14. If af/ay is continuous in the rectangle D, show that there is a positive constant K such that If(t, y₁) f(t, y2)| ≤ Kly1 - y2l, (31) where (1, y₁) and (1, 2) are any two points in D having the same f coordinate. This inequality is known as a Lipschitz22 condition. Hint: Hold t fixed and use the mean value theorem on f as a function of y only. Choose K to be the maximum value of laf/ayl in D. 15. If on-1(1) and on(1) are members of the sequence (on(1)), use the result of Problem 14 to show that |f(t, on(t)) = f(t, on-1(1))| ≤ Kon(1) - On-1(1). 16. a. Show that if th, then to 01 (1)| ≤ Mtb st where M is chosen so that If(t, y)| ≤ M for (1, y) in D. b. Use the results of Problem 15 and part a of Problem 16 to show that 102(1) - 01(1)| ≤ MK|t|2 2 c. Show, by mathematical induction, that lon(1)n-1(1)| ≤ MK"-1|t|" MK"-1" n! n! ≤ 17. Note that stooge nichos a lo noitelugog all ind on (1) = $1(1) + (02(1) — 1(1)) +...+ (On(t) — On-1(t)). - a. Show that lon(1)| ≤ 01 (1)|+|6₂(1) −61(1)| ++ln(1)n-1(t)\. b. Use the results of Problem 16 to show that Kh+ (Kh)² 2! M |ón (1)| ≤ K c. Show that the sum in part b converges as n→ ∞o and, hence, the sum in part a also converges as noo. Conclude therefore that the sequence ((1)) converges since it is the sequence of partial sums of a convergent infinite series. 18. In this problem we deal with the question of uniqueness of the solution of the integral equation (3) (1) = b. Show that +...+ - Ső · f(s, o (s)) ds. (1)-(t) = a. Suppose that and are two solutions of equation (3). Show that, for t≥ 0, (Kh)" n! =f(f(s. o(s)) - f(s, 4(s))) ds. HIRVICE 10(1)(1)| ≤ = √(√(5, 6(S) (f(s, o(s))-f(s, (s))) ds. 10 (1)-(1)| ≤ K c. Use the result of Problem 14 to show that *[ lo(s)-(s)|ds, 22 The German mathematician Rudolf Lipschitz (1832-1903), professor at the where K is an upper bound for laf/ayl in D. This is the san University of Bonn for many years, worked in several areas of mathematics. noi as equation (30), and the rest of the proof may be constructed The inequality (i) can replace the hypothesis that af/ay is continuous in Theorem 2.8.1; this results in a slightly stronger theorem. bovlos orindicated in the text.mil