2013
DOI: 10.1134/s0012266113100042
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On a boundary value problem of N. I. Ionkin type

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Cited by 17 publications
(21 citation statements)
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“…If t ∈ [ρ, π − ρ], then (7) follows from (17). Thus (7) folds for t ∈ [0, π] and n > N. In the same way we prove it for t ∈ (−π, 0) and n < −N .…”
Section: On the Spectrality Of Hsupporting
confidence: 60%
See 1 more Smart Citation
“…If t ∈ [ρ, π − ρ], then (7) follows from (17). Thus (7) folds for t ∈ [0, π] and n > N. In the same way we prove it for t ∈ (−π, 0) and n < −N .…”
Section: On the Spectrality Of Hsupporting
confidence: 60%
“…For this assuming that | a |=| b | and (8) holds, we prove (7) for large n. If t ∈ [n −3 , ρ], then by Proposition 2, (7) holds. Using (115), (118), and Lemma 2 we get (7) in the case t ∈ [0, n −3 ]. Hence (7) for t ∈ [0, ρ] is proved.…”
Section: On the Spectrality Of Hmentioning
confidence: 99%
“…Il'in [3] dubbed such problems essentially nonselfadjoint and pointed out their typical instability both under the choice of associated functions and under small perturbations of the operator. For details, see [4] and also [5][6][7][8][9].…”
Section: Introductionmentioning
confidence: 99%
“…investigate D k (−t) for 1 ≤ k ≤ 2n + p − 1 and p = 0, −2n. By (44),(7) and by (15) and (20) we have 2n + p − n(k) > 0 and n(s) < n(k) for s < k . Therefore the multiplicand n(s) − 2n − p of the denominator of the fraction in (44) is a negative integer:n(s) − 2n − p < 0(50)for s = 1, 2, ..., k. To investigate the other multiplicand p − n(s) consider the cases:Case 1: −2n < p < 0.…”
mentioning
confidence: 87%
“…Note that the operators L 0 (q) and L π (q) in the case q(x) = Ae 2πirx , where A ∈ C and r ∈ Z, was investigated in detail by N. B. Kerimov [7]. He found a necessary and sufficient condition for a system of root functions of these operators to be a basis in L p [0, 1] for arbitrary p ∈ (1, ∞).…”
Section: On the Bloch Eigenvalues And Bloch Functionsmentioning
confidence: 99%