Integral Equations

and the integral equationwhereΦ N (x) = Nϕ =Lϕ + Nϕ = f, ϕ ∈ ˜W 1 2 , (367)∫ 1−1dsN(q)ϕ(s)πp 1 (x) ,q = (x, s),and N(q) is an arbitrary complex-valued function satisfying the continuity conditionsandN(q) ∈ C 1 (Π 1 );Write the Fourier–Chebyshev expansionswhere the coefficientsϕ(s) = 1 √2ξ 0 T 0 +f n = 2 π∂ 2 N∂x 2 (q) ∈ L 2[Π 1 , p −11 (x)p −11 (s)],Π 1 = ([−1, 1] × [−1, 1]).∞∑ξ n T n (s), ξ = (ξ 0 , ξ 1 , ..., ξ n , ...) ∈ h 2 , (368)n=1f(x) = f 02 T 0(x) +∫ 1−1∞∑f n T n (x),n=1dxf(x)T n (x) √ , n = 0, 1, . . . .1 − x2Function Φ N (x) is continuous and may be thus also expanded in the Fourier–Chebyshev serieswhereΦ N (x) = 1 ∞∑2 b 0T 0 (x) + b n T n (x), (369)n=1∫ ∫b n = 2Π 1dxdsN(q)T n (x)ϕ(s)π 2 p 1 (x)p 1 (s) . (370)According to Lemma 6 and formulas (368) and (369), equation (367) is equivalent in ˜W 2 1 to theequationln 2 ξ 0∞∑√ T 0 (x) + n −1 ξ n T n (x) + b 0∞∑2 n=12 T 0(x) + b n T n (x)n=1= f 0∞∑(371)2 T 0(x) + f n T n (x), |x| ≤ 1.n=1Note that since Φ N (x) is a differentiable function of x in [−1, 1], the series in (371) convergeuniformly, and the equality in (371) is an identity in (−1, 1).Now, we substitute series (368) for ϕ(x) into (370). Taking into account that {T n (x)} is abasis and one can change the integration and summation and to pass to double integrals (thesestatements are verified directly), we obtain an infinite system of linear algebraic equations{ √2 ln 2ξ0 + b 0 = f 0 ,(372)n −1 ξ n + b n = f n , n ≥ 1.77