Integral Equations

and so on, obtaining for the (n + 1)st approximation∫ bφ n+1 (x) = f(x) + λ K(x, y)φ n (y)dy. (74)aAssume that the kernel K(x, y) is a bounded function in the square Π = {(x, y) : a ≤ x ≤b, a ≤ y ≤ b}, so that|K(x, y)| ≤ M, (x, y) ∈ Π.or even that there exists a constant C 1 such that∫ bThe the following statement is valid.a|K(x, y)| 2 dy ≤ C 1 ∀x ∈ [a, b], (75)Theorem 5 Assume that condition (75) holds Then the sequence φ n of successive approximations(74) converges uniformly for all λ satisfyingλ ≤ 1 B ,B = ∫ bThe limit of the sequence φ n is the unique solution to IE (70).a∫ ba|K(x, y)| 2 dxdy. (76)Proof. Write two first successive approximations (74):Set∫ bf(x) + λa∫ bφ 1 (x) = f(x) + λ K(x, y)f(y)dy, (77)a∫ bφ 2 (x) = f(x) + λK(x, y)f(y)dy + λ 2 ∫ bK 2 (x, y) =∫ bto obtain (by changing the order of integration)∫ bφ 2 (x) = f(x) + λaIn the same manner, we obtain the third approximationaaK(x, y)φ 1 (y)dy = (78)a∫ bK(x, t)dt K(t, y)f(y)dy. (79)aK(x, t)K(t, y)dt (80)∫ bK(x, y)f(y)dy + λ 2 K 2 (x, y)f(y)dy. (81)awhere∫ bφ 3 (x) = f(x) + λ K(x, y)f(y)dy +a∫ b∫ b+ λ 2 K 2 (x, y)f(y)dy + λ 3 K 3 (x, y)f(y)dy, (82)aK 3 (x, y) =∫ baaK(x, t)K 2 (t, y)dt. (83)15