By Muhammer Taşkıran, Cünet Kılıç
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Extra info for Analytic Analysis of Lines and Circles (Zambak)
Hence, ST eiλx = eiλx , that is, ST ϕ = ϕ, ϕ ∈ C (2) . 20). Suppose that this equation has more than one solution. Then there is a non-trivial solution of Sf = 0. 21) that S ∗ U f = 0, where U f = f (ω − x). 2, we have eixλ , U f = SB(x, λ), U f . 10), we obtain eixλ , U f = 0, that is, f = 0. This proves the theorem.
X+t Then the equality Q(x, t) = r(x, t) + N2 (x) holds. 13) ω N2 (t)ϕ(x − t + ω) dt x = ϕ(0)N2 (x). 14) we deduce the ﬁnal formula for T : ω d Tϕ = − dx Φ(x, t)ϕ (t) dt + ϕ(0)N2 (x). 2. Solutions of equations of the ﬁrst kind 37 3. Now, let us consider the case where γ = SN1 , U N2 − N1 , S ∗ U N2 = 0. 16) We put λ0 = −1/iγ. 6) at λ0 , we see that SB(x, λ0 ) = 0. 6) as follows: S B(x, λ) − B(x, λ0 ) iλγ + 1 = eiλx . 18) We introduce the operators x 1 Cγ ϕ = γ ϕ(t)eiλ0 (x−t) dt, Tγ = T Cγ . 19) 0 It is easy to see that Cγ eiλx = eiλx − eiλ0 x .
4), we see that a (λ0 )b(λ) − b (λ0 )a(λ) γ + B0 (x), S ∗ U Bγ (x, λ) = 0. 6) yield ω B0 (x)ei ∗ B0 (x), S U Bγ (x, λ) = ω−x λ dx. 52) lead to ∗ B0 (x), S U Bγ (x, λ) = a (λ0 ) b(λ) + eiλ0 ω − eiλω − b (λ0 )a(λ) i(λ0 − λ) . 51): a (λ0 )b(λ) − b (λ0 )a(λ) γ+ 1 i(λ0 − λ) + a (λ0 ) eiλ0 ω − eiλω = 0. 54) to the limit as λ → λ0 : a (λ0 ) = 0. 55) that b (λ0 )a(λ) γ 1 i(λ0 − λ) = 0. Hence, b (λ0 ) = 0. 56) indicate that u0 (x) p = 0. 49). 44) does not hold and B(x, λ0 ) p = 0. 17). This proves the theorem.
Analytic Analysis of Lines and Circles (Zambak) by Muhammer Taşkıran, Cünet Kılıç