1998
DOI: 10.1007/bf02314629
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On the Riesz basis property of the root functions in certain regular boundary value problems

Abstract: ABSTRACT. The differential operator ly = y" + q(z)y with periodic (antiperiodic) boundary conditions that are not strongly regular is studied. It is assumed that q(z) is a complex-valued function of class C(4) [0,1] and q(0) # q(1). We prove that the system of root functions of this operator forms a Riesz basis in the space L2(0, 1).KEY WORDS: second-order differential operator, Riesz basis, periodic (antiperiodic) boundary condition.It is well known [1][2][3] that the system of root functions of an ordinary d… Show more

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Cited by 42 publications
(39 citation statements)
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“…In this way one can easily write a similar result for the anti-periodic problem. In addition to all the above results, we note that if either the first condition of (8) and (9), or the second condition of (8) and (9) hold then all the periodic eigenvalues are asymptotically simple. We can write a similar result for the anti-periodic problem.…”
Section: Proof Of Theoremsupporting
confidence: 56%
“…In this way one can easily write a similar result for the anti-periodic problem. In addition to all the above results, we note that if either the first condition of (8) and (9), or the second condition of (8) and (9) hold then all the periodic eigenvalues are asymptotically simple. We can write a similar result for the anti-periodic problem.…”
Section: Proof Of Theoremsupporting
confidence: 56%
“…Recently, i.e., in the 2000s, many authors [36,[44][45][46]40,47,50,67,26,49,51] focused on the problem of convergence of eigenfunction (or more generally root function) decompositions in the case of regular but not strictly regular bc.…”
Section: 1mentioning
confidence: 99%
“…Therefore, in general, the root functions of P and A do not form a Riesz basis; they form a basis with bracket (see [8,9]). In this paper we prove that if m is an even integer and The case m = 2 is investigated in [1,3,5]. In this paper we consider the more complicated case m = 2k > 2.…”
mentioning
confidence: 98%