Integral Equations

In some cases one can determine the sum of series (344) explicitly and obtain the solutionin a closed form. One can show that the approximation ϕ N (x) of the solution to equation (338)defined by (340) converges to the exact solution ϕ(x) as N → ∞ in the norm of space L (1)2 .Consider integral equation (339). Expand the kernel K(x, y) in the double Fourier–Chebyshevseries, take its partial sumN∑ N∑K(x, y) ≈ K N (x, y) = k ij T i (x)T j (y), (345)i=0 j=0and repeat the above procedure. A more detailed description is given below.Example 14 Consider the integral equation∫1−1(ln1|x − y| + xy) ϕ(x) dx√√1 = y + − y 2 , −1 < y < 1.1 − x2Use the method of solution described above and setto obtainy +ϕ N (x) = a 02 T 0(x) +N∑a n T n (x)n=1√1 − y 2 = 2 π T 0(y) + T 1 (y) − 4 πK(x, y) = T 1 (x)T 1 (y).∞∑k=1T 2k (y)4k 2 − 1 ,Substituting these expressions into the initial integral equation, we obtaina 02 ln 2T 0(y) +N∑n=1a nn T n(y) + π 2 a 1T 1 (y) = 2 π T 0(y) + T 1 (y) − 4 [N/2]∑ T 2k (y)πk=14k 2 − 1 ,here [ N 2 ] denotes the integer part of a number. Equating the coefficients that multiply T n(y) withthe same indices n we determine the unknownsa 0 = 4π ln 2 , a 1 = 2 π , a 2n+1 = 0, a 2n = − 8 nπ 4n 2 − 1 , n ≥ 1.The approximate solution has the formϕ N (x) = 4π ln 2 + 2 π x − 8 [N/2]∑ nπn=14n 2 − 1 T 2n(x).The exact solution is represented as the Fourier–Chebyshev seriesϕ(x) = 4π ln 2 + 2 π x − 8 π∞∑n=1n4n 2 − 1 T 2n(x).The latter series converges to the exact solution ϕ(x) as N → ∞ in the norm of space L (1)and its sum ϕ ∈ L (1)2 .652 ,