By G. Freiling, Bill Tooker

This publication provides the most effects and strategies on inverse spectral difficulties for Sturm-Liouville differential operators and their functions. Inverse difficulties of spectral research consist in convalescing operators from their spectral features. Such difficulties usually look in arithmetic, mechanics, physics, electronics, geophysics, meteorology and different branches of traditional sciences. Inverse difficulties additionally play an incredible function in fixing nonlinear evolution equations in mathematical physics. curiosity during this topic has been expanding completely as a result visual appeal of recent very important functions, leading to in depth research of inverse challenge conception around the globe.

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**INVERSE STURM-LIOUVILLE PROBLEMS AND THEIR APPLICATIONS**

This e-book offers the most effects and techniques on inverse spectral difficulties for Sturm-Liouville differential operators and their functions. Inverse difficulties of spectral research consist in getting better operators from their spectral features. Such difficulties frequently look in arithmetic, mechanics, physics, electronics, geophysics, meteorology and different branches of average sciences.

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**Extra resources for INVERSE STURM-LIOUVILLE PROBLEMS AND THEIR APPLICATIONS**

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27) On the other hand, integrating by parts twice, we obtain λϕ(x, λ) = ρ2 cos ρx + ρ2 + ∂G(x, t) ∂t t=x x 0 G(x, t) cos ρt dt = ρ2 cos ρx + G(x, x)ρ sin ρx cos ρx − ∂G(x, t) ∂t t=0 − x 0 Gtt (x, t) cos ρt dt. 27) this gives ϕ (x, λ) + λϕ(x, λ) − q(x)ϕ(x, λ) = 2 + x 0 dG(x, x) ∂G(x, t) − q(x) cos ρx − dx ∂t t=0 (Gxx (x, t) − Gtt (x, t) − q(x)G(x, t)) cos ρt dt. 17). 26) for x = 0. 9) hold. 4, a(x) ∈ W21 (0, 2π). Denote by ϕ(x, the conditions ϕ(0, ˜ λ) = 1, ϕ˜ (0, λ) = h. Our goal is to prove that ϕ(x, ˜ λ) ≡ ϕ(x, λ).

14) that Res ξ=λkj ϕ(x, ˜ λ), ϕ˜kj (x) ϕ(x, ˜ λ), ϕ(x, ˜ ξ) ˆ M (ξ)ϕ(x, ξ) = ϕkj (x). 10). 2) Since 1 1 1 1 − = , λ−µ λ−ξ µ−ξ (λ − ξ)(ξ − µ) we have by Cauchy‘s integral formula Pjk (x, λ) − Pjk (x, µ) 1 = λ−µ 2πi 0 γN Pjk (x, ξ) dξ, (λ − ξ)(ξ − µ) k, j = 1, 2; 0 λ, µ ∈ int γN . 14) where lim εN jk (x, λ, µ) = 0, j, k = 1, n. 16) for any y(x) ∈ C 1 [0, 1]. 16) into account, we calculate P (x, λ) − P (x, µ) λ−µ y(x), ϕ(x, ˜ ξ) y(x) y (x) = 1 2πi ˜ ξ) y(x), Φ(x, γN dξ + ε0N (x, λ, µ), (λ − ξ)(ξ − µ) Φ(x, ξ) Φ (x, ξ) ϕ(x, ξ) ϕ (x, ξ) − lim ε0N (x, λ, µ) = 0.

11). 13) q(t) dt. 2 was proved above, here we prove the sufficiency. 9) be given. 11). 7. 11) has a unique solution G(x, t) in L2 (0, x). Proof. 15) 0 has only the trivial solution g(t) = 0. 15). Then x 0 or x 0 g 2 (t) dt + g 2 (t) dt + ∞ 1 n=0 αn x 0 x x 0 0 F (s, t)g(s)g(t) dsdt = 0 g(t) cos ρn t dt 2 ∞ − 1 0 n=0 αn x 0 g(t) cos nt dt Using Parseval‘s equality x 0 g 2 (t) dt = ∞ 1 0 n=0 αn x 0 2 g(t) cos nt dt , for the function g(t), extended by zero for t > x, we obtain ∞ 1 n=0 αn x 0 g(t) cos ρn t dt 2 = 0.