Abstract
In this paper, we are concerned with the inverse spectral problems for differential pencils defined on [0,pi ] with an interior discontinuity. We prove that two potential functions are determined uniquely by one spectrum and a set of values of eigenfunctions at some interior point bin (0,pi ) in the situation of b=pi /2 and bneq pi /2. For the latter, we need the knowledge of a part of the second spectrum.
Highlights
Remark 2.2 Note that if y(x) and z(x) are two continuously differentiable functions on [0, π/2) ∪ (π/2, π] and satisfy the same discontinuous condition (1.3), a direct calculation yields that y z – yz (π/2 – 0) = y z – yz (π/2 + 0)
We consider the quadratic pencils of Sturm–Liouville operator L(p, q; h, H; a) of the formLy := –y + 2λp(x) + q(x) y = λ2y, x ∈ [0, π/2) ∪ (π/2, π], (1.1)with the boundary conditions ⎧⎨y (0) – hy(0) = 0,⎩y (π) + Hy(π) = 0, (1.2)and with the discontinuous conditions⎨y(π/2 + 0) = ay(π/2 – 0),⎩y (π /2 + 0) = a–1y (π /2 – 0)
Inverse spectral problems consist in recovering operators from their spectral characteristics
Summary
Remark 2.2 Note that if y(x) and z(x) are two continuously differentiable functions on [0, π/2) ∪ (π/2, π] and satisfy the same discontinuous condition (1.3), a direct calculation yields that y z – yz (π/2 – 0) = y z – yz (π/2 + 0). Let the function φ(x, λ) be the solution of equation (1.1) with the initial-valued conditions φ(0, λ) = 1, φ (0, λ) = h, (3.1) Proof of Theorem 2.1 Let φ(x, λ) be the solution of equation (1.1) satisfying the initialvalued conditions (3.1) and the discontinuity conditions (1.3).
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