2020
DOI: 10.1007/s00025-020-1165-x
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Traces for Sturm–Liouville Operators with Frozen Argument on Star Graphs

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Cited by 16 publications
(14 citation statements)
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“…Put Y n (z) := q n (2z), n = 0, k − 1. By virtue of (25), these polynomials Y n (z) satisfy the recurrent relations (20). Substituting q n (z) = Y n (z/2) into (37), we get…”
Section: Corollary 1 the Determinant Of The Matrixmentioning
confidence: 98%
See 2 more Smart Citations
“…Put Y n (z) := q n (2z), n = 0, k − 1. By virtue of (25), these polynomials Y n (z) satisfy the recurrent relations (20). Substituting q n (z) = Y n (z/2) into (37), we get…”
Section: Corollary 1 the Determinant Of The Matrixmentioning
confidence: 98%
“…Put Y n (z) := i −n q n (2iz), n = 0, k − 1. Using (25), one can easily check that the polynomials Y n (z) satisfy the recurrent relations (20). Substituting q n (z…”
Section: Corollary 1 the Determinant Of The Matrixmentioning
confidence: 99%
See 1 more Smart Citation
“…Indeed, for 𝜈 = 1, formula ( 26) is obvious. Further, let it hold for any 𝜈 ≤ n. Then expanding the determinant in (26) for 𝜈 = n + 1 with respect to the elements of the last row, we obtain the last equality in (25). Finally, expanding the determinant…”
Section: Conflict Of Interestmentioning
confidence: 99%
“…Other aspects of recovering the operator 𝓁 as well as its spectral properties were studied in previous research. [23][24][25][26] Returning to Inverse Problem 1, we remind that in earlier work, [15][16][17] it was reduced to some linear functional equation with respect to the potential q(x) (see Equation (10) in the next section), which was referred to as main equation of the inverse problem. For rational values of a, the main equation can be represented as linear system (13) with a special k × k-matrix A (𝛼,𝛽) 𝑗,k , whose rank appeared to be ranging between k − 1 and k. These two possibilities, in turn, correspond to the degenerate and non-degenerate cases, respectively.…”
Section: Introductionmentioning
confidence: 99%