Integral Equations

Example 7 The space C[a, b] of continuous functions defined in a closed interval a ≤ x ≤ bwith the metricρ(φ 1 , φ 2 ) = maxa≤x≤b |φ 1(x) − φ 2 (x)|is complete. Indeed, every fundamental (convergent) functional sequence {φ n (x)} of continuousfunctions converges to a continuous function in the C[a, b]-metric.Example 8 The space C (n) [a, b] of componentwise continuous n-dimensional vector-functionsf = (f 1 , . . . , f n ) defined in a closed interval a ≤ x ≤ b with the metricρ(f 1 , f 2 ) = || maxa≤x≤b |f 1 i (x) − f 2 i (x)||| 2is complete. Indeed, every componentwise-fundamental (componentwise-convergent) functionalsequence { ¯φ n (x)} of continuous vector-functions converges to a continuous function in theabove-defined metric.Definition 2 Let R be an arbitrary metric space. A mapping A of the space B into itself issaid to be a contraction if there exists a number α < 1 such thatρ(Ax, Ay) ≤ αρ(x, y) for any two points x, y ∈ R. (49)Every contraction mapping is continuous:x n → x yields Ax n → Ax (50)Theorem 2 Every contraction mapping defined in a complete metric space R has one and onlyone fixed point; that is the equation Ax = x has only one solution.We will use the principle of contraction mappings to prove Picard’s theorem.Theorem 3 Assume that a function f(x, y) satisfies the Lipschitz condition|f(x, y 1 ) − f(x, y 2 )| ≤ M|y 1 − y 2 | in G. (51)Then on a some interval x 0 − d < x < x 0 + d there exists a unique solution y = φ(x) to theinitial value problemdydx = f(x, y), y(x 0) = y 0 . (52)Proof. Since function f(x, y) is continuous we have |f(x, y)| ≤ k in a region G ′ ⊂ G whichcontains (x 0 , y 0 ). Now choose a number d such that(x, y) ∈ G ′ if |x 0 − x| ≤ d, |y 0 − y| ≤ kdMd < 1.Consider the mapping ψ = Aφ defined byψ(x) = y 0 +∫ xx 0f(t, φ(t))dt, |x 0 − x| ≤ d. (53)11