Spectrums of Solvable Pantograph Differential-Operators for First Order

and Applied Analysis 3 Theorem 1. Each solvable extension ?̃? of the minimal operator L 0 in L2(H, (0, 1)) is generated by the pantograph differentialoperator expression (1) and boundary condition (K + E) u (0) = KU (0, 1) u (1) , (16) whereK ∈ L(H) and E is an identity operator inH.The operator K is determined uniquely by the extension ?̃?; that is, ?̃? = L K . On the contrary, the restriction of the maximal operator L 0 to the manifold of vector-functions satisfies condition (16) for some bounded operatorK ∈ L(H) is a solvable extension of the minimal operator L 0 in the L2(H, (0, 1)). Proof. Firstly, all solvable extensions ?̃? of theminimal operator M 0 in L2(H, (0, 1)) in terms of boundary values are described. Consider the following so-called Cauchy extensionM


Introduction
The quantitative and qualitative theory of linear pantograph differential equations, sometimes known as pantograph-type delay differential equations, was first studied in detail by T. Kato and J. B. McLeod [1], L. Fox et al. [2], and A. Iserles [3] in the nineteen seventies.
These equations arose as a mathematical model of an industrial problem involving wave motion in the overhead supply line to an electrified railway system, so they are often called pantograph equations.
In industrial applications in works [2,4] and in studies on biology and economics, control and electrodynamics in works [5][6][7] have been researched (for more comprehensive list of features see [3]).
Since an analytical computation of solutions, eigenvalues, and eigenfunctions of corresponding problems is very difficult theoretically and technically, then in this theory methods of numerical analysis play a significant role (for more information see [8][9][10][11][12][13]).
In this work, by using methods of operator theory all solvable extensions of minimal operator generated by pantograph-type delay differential-operator expression for first order in the Hilbert space of vector-functions at a finite interval have been described in terms of boundary values in Section 2. Consequently, the resolvent operators of these extensions can be written clearly.
The exact formula for the spectrums of these extensions has been given in Section 3. Applications of obtained results to the concrete models have been illustrated in Section 4.
In this situation the following defined operator is a linear bounded operator in  2 (, (0, 1)).
Let us introduce the operator In this case it is easy to see that, for the differentiable vectorfunction  ∈  2 (, (0, 1)),  : [0, 1] →  satisfies the following relation: From this  −1 () = ().Hence it is clear that if L is some extension of the minimal operator  0 , that is,  0 ⊂ L ⊂ , then For example, prove the validity of the last relation.It is known that If  ∈ (), then () = () ∈  2 (, (0, 1)); that is,  ∈ ().From the last relation  ⊂  −1 .Contrarily, if a vector-function  ∈ (), then that is, Theorem 1.Each solvable extension L of the minimal operator  0 in  2 (, (0, 1)) is generated by the pantograph differentialoperator expression (1) and boundary condition where  ∈ () and  is an identity operator in .The operator  is determined uniquely by the extension L; that is, L =   .On the contrary, the restriction of the maximal operator  0 to the manifold of vector-functions satisfies condition (16) for some bounded operator  ∈ () is a solvable extension of the minimal operator  0 in the  2 (, (0, 1)).
Proof.Firstly, all solvable extensions M of the minimal operator  0 in  2 (, (0, 1)) in terms of boundary values are described.

(23)
From the properties of the family of evolution operators (, ), ,  ∈ [0, 1], it is implied that an operator  is linear bounded and has a bounded inverse and On the other hand, from the relations it is implied that an operator  is one-to-one between sets of solvable extensions of minimal operators  0 and  0 in  2 (, (0, 1)).
Remark 3. Note that in the general case   ̸ =   , for any  ∈ ().

Corollary 6. It can be proved that all the solvable extensions of the minimal operator are generated by pantograph-type differential-operator expressions for first order
in  2 (, (0, 1)) generated by (⋅) and boundary condition in  2 (, (0, 1)).

Spectrum of Solvable Extensions
In this section, the spectrum structure of solvable extensions of minimal operator  0 in  2 (, (0, 1)) will be investigated.Firstly, prove the following fact.
Theorem 9.If L is a solvable extension of a minimal operator  0 and M =  −1 L corresponds to the solvable extension of a minimal operator  0 , then the spectrum of these extensions is true ( L) = ( M).
Proof.Consider a problem to the spectrum for a solvable extension   of a minimal operator  0 generated by pantograph differential-operator expression (1); that is, From this it is obtained that (44) Proof.Firstly, the spectrum of the solvable extension   =  −1    of the minimal operator  0 in  2 (, (0, 1)) will be investigated.