation for we have

Then if and only if

In this case since

Then

Remark 4.3. In finite interval case similar problems have been investigated in ⁵.

Example 4.4. All boundedly solvable extensions of the minimal operator in generated by differential expression

are generated by differential expression and boundary condition

where are the corresponding evolution operators. In this case the spectrum of the extension is in form

References