Integral Equations

the behaviour (singularty) of the solution in the vicinity of the endpoints −1 and +1. Linearintegral operators L and L + K are defined by the left-hand sides of (338) and (339).Equation (338) is a particular case of (339) for K(x, y) ≡ 0 and can be solved explicitly.Assume that the right-hand side f(y) ∈ ˜W 21 (for simplicity, one may assume that f is acontinuously differentiable function in the segment [−1, 1]) and look for the solution ϕ(x) inthe space L (1)2 . This implies that integral operators L and L + K are considered as boundedlinear operatorsL : L (1)2 → ˜W 2 1 and L + K : L (1)2 → ˜W 2 1 .Note that if K(x, y) is a sufficiently smooth function, then K is a compact (completely continuous)operator in these spaces.17.2 Solution via Fourier–Chebyshev seriesLet us describe the method of solution to integral equation (338). Look for an approximationϕ N (x) to the solution ϕ(x) of (338) in the form of a partial sum of the Fourier–Chebyshev seriesϕ N (x) = a 02 T 0(x) +N∑a n T n (x), (340)n=1where a n are unknown coefficients. Expand the given function f(y) in the Fourier–Chebyshevseriesf N (y) = f 0N∑2 T 0(y) + f n T n (y), (341)where f n are known coefficients determined according to (329). Substituting (340) and (341)into equation (338) and using formula (337), we obtaina 02 ln 2T 0(y) +N∑n=1n=11n a nT n (y) = f 02 T 0(y) +Now equate coefficients multiplying T n (y) with the same indices n:a 02 ln 2 = f 02 , 1n a n = f n ,N∑f n T n (y). (342)n=1which gives unknown coefficients a na 0 = f 0ln 2 ,and the approximationϕ N (x) = f 02 ln 2 T 0(x) +a n = nf nN∑f n T n (x). (343)n=1The exact solution to equation (338) is obtained by a transition to the limit N → ∞ in (343)ϕ(x) = f 02 ln 2 T 0(x) +64∞∑nf n T n (x). (344)n=1